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ldquoBaseball is 90 mental The other half is physicalrdquo
Yogi Berra
Probabilitybull Denoted by P(Event)
outcomes total
outcomes favorable)( EP
This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely
Experimental Probability
bull The relative frequency at which a chance experiment occursndashFlip a fair coin 30 times amp get 17
heads
30
17
Law of Large Numbers
bull As the number of repetitions of a chance experiment increase the difference between the relative frequency of occurrence for an event and the true probability approaches zero
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
A probabilityis a number between 0 and 1
The probability of rain must be 110
Rule 2 Sample spaceIf S is the sample space P(S) = 1
ldquoSomething Has to Happen Rulerdquo
The probability of the set of all possible outcomes must be 1
Irsquom 100 sure you are going to have a boyhellip or a girl
If the probability that you get to class on time is 8 then the probability that you do not get to class on time is 2
Rule 3 ComplementFor any event E
P(E) + P(not E) = 1 P(E) = 1 ndash P(not E)
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilitybull Denoted by P(Event)
outcomes total
outcomes favorable)( EP
This method for calculating probabilities is only appropriate when the outcomes of the sample space are equally likely
Experimental Probability
bull The relative frequency at which a chance experiment occursndashFlip a fair coin 30 times amp get 17
heads
30
17
Law of Large Numbers
bull As the number of repetitions of a chance experiment increase the difference between the relative frequency of occurrence for an event and the true probability approaches zero
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
A probabilityis a number between 0 and 1
The probability of rain must be 110
Rule 2 Sample spaceIf S is the sample space P(S) = 1
ldquoSomething Has to Happen Rulerdquo
The probability of the set of all possible outcomes must be 1
Irsquom 100 sure you are going to have a boyhellip or a girl
If the probability that you get to class on time is 8 then the probability that you do not get to class on time is 2
Rule 3 ComplementFor any event E
P(E) + P(not E) = 1 P(E) = 1 ndash P(not E)
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Experimental Probability
bull The relative frequency at which a chance experiment occursndashFlip a fair coin 30 times amp get 17
heads
30
17
Law of Large Numbers
bull As the number of repetitions of a chance experiment increase the difference between the relative frequency of occurrence for an event and the true probability approaches zero
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
A probabilityis a number between 0 and 1
The probability of rain must be 110
Rule 2 Sample spaceIf S is the sample space P(S) = 1
ldquoSomething Has to Happen Rulerdquo
The probability of the set of all possible outcomes must be 1
Irsquom 100 sure you are going to have a boyhellip or a girl
If the probability that you get to class on time is 8 then the probability that you do not get to class on time is 2
Rule 3 ComplementFor any event E
P(E) + P(not E) = 1 P(E) = 1 ndash P(not E)
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Law of Large Numbers
bull As the number of repetitions of a chance experiment increase the difference between the relative frequency of occurrence for an event and the true probability approaches zero
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
A probabilityis a number between 0 and 1
The probability of rain must be 110
Rule 2 Sample spaceIf S is the sample space P(S) = 1
ldquoSomething Has to Happen Rulerdquo
The probability of the set of all possible outcomes must be 1
Irsquom 100 sure you are going to have a boyhellip or a girl
If the probability that you get to class on time is 8 then the probability that you do not get to class on time is 2
Rule 3 ComplementFor any event E
P(E) + P(not E) = 1 P(E) = 1 ndash P(not E)
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Basic Rules of ProbabilityRule 1 Legitimate Values
For any event E 0 lt P(E) lt 1
A probabilityis a number between 0 and 1
The probability of rain must be 110
Rule 2 Sample spaceIf S is the sample space P(S) = 1
ldquoSomething Has to Happen Rulerdquo
The probability of the set of all possible outcomes must be 1
Irsquom 100 sure you are going to have a boyhellip or a girl
If the probability that you get to class on time is 8 then the probability that you do not get to class on time is 2
Rule 3 ComplementFor any event E
P(E) + P(not E) = 1 P(E) = 1 ndash P(not E)
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Rule 2 Sample spaceIf S is the sample space P(S) = 1
ldquoSomething Has to Happen Rulerdquo
The probability of the set of all possible outcomes must be 1
Irsquom 100 sure you are going to have a boyhellip or a girl
If the probability that you get to class on time is 8 then the probability that you do not get to class on time is 2
Rule 3 ComplementFor any event E
P(E) + P(not E) = 1 P(E) = 1 ndash P(not E)
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
If the probability that you get to class on time is 8 then the probability that you do not get to class on time is 2
Rule 3 ComplementFor any event E
P(E) + P(not E) = 1 P(E) = 1 ndash P(not E)
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Independent
bull Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occursndash A randomly selected student is female - What is
the probability she plays soccer for SHS
ndash A randomly selected student is female - What is the probability she plays football for SHS
Independent
Dependent
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Rule 4 Multiplication
If two events A amp B are independent
General rule
P(B) P(A) B) ampP(A
A)|P(B P(A) B) ampP(A If the probability of rolling a 5 on a fair dice is 16 what is the probability of rolling a 5 three times in a row
P( three 5rsquos in a row) = (16) x (16) x (16) = 1216 or 004629
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
)( BAP
Independent)( BAP
Yes
)()( BPAP
What does this mean
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 1) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that both bulbs are defective
Can you assume they are independent
00250505D) amp P(D
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
)( BAP
Independent)( BAP
Yes No
)()( BPAP )|()( ABPAP
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 2) There are seven girls and eight boys in a math class The teacher selects two students at random to answer questions on the board What is the probability that both students are girls
Are these events independent
2146
157
G) amp P(G
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 6) Suppose I will pick two cards from a standard deck without replacement What is the probability that I select two spades
Are the cards independent NO
P(A amp B) = P(A) P(B|A)
Read ldquoprobability of B given that A occursrdquo
P(Spade amp Spade) = 14 1251 = 117
The probability of getting a spade given that a spade has already been drawn
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Rule 5 AdditionIf two events E amp F are disjoint
P(E or F) = P(E) + P(F)
If the probability that a randomly selected student is a junior (A) is 2 and the probability that the student is a senior (B) is 5 what is the probability that the student is either a junior or a senior
P(A υ B) = P(A) + P(B) if A and B are disjoint
P(A υ B) = 2 + 5 = 7
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Rule 5 AdditionIf two events E amp F are disjoint P(E or F) = P(E) + P(F)
(General) If two events E amp F are not disjoint
P(E or F) = P(E) + P(F) ndash P(E amp F)Probability of owning a MP3 player 50Probability of owning a computer 90
So the probability of owning a MP3 player or a computer
is 140
Not disjoint events
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
)( FEP What does this mean
)( FEP Mutually exclusive
)()( FPEP
Yes
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 3) A large auto center sells cars made by many different manufacturers Three of these are Honda Nissan and Toyota (Note these are not simple events since there are many types of each brand) Suppose that P(H) = 25 P(N) = 18 P(T) = 14
Are these disjoint events
P(H or N or T) =
P(not (H or N or T) =
yes
25 + 18+ 14 = 57
1 - 57 = 43
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 4) Musical styles other than rock and pop are becoming more popular A survey of college students finds that the probability they like country music is 40 The probability that they liked jazz is 30 and that they liked both is 10 What is the probability that they like country or jazz
P(C or J) = 4 + 3 -1 = 6
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
)( FEP
)( FEP Mutually exclusive
)()( FPEP
Yes No
)()()( FEPFPEP
Independent Yes )()( FPEP
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 5)
If P(A) = 045 P(B) = 035 and A amp B are independent find P(A or B)Is A amp B disjoint
If A amp B are disjoint are they independent
Disjoint events do not happen at the same time
So if A occurs can B occur
Disjoint events are dependent
NO independent events cannot be disjoint
P(A or B) = P(A) + P(B) ndash P(A amp B)
How can you find the
probability of A amp B
P(A or B) = 45 + 35 - 45(35) = 06425
If independent
multiply
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6
If a coin is flipped amp a die rolled at the same time what is the probability that you will get a tail or a number less than 3
T1
T2
T3
T4
T5
T6 H1
H3
H2
H5
H4
H6
Coin is TAILS Roll is LESS THAN 3
P (tails or even) = P(tails) + P(less than 3) ndash P(tails amp less than 3)
= 12 + 13 ndash 16= 23
Flipping a coin and rolling a die are independent events
Independence also implies the events are NOT disjoint (hence the overlap)
Count T1 and T2 only once
Sample Space
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 7) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that exactly one bulb is defective
P(exactly one) = P(D amp DC) or P(DC amp D)
= (05)(95) + (95)(05)
= 095
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 8) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store What is the probability that at least one bulb is defective
P(at least one) = P(D amp DC) or P(DC amp D) or (D amp D)
= (05)(95) + (95)(05) + (05)(05)
= 0975
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Rule 6 At least one
The probability that at least one outcome happens is 1 minus the probability that no outcomes happen
P(at least 1) = 1 ndash P(none)
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 8 revisited) A certain brand of light bulbs are defective five percent of the time You randomly pick a package of two such bulbs off the shelf of a store
What is the probability that at least one bulb is defective
P(at least one) = 1 - P(DC amp DC)
0975
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 9) For a sales promotion the manufacturer places winning symbols under the caps of 10 of all Dr Pepper bottles You buy a six-pack What is the probability that you win something
P(at least one winning symbol) =
1 ndash P(no winning symbols) 1 - 96 = 4686
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is not US made
P(not US made) = 1 ndash P(US made) = 1 - 4 = 6
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that it is made in Japan or Germany
P(Japanese or German) = P(Japanese) + P(German) = 3 + 1 = 4
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that you see two in a row from Japan
P(2 Japanese in a row) = P(Japanese) and P(Japanese) = P(J) x P(J) = 3 x 3 = 09
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that none of three cars came from Germany
P(no Germany in three) = P(not G) x P(not G) x P(not G) = 9 x 9 x 9 = 729
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that at least one of three cars is US made
P(at least one US in three) = 1 ndash P(no US in three) = 1 ndash (6)(6)(6) = 784
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Suppose that 40 of cars in Fort Smith are manufactured in the United States 30 in Japan 10 in Germany and 20 in other countries
If cars are selected at random what is the probability that the first Japanese car is the fourth one you choose
P(first J is the fourth car) = P(not J) x P(not J) x P(not J) x P(J) = (7)3 (3) = 1029
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Watch out for
bull probabilities that donrsquot add up to 1
bull donrsquot add probabilities of events if they are not disjoint
bull donrsquot multiply probabilities of events if they are not independent
bull donrsquot confuse disjoint and independent
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Rule 7 Conditional Probability
bull A probability that takes into account a given condition
P(A)
B)P(AA)|P(B
P(given)
P(and)A)|P(B
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 10) In a recent study it was found that the probability that a randomly selected student is a girl is 51 and is a girl and plays sports is 10 If the student is female what is the probability that she plays sports
196151
1
P(F)
F)P(S F)|P(S
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Ex 11) The probability that a randomly selected student plays sports if they are male is 31 What is the probability that the student is male and plays sports if the probability that they are male is 49
1519
4931
P(M)
M)P(SM)|P(S
x
x
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
12) What is the probability that the driver is a student
359195
)( StudentP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
13) What is the probability that the driver drives a European car
35945
)( EuropeanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
14) What is the probability that the driver drives an American or Asian car
Disjoint359102212
)(
AsianorAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
15) What is the probability that the driver is staff or drives an Asian car
Disjoint35947102164
)(
AsianorStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
16) What is the probability that the driver is staff and drives an Asian car
35947
)( AsianandStaffP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
17) If the driver is a student what is the probability that they drive an American car
Condition195107
)|( StudentAmericanP
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Probabilities from two way tables
Stu Staff TotalAmerican 107 105 212European 33 12 45Asian 55 47 102Total 195 164 359
18) What is the probability that the driver is a student if the driver drives a European car
Condition 4533
)|( EuropeanStudentP
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
Example 19Management has determined that customers return 12 of the items assembled by inexperienced employees whereas only 3 of the items assembled by experienced employees are returned Due to turnover and absenteeism at an assembly plant inexperienced employees assemble 20 of the items Construct a tree diagram or a chart for this data
What is the probability that an item is returned
If an item is returned what is the probability that an inexperienced employee assembled it
P(returned) = 48100 = 0048
P(inexperienced|returned) = 2448 = 05
Returned Not returned
Total
Experienced 24 776 80
Inexperienced 24 176 20
Total 48 952 100
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
ldquoSlump I ainrsquot in no slump I just ainrsquot hittinrdquo
Yogi Berra
![The Wit and Wisdom of Yogi Berra by Phil Pepe [Excerpt]](https://img.pdfslide.us/doc/110x75/5452b4a4b1af9f83248b5290/the-wit-and-wisdom-of-yogi-berra-by-phil-pepe-excerpt.jpg)
