A Journey to Probability Concept

7/29/2019 A Journey to Probability Concept

http://slidepdf.com/reader/full/a-journey-to-probability-concept 1/22

Page | 1

Probability

Problem: A spinner has 4 equal sectors colored yellow, blue, green and red.What are the chances of landing on blue after spinning thespinner? What are the chances of landing on red?

Solution: The chances of landing on blue are 1 in 4, or one fourth.

The chances of landing on red are 1 in 4, or one fourth.

This problem asked us to find some probabilities involving a spinner. Let's look at somedefinitions and examples from the problem above.

Definition Example

An experiment is a situation involving chance or probability that leads to results called outcomes.

In the problem above, theexperiment is spinning the spinner.

An outcome is the result of a single trial of anexperiment.

The possible outcomes are landingon yellow, blue, green or red.

An event is one or more outcomes of an experiment. One event of this experiment islanding on blue.

Probability is the measure of how likely an event is. The probability of landing on blue isone fourth.

In order to measure probabilities, mathematicians have devised the following formula for findingthe probability of an event.

Probability Of An Event

P(A) =The Number Of Ways Event A Can Occur The total number Of Possible Outcomes

The probability of event A is the number of ways event A can occur divided by thetotal number of possible outcomes. Let's take a look at a slight modification of the problemfrom the top of the page.

Experiment 1: A spinner has 4 equal sectors colored yellow, blue, greenand red. After spinning the spinner, what is the probabilityof landing on each color?

Outcomes: The possible outcomes of this experiment are yellow,blue, green, and red.

Probabilities:P(yellow) =

# of ways to land on yellow=

1total # of colors 4



Page | 2

P(blue) =# of ways to land on blue

=1

total # of colors 4

P(green) =# of ways to land on green

=1

total # of colors 4

P(red) = # of ways to land on red = 1total # of colors 4

Experiment 2: A single 6-sided die is rolled. What isthe probability of each outcome?What is the probability of rolling aneven number? of rolling an oddnumber?

Outcomes: The possible outcomes of this

experiment are 1, 2, 3, 4, 5 and 6.

Probabilities:P(1) =

# of ways to roll a 1=

1total # of sides 6

P(2) =# of ways to roll a 2

=1

total # of sides 6


=1

total # of sides 6


=1

total # of sides 6


=1

total # of sides 6


=1

total # of sides 6

P(even) =# ways to roll an even number

=3

=1

total # of sides 6 2

P(odd) =# ways to roll an odd number

=3

=1

total # of sides 6 2

Experiment 2 illustrates the difference between an outcome and an event. A single outcome of this experiment is rolling a 1, or rolling a 2, or rolling a 3, etc. Rolling an even number (2, 4 or 6)is an event, and rolling an odd number (1, 3 or 5) is also an event.

In Experiment 1 the probability of each outcome is always the same. The probability of landingon each color of the spinner is always one fourth. In Experiment 2, the probability of rolling eachnumber on the die is always one sixth. In both of these experiments, the outcomes are equally

http://x1096653463%28%27equally_likely%27%29/





Page | 3

likely to occur. Let's look at an experiment in which the outcomes are not equally likely.

Experiment 3: A glass jar contains 6 red, 5 green, 8 blue and 3yellow marbles. If a single marble is chosen at randomfrom the jar, what is the probability of choosing a redmarble? a green marble? a blue marble? a yellowmarble?

Outcomes: The possible outcomes of this experiment are red,green, blue and yellow.

Probabilities:P(red) =

# of ways to choose red=

6=

3total # of marbles 22 11

P(green) =# of ways to choose green

=5

total # of marbles 22

P(blue) =# of ways to choose blue

=8

=4

total # of marbles 22 11

P(yellow) =# of ways to choose yellow

=3

total # of marbles 22

The outcomes in this experiment are not equally likely to occur. You are more likely to choose ablue marble than any other color. You are least likely to choose a yellow marble.

Experiment 4: Choose a number at random from 1 to 5. What is the probability of eachoutcome? What is the probability that the number chosen is even? What is theprobability that the number chosen is odd?

Outcomes: The possible outcomes of this experiment are 1, 2, 3, 4 and 5.


# of ways to choose a 1=

1total # of numbers 5

P(2) =# of ways to choose a 2

=1

total # of numbers 5


=1



=1



=1


P(even) =# of ways to choose an even number

=2


P(odd) = # of ways to choose an odd number = 3






Page | 4


The outcomes 1, 2, 3, 4 and 5 are equally likely to occur as a result of this experiment.However, the events even and odd are not equally likely to occur, since there are 3 oddnumbers and only 2 even numbers from 1 to 5.

Summary: The probability of an event is the measure of the chance that the event will occur as a result of an experiment. The probability of an event A is the number of waysevent A can occur divided by the total number of possible outcomes. Theprobability of an event A, symbolized by P(A), is a number between 0 and 1,inclusive, that measures the likelihood of an event in the following way:

If P(A) > P(B) then event A is more likely to occur than event B. If P(A) = P(B) then events A and B are equally likely to occur.

Certain and Impossible Events:

Experiment 1: A spinner has 4 equal sectors colored yellow, blue, green,and red. What is the probability of landing on purple after spinning the spinner?

Probability: It is impossible to land on purple since the spinner does notcontain this color.

P(purple) =0

= 04

Experiment 2: A teacher chooses a student at random from a class of 30girls. What is the probability that the student chosen is agirl?

Probability: Since all the students in the class are girls, the teacher iscertain to choose a girl.

P(girl) =30

= 130

In the first experiment, it was not possible to land on purple. This is an example of an impossible event. In the second experiment, choosing a girl was certain to occur. This is an

example of a certain event.

The next experiment will involve a standard deck of 52 playing cards, whichconsists of 4 suits: hearts, clubs, diamonds and spades. Each suit has 13 cardsas follows: ace, deuce, three, four, five, six, seven, eight, nine, ten, jack, queen,and king. Picture cards include jacks, queens and kings. There are no joker cards.There are only 4 of a kind, for example, 4 tens.

http://popupwindow%28%27impossible%27%29/



http://popupwindow%28%27certain%27%29/







Page | 5

Experiment 3: A single card is chosen at random from a standarddeck of 52 playing cards. What is the probability thatthe card chosen is a joker card?

Probability: It is impossible to choose a joker card since a

standard deck of cards does not contain any jokers.This is an impossible event.

P(joker) =0

= 052

Experiment 4: A single 6-sided die is rolled. What isthe probability of rolling a number lessthan 7?

Probability: Rolling a number less than 7 is acertain event since a single die has 6sides, numbered 1 through 6.

P(number < 7) = 6 = 16

Experiment 5: A total of five cards are chosen at random from astandard deck of 52 playing cards. What is theprobability of choosing 5 aces?

Probability: It is impossible to choose 5 aces since a standarddeck of cards has only 4 of a kind. This is animpossible event.

P(5 aces) =0

= 052

Experiment 6: A glass jar contains 15 red marbles. If a single marble ischosen at random from the jar, what is the probability thatit is red?

Probability: Choosing a red marble is certain to occur since all 15marbles in the jar are red. This is a certain event.

P(red) =15

= 115

Summary: The probability of an event is the measure of the chance that the event will occur as a result of the experiment. The probability of an event A, symbolized by P(A), is

a number between 0 and 1, inclusive, that measures the likelihood of an event inthe following way:

If P(A) > P(B) then event A is more likely to occur than event B. If P(A) = P(B) then events A and B are equally likely to occur. If event A is impossible, then P(A) = 0. If event A is certain, then P(A) = 1.



Page | 6

Sample Spaces:

Experiment 1: What is the probability of each outcome when adime is tossed?

Outcomes: The outcomes of this experiment are head and tail.

Probabilities:P(head) =

12

P(tail) =12

Definition: The sample space of an experiment is the set of all possible outcomes of thatexperiment.

The sample space of Experiment 1 is: {head, tail}

Experiment 2: A spinner has 4 equal sectors colored yellow, blue,green and red. What is the probability of landing on eachcolor after spinning this spinner?

Sample Space: {yellow, blue, green, red}

Probabilities:P(yellow) =

14

P(blue) =1

4

P(green) =14

P(red) =14

Experiment 3: What is the probability of eachoutcome when a single 6-sideddie is rolled?

Sample Space: {1, 2, 3, 4, 5, 6}


16

P(2) =16

P(3) = 1



Page | 7

6

P(4) =16

P(5) =

1

6

P(6) =16

Experiment 4: A glass jar contains 1 red, 3 green, 2 blue and 4yellow marbles. If a single marble is chosen atrandom from the jar, what is the probability of eachoutcome?

Sample Space: {red, green, blue, yellow}

Probabilities: P(red) = 110

P(green) =3

10

P(blue) =2

=1

10 5

P(yellow) =4

=2

10 5

Summary: The sample space of an experiment is the set of all possible outcomes for thatexperiment. You may have noticed that for each of the experiments above, thesum of the probabilities of each outcome is 1. This is no coincidence. The sum ofthe probabilities of the distinct outcomes within a sample space is 1.

The sample space for choosing a single card at random from a deck of 52 playingcards is shown below. There are 52 possible outcomes in this sample space.



Page | 8

The probability of each outcome of this experiment is:

P(card) =152

The sum of the probabilities of the distinct outcomes within this sample space is:52

= 152

Complement of an Event:

Experiment 1: A spinner has 4 equal sectors colored yellow, blue, greenand red. What is the probability of landing on a sector thatis not red after spinning this spinner?


Probability: The probability of each outcome in this experiment is one

fourth. The probability of landing on a sector that is not redis the same as the probability of landing on all the other colors except red.

P(not red) =1

+1

+1

=3

4 4 4 4

In Experiment 1, landing on a sector that is not red is the complement of landing on a sector that isred.

Definition: The complement of an event A is the set of all outcomes in the sample spacethat are not included in the outcomes of event A. The complement of event A is

represented by (read as A bar).

Rule: Given the probability of an event, the probability of its complement can be foundby subtracting the given probability from 1.

P( ) = 1 - P(A)

You may be wondering how this rule came about. In the last lesson, we learned that the sum of the probabilities of the distinct outcomes within a sample space is 1. For example, the



Page | 9

probability of each of the 4 outcomes in the sample space above is one fourth, yielding a sumof 1. Thus, the probability that an outcome does not occur is exactly 1 minus the probability thatit does. Let's look at Experiment 1 again, using this subtraction principle.

Experiment 1: A spinner has 4 equal sectors colored yellow, blue, greenand red. What is the probability of landing on a sector that

is not red after spinning this spinner?


Probability: P(not red) = 1 - P(red)

= 1 -14

=34

Experiment 2: A single card is chosen at random from a standard

deck of 52 playing cards. What is the probability of choosing a card that is not a king?

Probability: P(not king) = 1 - P(king)

= 1 -452

=4852

=1213

Experiment 3: A single 6-sided die is rolled. What isthe probability of rolling a number thatis not 4?

Probability: P(not 4) = 1 - P(4)

= 1 -16

=56

Experiment 4: A single card is chosen at random from a standarddeck of 52 playing cards. What is the probability of choosing a card that is not a club?

Probability: P(not club) = 1 - P(club)

= 1 -1352

http://x1096653463%28%27cards%27%29/

http://x1096653463%28%27cards%27%29/

http://x1096653463%28%27cards%27%29/

http://x1096653463%28%27cards%27%29/

http://x1096653463%28%27cards%27%29/

http://x1096653463%28%27cards%27%29/



Page | 10

=3952

=34

Experiment 5: A glass jar contains 20 red marbles. If a marble is chosenat random from the jar, what is the probability that it is notred?

Probability: P(not red) = 1 - P(red)

= 1 - 1

= 0

Note: This is an impossible event.

Summary: The probability of an event is the measure of the chance that the event will occur as a result of the experiment. The probability of an event A, symbolized by P(A),is a number between 0 and 1, inclusive, that measures the likelihood of an eventin the following way:

If P(A) > P(B) then event A is more likely to occur than event B. If P(A) = P(B) then events A and B are equally likely to occur. If event A is impossible, then P(A) = 0. If event A is certain, then P(A) = 1.

The complement of event A is . P( ) = 1 - P(A)

Mutually Exclusive Events:

Experiment 1: A single card is chosen at random from a standard deckof 52 playing cards. What is the probability of choosing a5 or a king?

Possibilities: 1. The card chosen can be a 5.2. The card chosen can be a king.

Experiment 2: A single card is chosen at random from a standard deckof 52 playing cards. What is the probability of choosing aclub or a king?

Possibilities: 1. The card chosen can be a club.2. The card chosen can be a king.3. The card chosen can be a king and a club (i.e., the

king of clubs).

In Experiment 1, the card chosen can be a five or a king, but not both at the same time . Theseevents are mutually exclusive. In Experiment 2, the card chosen can be a club, or a king, or both at the same time. These events are not mutually exclusive.



Page | 11

Definition: Two events are mutually exclusive if they cannot occur at the same time (i.e., theyhave no outcomes in common).

Experiment 3: A single 6-sided die is rolled. What is the

probability of rolling an odd number or an even number?

Possibilities: 1. The number rolled can be an oddnumber.

2. The number rolled can be an evennumber.

Events: These events are mutually exclusivesince they cannot occur at the sametime.

Experiment 4: A single 6-sided die is rolled. What is the probability of rolling a 5 or an odd number?

Possibilities: 1. The number rolled can be a 5.2. The number rolled can be an odd number (1, 3 or 5).3. The number rolled can be a 5 and odd.

Events: These events are not mutually exclusive since they can occur at the sametime.

Experiment 5: A single letter is chosen at random from the word SCHOOL. What is theprobability of choosing an S or an O?

Possibilities: 1. The letter chosen can be an S

2. The letter chosen can be an O.

Events: These events are mutually exclusive since they cannot occur at the same time.

Experiment 6: A single letter is chosen at random from the word SCHOOL. What is theprobability of choosing an O or a vowel?

Possibilities: 1. The letter chosen can be an O2. The letter chosen can be a vowel.3. The letter chosen can be an O and a vowel.

Events: These events are not mutually exclusive since they can occur at the sametime.

Summary: In this lesson, we have learned the difference between mutually exclusive and non-mutually exclusive events. We can use set theory and Venn Diagrams to illustrate thisdifference.

Mutually Exclusive Events Non-Mutually Exclusive Events

Two events are mutually exclusive if Two events are non-mutually exclusive



Page | 12

they cannot occur at the same time(i.e., they have no outcomes incommon).

if they have one or more outcomes incommon.

In the Venn Diagram above, theprobabilities of events A and B arerepresented by two disjoint sets (i.e.,they have no elements in common).

In the Venn Diagram above, theprobabilities of events A and B arerepresented by two intersecting sets(i.e., they have some elements incommon).

Note: In each Venn diagram above, the sample space of the experiment isrepresented by S, with P(S) = 1

Addition Rules for Probability:

Experiment 1: A single 6-sided die is rolled. What is theprobability of rolling a 2 or a 5?

Possibilities: 1. The number rolled can be a 2.

2. The number rolled can be a 5.Events: These events are mutually exclusive

since they cannot occur at the sametime.

Probabilities: How do we find the probabilities of these mutually exclusive events? We needa rule to guide us.

Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event.

P(A or B) = P(A) + P(B)

Let's use this addition rule to find the probability for Experiment 1.

Experiment 1: A single 6-sided die is rolled. What is the probability of rolling a 2 or a 5?


16

P(5) = 1

http://popupwindow%28%27sample_space%27%29/






Page | 13

6

P(2 or 5) = P(2) + P(5)

=1

+1

6 6

=

2

6

=13

Experiment 2: A spinner has 4 equal sectors colored yellow, blue, green,and red. What is the probability of landing on red or blueafter spinning this spinner?


14

P(blue) =14

P(red or blue) = P(red) + P(blue)

=1

+1

4 4

=24

=12

Experiment 3: A glass jar contains 1 red, 3 green, 2 blue, and 4 yellowmarbles. If a single marble is chosen at random from the

jar, what is the probability that it is yellow or green?Probabilities:

P(yellow) =410

P(green) =310

P(yellow or green) = P(yellow) + P(green)

=4

+3

10 10

=710

In each of the three experiments above, the events are mutually exclusive. Let's look at someexperiments in which the events are non-mutually exclusive.

Experiment 4: A single card is chosen at random from a standarddeck of 52 playing cards. What is the probability of choosing a king or a club?



Page | 14

Probabilities: P(king or club) = P(king) + P(club) - P(king of clubs)

=4

+13

-1

52 52 52

=1652

= 413

In Experiment 4, the events are non-mutually exclusive. The addition causes the king of clubsto be counted twice, so its probability must be subtracted. When two events are non-mutuallyexclusive, a different addition rule must be used.

Addition Rule 2: When two events, A and B, are non-mutually exclusive, the probability that A or B will occur is:

P(A or B) = P(A) + P(B) - P(A and B)

In the rule above, P(A and B) refers to the overlap of the two events. Let's apply this rule tosome other experiments.

Experiment 5: In a math class of 30 students, 17 are boys and 13 aregirls. On a unit test, 4 boys and 5 girls made an A grade.If a student is chosen at random from the class, what isthe probability of choosing a girl or an A student?

Probabilities: P(girl or A) = P(girl) + P(A) - P(girl and A)

=13

+9

-5

30 30 30

=1730

Experiment 6: On New Year's Eve, the probability of a personhaving a car accident is 0.09. The probability of aperson driving while intoxicated is 0.32 andprobability of a person having a car accident whileintoxicated is 0.15. What is the probability of aperson driving while intoxicated or having a car accident?

Probabilities:P(intoxicated or accident) = P(intoxicated) + P(accident) - P(intoxicated and accident)

= 0.32 + 0.09 - 0.15= 0.26

Summary: To find the probability of event A or B, we must first determine whether the eventsare mutually exclusive or non-mutually exclusive. Then we can apply theappropriate Addition Rule:



Page | 15

Addition Rule 1: When two events, A and B, are mutually exclusive, theprobability that A or B will occur is the sum of the probability of each event.P(A or B) = P(A) + P(B)

Addition Rule 2:: When two events, A and B, are non-mutually exclusive, there is

some overlap between these events. The probability that A or B will occur is the sum of the probability of each event, minusthe probability of the overlap.P(A or B) = P(A) + P(B) - P(A and B)

Independent Events:Experiment 1: A dresser drawer contains one pair of socks with

each of the following colors: blue, brown, red, whiteand black. Each pair is folded together in a matchingset. You reach into the sock drawer and choose apair of socks without looking. You replace this pair and then choose another pair of socks. What is the

probability that you will choose the red pair of socksboth times?

There are a couple of things to note about this experiment. Choosing a pairs of socks from thedrawer, replacing it, and then choosing a pair again from the same drawer is a compoundevent. Since the first pair was replaced, choosing a red pair on the first try has no effect on theprobability of choosing a red pair on the second try. Therefore, these events are independent.

Definition: Two events, A and B, are independent if the fact that A occurs does not affect theprobability of B occurring.

Some other examples of independent events are:

Landing on heads after tossing a coin AND rolling a 5 on a single 6-sided die. Choosing a marble from a jar AND landing on heads after tossing a coin. Choosing a 3 from a deck of cards, replacing it, AND then choosing an ace as the second

card. Rolling a 4 on a single 6-sided die, AND then rolling a 1 on a second roll of the die.

To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule isdefined symbolically below. Note that multiplication is represented by AND.

Multiplication Rule 1: When two events, A and B, are independent, the probability of bothoccurring is:

P(A and B) = P(A) · P(B)

(Note: Another multiplication rule will be introduced in the next lesson.) Now we can apply thisrule to find the probability for Experiment 1.

http://x1096653463%28%27compound_event%27%29/








Page | 16

Experiment 1: A dresser drawer contains one pair of socks witheach of the following colors: blue, brown, red, whiteand black. Each pair is folded together in a matchingset. You reach into the sock drawer and choose apair of socks without looking. You replace this pair and then choose another pair of socks. What is the

probability that you will choose the red pair of socksboth times?


15

P(red and red) = P(red) · P(red)

=1

· 1

5 5

=125

Experiment 2: A coin is tossed and a single 6-sided die is rolled.Find the probability of landing on the head side of the coin and rolling a 3 on the die.

Probabilities:P(head) =

12

P(3) =16

P(head and 3) = P(head) · P(3)

=1

· 1

2 6

=1

12

Experiment 3: A card is chosen at random from a deck of 52 cards. Itis then replaced and a second card is chosen. What isthe probability of choosing a jack and then an eight?

Probabilities:P(jack) =

452

P(8) =452

P(jack and 8) = P(jack) · P(8)

=

4

·

4

52 52

=16

2704

=1

169

Experiment 4: A jar contains 3 red, 5 green, 2 blue and 6 yellow



Page | 17

marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. What is theprobability of choosing a green and then a yellow marble?

Probabilities:P(green) =

516

P(yellow) = 616

P(green and yellow) = P(green) · P(yellow)

=5

· 6

16 16

=30256

=15128

Each of the experiments above involved two independent events that occurred in sequence. Insome cases, there was replacement of the first item before choosing the second item; thisreplacement was needed in order to make the two events independent. Multiplication Rule 1can be extended to work for three or more independent events that occur in sequence. This isdemonstrated in Experiment 5 below.

Experiment5:

A school survey found that 9 out of 10 students like pizza. If threestudents are chosen at random with replacement, what is theprobability that all three students like pizza?

Probabilities:

P(student 1 likes pizza)=

9 1

0


9 10


9 10

P(student 1 and student 2 and student 3 like

pizza)

=

9 ·

9 ·

9=

729

10 10 10 1000

All of the experiments above involved independent events with a small population (e.g. A 6-sided die, a 2-sided coin, a deck of 52 cards). When a small number of items are selected froma large population without replacement , the probability of each event changes so slightly thatthe amount of change is negligible. This is illustrated in the following problem.



Page | 18

Problem: A nationwide survey found that 72% of people in the UnitedStates like pizza. If 3 people are selected at random, what isthe probability that all three like pizza?

Solution: Let L represent the event of randomly choosing a person wholikes pizza from the U.S.

P(L) · P(L) · P(L) = (0.72)(0.72)(0.72) = 0.37 = 37%

In the next lesson, we will address how to handle non-replacement in a small population.

Summary: The probability of two or more independent events occurring in sequence can befound by computing the probability of each event separately, and then multiplyingthe results together.

Dependent Events:Experiment 1: A card is chosen at random from a standard deck of 52

playing cards. Without replacing it, a second card ischosen. What is the probability that the first card chosenis a queen and the second card chosen is a jack?

Analysis: The probability that the first card is a queen is 4 out of 52. However, if the first card is not replaced, then thesecond card is chosen from only 51 cards. Accordingly,the probability that the second card is a jack given thatthe first card is a queen is 4 out of 51.

Conclusion: The outcome of choosing the first card has affected the outcome of choosingthe second card, making these events dependent.

Definition: Two events are dependent if the outcome or occurrence of the first affects the

outcome or occurrence of the second so that the probability is changed.

Now that we have accounted for the fact that there is no replacement, we can find theprobability of the dependent events in Experiment 1 by multiplying the probabilities of eachevent.

Experiment1:

A card is chosen at random from a standard deck of 52playing cards. Without replacing it, a second card is chosen.What is the probability that the first card chosen is a queenand the second card chosen is a jack?

Probabilities:

P(queen on first pick) =

4

52

P(jack on 2nd pick given queen on 1stpick)

=

4 51



Page | 19

P(queen and jack)=

4 ·

4=

16=

4

52

51

2652

663

Experiment 1 involved two compound, dependent events. The probability of choosing a jack onthe second pick given that a queen was chosen on the first pick is called a conditional probability .

Definition: The conditional probability of an event B in relationship to an event A is theprobability that event B occurs given that event A has already occurred. Thenotation for conditional probability is P(B|A) [pronounced as The probability of event B given A].

The notation used above does not mean that B is divided by A. It means the probability of event B given that event A has already occurred . To find the probability of the two dependent events,we use a modified version of Multiplication Rule 1, which was presented in the last lesson.

Multiplication Rule 2: When two events, A and B, are dependent, the probability of both occurringis:

P(A and B) = P(A) · P(B|A)

Let's look at some experiments in which we can apply this rule.

Experiment 2: Mr. Parietti needs two students to help him with a sciencedemonstration for his class of 18 girls and 12 boys. He

randomly chooses one student who comes to the front of the room. He then chooses a second student from thosestill seated. What is the probability that both studentschosen are girls?

Probabilities: P(Girl 1 and Girl 2) = P(Girl 1) and P(Girl 2|Girl 1)

=18

· 17

30 29

=306870

=51145

Experiment 3: In a shipment of 20 computers, 3 are defective.Three computers are randomly selected and tested.What is the probability that all three are defective if the first and second ones are not replaced after being tested?

Probabilities: P(3 defectives) =

http://popupwindow%28%27rule1%27%29/






Page | 20

3·

2·

1=

6=

120 19 18 6840 1140

Experiment 4: Four cards are chosen at random from a deck of 52cards without replacement. What is the probability of

choosing a ten, a nine, an eight and a seven inorder?

Probabilities: P(10 and 9 and 8 and 7) =

4·

4·

4·

4=

256=

3252 51 50 49 6,497,400 812,175

Experiment 5: Three cards are chosen at random from a deck of 52cards without replacement. What is the probability of choosing 3 aces?

Probabilities: P(3 aces) =

4 · 3 · 2 = 24 = 152 51 50 132,600 5,525

Summary: Two events are dependent if the outcome or occurrence of the first affects theoutcome or occurrence of the second so that the probability is changed. Theconditional probability of an event B in relationship to an event A is the probabilitythat event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A). When two events, A and B, are dependent, theprobability of both occurring is: P(A and B) = P(A) · P(B|A)

Conditional Probability:

Problem: A math teacher gave her class two tests. 25% of the classpassed both tests and 42% of the class passed the first test.What percent of those who passed the first test also passedthe second test?

Analysis: This problem describes a conditional probability since it asksus to find the probability that the second test was passedgiven that the first test was passed. In the last lesson, thenotation for conditional probability was used in the statementof Multiplication Rule 2.

Multiplication Rule 2: When two events, A and B, are dependent, the probability of both occurringis:

The formula for the Conditional Probability of an event can be derived from Multiplication Rule 2 asfollows:

Start with Multiplication Rule 2.

http://x1096653463%28%27conditional%27%29/






Page | 21

Divide both sides of equation by P(A).

Cancel P(A)s on right-hand side of equation.

Commute the equation.

We have derived the formula for conditional probability.

Now we can use this formula to solve the problem at the top of the page.

Problem: A math teacher gave her class two tests. 25% of the class

passed both tests and 42% of the class passed the first test.What percent of those who passed the first test also passed thesecond test?

Solution: P(Second|First) =P(First and Second)

=0.25

= 0.60 = 60%P(First) 0.42

Let's look at some other problems in which we are asked to find a conditional probability.

Example 1: A jar contains black and white marbles. Two marbles are chosen withoutreplacement. The probability of selecting a black marble and then a white marbleis 0.34, and the probability of selecting a black marble on the first draw is 0.47.What is the probability of selecting a white marble on the second draw, given thatthe first marble drawn was black?

Solution:P(White|Black) =

P(Black and White)=

0.34= 0.72 = 72%

P(Black) 0.47

Example 2: The probability that it is Friday and that a student is absent is 0.03. Since thereare 5 school days in a week, the probability that it is Friday is 0.2. What is theprobability that a student is absent given that today is Friday?

Solution:P(Absent|Friday) =

P(Friday and Absent)=

0.03= 0.15 = 15%

P(Friday) 0.2

Example 3: At Kennedy Middle School, the probability that a student takes Technology andSpanish is 0.087. The probability that a student takes Technology is 0.68. What isthe probability that a student takes Spanish given that the student is takingTechnology?

Solution:P(Spanish|Technology) =

P(Technology and Spanish)=

0.087= 0.13 = 13%

P(Technology) 0.68



Page | 22

Summary: The conditional probability of an event B in relationship to an event A is theprobability that event B occurs given that event A has already occurred. Thenotation for conditional probability is P(B|A), read as the probability of B given A. The formula for conditional probability is:

The Venn Diagram below illustrates P(A), P(B), and P(A and B). What twosections would have to be divided to find P(B|A)? Answer

http://x1096653463%28%27answer%27%29/

http://x1096653463%28%27answer%27%29/

http://x1096653463%28%27answer%27%29/

Documents

A Journey to Probability Concept