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Counting Principles(Permutations and Combinations )
© 2012 Pearson Education, Inc.. All rights reserved.
Example
A combination lock can be set to open to any 4-digit
sequence.
(a) How many sequences are possible?
(b) How many sequences are possible if no digit is repeated?
© 2012 Pearson Education, Inc.. All rights reserved.
Solution: (a) Since there are 10 digits namely 0, 1, 2…..9, thereare 10 choices for each of the digit. By the multiplicationprinciple, there are 10 ∙10 ∙10 ∙10 =10,000 different sequences.
(b)There are 10 choices for the first digit. It cannot be usedagain, so there are 9 choices for the second digit, 8 choices forthe third digit, and then 7 choices for the fourth digit.Consequently, the number of such sequences is 10 ∙ 9 ∙ 8 ∙7 =5040 different sequences.
© 2012 Pearson Education, Inc.. All rights reserved.
Example
How many different ways can you choose a bagel, muffin or donut to eat and coffee or juice to drink?
Tree diagram
Example
A teacher is lining up 8 students for a spelling bee. How many
different line-ups are possible?
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Solution: Eight choices will be made, one for each space that willhold a student. Any of the students could be chosen for the first space. There are 7 choices for the second space, since 1 studenthas already been placed in the first space; there are 6 choices forthe third space, and so on. By the multiplication principle, the number of different possiblearrangements is 8 ∙7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙1 = 40,320.
© 2012 Pearson Education, Inc.. All rights reserved.
TI-83/84 function
(order does matter)
Example
A teacher wishes to place 5 out of 8 different books on her
shelf. How many arrangements of 5 books are possible?
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Solution : The teacher has 8 ways to fill the first space, 7 waysto fill the second space, 6 ways to fill the third, and so on…Since the teacher wants to use only 5 books, only 5 spaces canbe filled (5 events) instead of 8, for 8 ∙7 ∙ 6 ∙ 5 ∙ 4 = 6720arrangements.
Example
Find the number of permutations of the letters L, M, N, O, P,
and Q, if just three of the letters are to be used.
Solution:
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!( , ) . Here 6 and 3.
( - )!n
P n r n rn r
6! 6!(6 -3)! 3!
6 5 4 3 2 1120
3 2 1
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How many permutations are there of two letters from the set {A,B,C}.
)!(
!),(
rn
nrnP
623!1
!3
)!23(
!3)2,3(
P
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(order does not matter)
TI-83/84 function
Example
How many committees of 4 people can be formed
from a group of 10 people?
Solution: A committee is an unordered group, so use the
combinations formula for C(10,4).
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10! 10!(10,4)
4!(10 4)! 4!6!C
10 9 8 7 6 5 4 3 2 14 3 2 1 6 5 4 3 2 1
10 9 8 74 3 2 1
210
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Example
From a class of 15 students, a group of 3 or 4 students will be
selected to work on a special project. In how many ways can a
group of 3 or 4 students be selected?
Solution: The number of ways to select group of 3 students
from a class of 15 students is C(15, 3) = 455.
The number of ways to select group of 4 students from a class
of 15 students is C(15, 4) = 1365.
The total number of ways to select a group of 3 or 4 students
will be the 1820.
© 2012 Pearson Education, Inc.. All rights reserved.
© 2012 Pearson Education, Inc.. All rights reserved.
Example
(a) How many 4-digit code numbers are possible if no digits are repeated?
Solution: Since changing the order of the 4 digits results in a
different code, permutations should be used.
(b) A sample of 3 light bulbs is randomly selected from a batch
of 15. How many different samples are possible?
Solution: The order in which the 3 light bulbs are selected is not
important. The sample is unchanged if the items are rearranged,
so combinations should be used.
© 2012 Pearson Education, Inc.. All rights reserved.
Example
(c) In a baseball conference with 8 teams, how many games
must be played so that each team plays every other team exactly once?
Solution: Selection of 2 teams for a game is an unordered
subset of 2 from the set of 8 teams. Use combinations again.
(d) In how many ways can 4 patients be assigned to 6 different
hospital rooms so that each patient has a private room?
Solution: The room assignments are an ordered selection of 4
rooms from the 6 rooms. Exchanging the rooms of any 2
patients within a selection of 4 rooms gives a different
assignment, so permutations should be used.
© 2012 Pearson Education, Inc.. All rights reserved.
Introduction to Probability
Write the elements belonging to the set {x | x is a state whose
name begins with the letter O}.
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Solution: The state names that begin with the letter O make upthe set {Ohio, Oklahoma, Oregon }.
Set – a collection of elements that satisfy a certain condition.
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Example 1
Decide if the statement is true or false.
Solution: The first set is a subset of the second because each element of first set belongs to second set. (The fact that the elements are listed in a different order does not matter.) Therefore, the statement is true.
{2,4,6} {6,2,4}
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© 2012 Pearson Education, Inc.. All rights reserved.
A) What is the sample space?
B) What is the probability of getting a 3?
C) What is the probability of getting an odd number?
Sample Space – The set of all possible outcomes of an experiment or event.
Example
Two coins are tossed, and a head or a tail is recorded for each
coin. Write the event E: the coins show exactly one head.
Solution: Tossing a coin is made up of the outcomes heads (H)
or tails (T). If S represents the sample space of tossing two
coins, then S = {HH, HT, TH, TT}.
Two outcomes satisfy this condition: so,
E = {HT, TH}.
© 2012 Pearson Education, Inc.. All rights reserved.
ExampleTwo coins are tossed, and a head or a tail is recorded for each
coin.
A) Give a sample space for this experiment.
B) What is the probability of getting at least one head.
Solution: A) Tossing a coin is made up of the outcomes heads (H)
or tails (T). If S represents the sample space of tossing two
coins then S = {HH, HT, TH, TT}.
Solution : B) E = {HH,HT,TH}
© 2012 Pearson Education, Inc.. All rights reserved.
Number of elements in the sets
4
3
)(
)()(
Sn
EnEP
384.0 152.0 824.056.0
616.0 712.0
179.0
0625.0
625.025.0
75.0
http://www.youtube.com/watch?v=mhlc7peGlGg&safe=active
Monty Hall probability puzzle
Probability of Multiple Events
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46
5
1
2
6FE
(no outcomes in common; if one event occurs the other cannot)
(E’ is called the complement of E)
Example
A study of workers earning the minimum wage grouped such
workers into various categories, which can be interpreted as
events when a worker is selected at random. Source:
Economic Policy Institute. Consider the following events:
E: worker is under 20;
F: worker is white;
G: worker is female.
Describe the following event in words:
Solution:
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.E F
is the event that the worker is not under 20
and the worker is not white.
E F
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Independent Events
The fact that one event has occurred, does not change the probability of the second event occurring.
** Events that are mutually exclusive must be dependent, but dependent events are not necessarily mutually exclusive**
Read as; the probability of event F occurring given E has occurred
Read as; the probability of event E occurring given F has occurred
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(General rule of addition)
For mutually exclusive events,
So the equation simplifies to
0)( FEP)()()( FPEPFEP
Example If a single card is drawn from an ordinary deck of cards, find
the probability of an ace or a club.
Solution: Let A represent the event “an ace ” and C the event
“club card.” There are 4 aces in the deck, so
There are 13 clubs in the deck, so
Since there is 1 ace of club in the deck,
By the union rule, the probability of the card being an ace or a
Club card is
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( ) 4 / 52.P A
( ) 13 / 52.P C
( ) 1/ 52.P A C
( ) ( ) ( ) ( ).P A C P A P C P A C
4 13 152 52 52
16 4.
52 13
ExampleSuppose two fair dice are rolled. Find the probability that the sum is 8, or both
die show the same number.
Solution:
1-1 1-2 1-3 1-4 1-5 1-6
2-1 2-2 2-3 2-4 2-5 2-6
3-1 3-2 3-3 3-4 3-5 3-6
4-1 4-2 4-3 4-4 4-5 4-6
5-1 5-2 5-3 5-4 5-5 5-6
6-1 6-2 6-3 6-4 6-5 6-6
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By the union rule, (sum is 8 or both die show same number)
5 6 1 10 5= .
36 36 36 36 18
P
Example
Find the probability that when two fair dice are rolled, the sum
is less than 11.
Solution: To calculate this probability directly, we must find the
probabilities that the sum is 2, 3, 4, 5, 6, 7, 8, 9 or 10 and then
add them. It is much simpler to first find the probability of the
complement, the event that the sum is greater than or equal to 11.
1-1 1-2 1-3 1-4 1-5 1-6
2-1 2-2 2-3 2-4 2-5 2-6
3-1 3-2 3-3 3-4 3-5 3-6
4-1 4-2 4-3 4-4 4-5 4-6
5-1 5-2 5-3 5-4 5-5 5-6
6-1 6-2 6-3 6-4 6-5 6-6© 2012 Pearson Education, Inc.. All rights reserved.
(sum 11) 1 (sum 11)P P
3 33 111 .
36 36 12
(General rule of multiplication)
When events are Independent;
So the equation simplifies to:
)()|( EPFEP )()|( FPEFP
Independent
Not Independent
Independent
Not Independent
1/6
0.54 2/7
9/34
9/25
Not mutually exclusive
mutually exclusive
Not mutually exclusive
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P(F); represented by large circle.
P(E and F); represented by overlapping section
Example
The following table shows national employment statistics. Use the table to findeach probability.
7. P(male | professional) 8. P(laborer | female) 9. P(female | sales)
10.P(professional | female) 11.P(sales | male) 12.P(male | laborer)
8937 5801 43839392
9729
19121
47.08937
4190
27.09729
2588 67.0
4383
2951
15.09392
1432 55.0
5801
3213
51.09392
4747
ExampleUse a tree diagram to solve the following problem.
16. A car insurance company compiled the following information from a recentsurvey. 75% of drivers carefully follow the speed limit Of the drivers who carefully follow the speed limit, 80% have never had an accident. Of the drivers who do not carefully follow the speed limit, 65% have never had an accident.
What is the probability that a driver does not carefully follow the speed limit and has never had an accident?
15.0)'( NACP0.75
0.65
0.20
0.80
0.25
0.35
C is the event careful driver.
NA is the event no accident
1625.0)'( NACP
6.0)( NACP
0875.0)''( NACP