Transcript Example 3: Probability and Combinations

§11.5, Probability with the
Fundamental Counting Principle,
Permutations, and Combinations
© 2010 Pearson Prentice Hall. All rights reserved.
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Learning Targets
I will compute probabilities with permutations.
I will compute probabilities with combinations
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© 2010 Pearson Prentice Hall. All rights reserved.
Example 1: Probability and Permutations
Six jokes about books by Groucho Marx, George Carlin,
Steven Wright, Greg Ray, Jerry Seinfeld, and Phyllis
Diller are each written on one of six cards. The cards are
placed in a hat and drawn one at a time. What is the
probability that a man’s joke will be delivered first and a
woman’s joke last?
Solution:
P(man first, woman last)=
number of permutations man's joke first, woman's joke last
total number of possible permutations
6  5  4  3  2 1  720 permutations
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© 2010 Pearson Prentice Hall. All rights reserved.
Example 1: Probability and Permutations
continued
• Use the Fundamental Counting Principle to find the
number of permutations with a man’s joke first and a
woman’s joke last:
5  4  3  2 11  120 permutations
P(man first, woman last)=
120 1

720 6
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Example 2: Probability and Combinations:
Winning the Lottery
Florida’s lottery game, LOTTO, is set up so that each player
chooses six different numbers from 1 to 53. With one LOTTO
ticket, what is the probability of winning this prize?
Solution:
Because the order of the six numbers does not matter, this
situation involves combinations.
P(winning)=
53 C6 
number of ways of winning
total number of possible combinations
53!
53! 53  52  51 50  49  48  47


 22,957, 480
(53  6)!6! 47!6!
47! 6  5  4  3  2 1
1
P(winning) 
 0.0000000436
22,957, 480
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Example 3: Probability and Combinations
A club consists of five men and seven
women. Three members are selected
at random to attend a conference.
Find the probability that the selected
group consists of 3 men.
Solution:
Order of selection does not matter, so this is a problem
involving combinations.
number of ways of selecting 3 men
P(3 men)=
total number of possible combinations
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Example 3: Probability and Combinations
continued
number of ways of selecting 3 men
P(3 men)=
total number of possible combinations
12!
12! 12 1110  9!


 220
12 C3 
(12  3)!3! 9!3!
9! 3  2 1
There are 220 possible three-person selections.
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© 2010 Pearson Prentice Hall. All rights reserved.
Example 3: Probability and Combinations
continued
5 C3 
P(3 men)=
5!
5! 5  4  3!


 10
(5  3)!3! 2!3! 3! 2 1
number of ways of selecting 3 men
10
1


total number of possible combinations 220 22
Homework: Pg 634 – 635, #1 – 18.
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© 2010 Pearson Prentice Hall. All rights reserved.