Transcript solution

Math in Our World
Section 11.5
Probability Using Permutations
and Combinations
Learning Objectives
 Compute probabilities using combinations.
 Compute probabilities using permutations.
Events with Large Possibilities
When the number of possibilities gets larger, the
combination and permutation rules from will be our
best friends.
Our general game plan will be to use these rules to
find the number of outcomes that satisfy a certain
event, as well as the total number of outcomes in
the sample space.
Then we can divide the first number by the second
to obtain the probability of the event occurring.
EXAMPLE 1
Using Combinations to
Compute Probability
Stacy has the option of selecting three books to
read for a humanities course. The suggested book
list consists of 10 biographies and five current
events books. She decides to select the three
books at random. Find the probability that all three
books selected will be current events books.
SOLUTION
Since there are five current events books and Stacy will
need to select three of them, then there are 5C3 ways of
doing this.
5  4  3!
5!
5!
5
C3 
(5  3)!3!

2!3!

2  1 3!
 10
EXAMPLE 1
Using Combinations to
Compute Probability
SOLUTION
There are 10 ways to choose 3 current events books.
The total number of outcomes in the sample space is 15C3
since she has to select three books from 15 books.
15! 515 714  13  12!
15!


 455
15 C3 
12! 3  2  1
(15  3)!3! 12!3!
The probability of selecting three current events books is
10
2

 0.022
455 91
EXAMPLE 2
Using Combinations to
Compute Probability
What is the probability of getting 4 aces when
drawing 5 cards from a standard deck of 52 cards?
SOLUTION
First, we’ll figure out how many five-card hands have four
aces.
Since there are only four aces in the deck, there’s only one
way to get all four of them in your hand. At that point there
are 48 cards left for the other card in the hand.
Using the fundamental counting principle, there are 1 x 48
ways to be dealt four aces.
EXAMPLE 2
Using Combinations to
Compute Probability
SOLUTION
So there are 48 ways to get a five-card hand with four aces.
The total number of hands with any 5 cards is the
combinations of 5 cards chosen from 52, or 52C5.
24
13 17
52  51 50  49  48  47!
52!
52!


52 C5 
47! 5  4  3  2  1
(52  5)!5! 47!5!
 2,598,960
The probability of getting four aces is
48
1

 0.0000185
2,598,960 54,145
EXAMPLE 3
Using Permutations to
Compute Probability
A combination lock has 40 numbers on it, from
zero to 39. Find the probability that if the
combination to unlock it consists of three numbers,
it will contain the numbers 1, 2, and 3 in some
order. Assume that numbers cannot be repeated in
the combination.
SOLUTION
This is a permutation since the order of the numbers is
important when you are unlocking the lock.
The number of combinations for the lock containing 1, 2,
3! 3  2  1
3!
and 3 is 3P3.
3
P3 
(3  3)!

0!

1
6
EXAMPLE 3
Using Permutations to
Compute Probability
SOLUTION
There are 6 ways to create a combination with 1, 2, and 3.
The total number of combinations is a permutation of the 40
numbers taken 3 at a time, or 40P3.
40! 40  39  38  37!
40!


 59,280
40 P3 
37!
(40  3)! 37!
The probability of the combination containing 1, 2, and 3 is
6
1

 0.000101
9,880
59,280
EXAMPLE 4
Using Combinations to
Compute Probability
A store has six different fitness magazines and
three different news magazines. If a customer
buys three magazines at random, find the
probability that the customer will pick two fitness
magazines and one news magazine.
SOLUTION
There are 6C2 or 15 ways to select two fitness magazines
from six fitness magazines, as shown:
6  5  4!
6!
6!


 15
6 C2 
(6  2)!2! 4!2! 2  1 4!
EXAMPLE 4
Using Combinations to
Compute Probability
SOLUTION
There are 3C1 or three ways to select one magazine from
three news magazines:
3!
3  2!
3!


3
3 C1 
(3  1)!1! 2!1! 2! 1
Using the fundamental counting principle, there are 15 x 3
or 45 ways to select two fitness magazines and one news
magazine.
EXAMPLE 4
Using Combinations to
Compute Probability
SOLUTION
Next, there are 9C3 or 84 ways to select three magazines
from nine magazines:
3
4
9!
9  8  7  6!
9!


 84
9 C3 
(9  3)!3! 6!3! 6! 3  2  1
The probability of selecting two fitness magazines and one
news magazine is
45
15

 0.536
84
28