Download Section 11.5 - Probability with the Fundamental Counting Principle

Section 11.5 Probability with the Fundamental
Counting Principle, Permutations,
and Combinations
1.
2.
10/2/2011
Example 1 - Probability with
Permutations.
Objectives
Compute probabilities with
permutations.
Compute probabilities with
combinations.
4 students, Corey (C), Matt (M), Shanity
(S), and Hope (H), have volunteered to
work problems at the board. The order will
be decided by drawing names out of a hat.
What is the probability that
a. They go in alphabetical order?
b. C is first and H is last?
Since order matters, we count permutations.
1
10/2/2011
Section 11.5
2
Section 11.5
Solution
Solution (continued)
Recall the theoretical probability of any
event, E, is
a) Since there is only one way to arrange the
students in alphabetical order (n(E) = 1)
P (alphabetical order) =
1
24
P(E) = number of outcomes in event E =
total number of possible outcomes
n( E )
n( S )
b) Use the fundamental counting principle to find
the number of possible outcomes with C first and H
last:
Count total number of possible outcomes
in both cases as 4P4 = 4! = 24 = n(S)
We still need to calculate n(E) for each
event.
Position:
1
2
3
4
n(C first, H last) =
Number of
ways to choose
1
2
1
1
1 . 2 .1 . 1 = 2
10/2/2011
Section 11.5
4
Example 2 – Choosing conference
attendees
Solution
So,
P(C first, H last) =
n(C first, H last)
2
1
=
=
total possible outcomes 24 12
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
a) 3 men
b) 1 man, 2 women
10/2/2011
Section 11.5
5
10/2/2011
Section 11.5
6
Example 2 – Choosing conference
attendees
Example 2 , continued
Solution:
a)
P(3 men) = number of ways of selecting 3 men
Out of the 12,
select 3 for
total possible
number ways
to select a
conference,
n(S).
total number of possible combinations
Order of selection does not matter, so this is
a problem involving combinations.
10/2/2011
Section 11.5
7
Example 2 continued
Example 2 continued
Number of ways to select 3 men, n(E):
5
C3 =
b) Choosing 1 man and 2 women is a 2step process (it doesn’t matter which step
is 1st).
5!
5!
=
= 10
(5 − 3)!3! 2!3!
Number of ways to pick any 3 club
members, n(S):
12!
12!
C3 =
=
= 220
12
(12 − 3)!3! 9!3!
Therefore: P(3 men ) = n( E ) = 10 = 1
n( S ) 220 22
10/2/2011
Out of 5 men, select 3
for n(E) where E is the
event that 3 men are
chosen for the
conference.
To count the number of possible outcomes
for the event, we will need to know how
many ways to complete each step, and
use the Fundamental Counting Principle.
≈ .05
Section 11.5
9
Example 2 continued
Time permitting
b) P(1 man, 2 women) =
number of ways of selecting 1 man, 2 women
total number of possible combinations
Calculate P( 2 men, 1 women) and
P(3 women) to verify that
Choose:
Number of
ways
1 man
5C1
=5
2 women
7C2=21
The number of was to choose 1 man, 2 women =
5 . 21 = 105, so
P(1 man, 2 women) =
105
≈ .48
220
sum of probabilities of every possible
outcome = 1
Example 3 continued
Example 3 - Winning the Lottery
In Virginia’s Cash 5 lottery game, players
choose 5 numbers from 1 to 34. The
prizes on a $1 ticket are:
$100,000 if you match all 5 in any order,
$100 for 4,
$5 for 3.
Solution:
The probability of each of this events is equal to:
number of ways of winning
total number of possible outcomes
For each case, the total number of possible
outcomes is the same. Because the order of the
5 numbers does not matter, the total possible
outcomes involves counting combinations, and
equals:
Calculate the probability of each of these
results.
n
10/2/2011
Section 11.5
13
C r = 34 C5 =
34!
34!
=
= 278,256
(34 − 5)!5! 29!5!
10/2/2011
Section 11.5
Example 3 continued
Example 3 continued
To find the P(match 4) and P(match 3), we need to
know the number of ways to match 4 and 3.
Solution (cont’d)
There is only one way of winning (choose all
5 correct numbers). So if the event, E, is
match all 5 numbers, n(E) = 1.
P(match all 5) =
These outcomes can be viewed as 2-step
processes. To match 4, we need to choose 4 of
the 5 winning numbers, and 1 of the 29 losing
numbers.
We will use the fundamental counting principle.
1
≈ .0000036
278,256
Choose:
Number of
ways
10/2/2011
14
Section 11.5
4 correct
5C4
=5
29C1
= 29
15
Example 3 continued
Example 3 continued
There are 5 . 29 = 145 possible outcomes
with 4 numbers correct
(n(4 correct) = 145), so
To match 3:
145
≈ .00052
P(match all 4) = 278,256
1 wrong
Choose:
Number of
ways
3 correct
5C3
= 10
2 wrong
29C2=406
There are 10 . 406 = 4060 combinations
with 3 numbers correct, so
P(match all 3) =
4060
≈ .015
278,256
Compare to Virginia Lottery Web
Page
Our results
Virginia Lottery
probabilities
5 out of 5
1
278,256
1
278,256
4 out of 5
145
≈ .00052
278,256
1
≈ .00052
1,919
3 out of 5
4060
≈ .015
278,256
1
≈ .014
69