Transcript Document
Thinking
Mathematically
Probability with the Fundamental
Counting Principle, Permutations,
and Combinations
Example Probability and Permutations
Five groups in a tour, Offspring, Pink Floyd,
Sublime, the Rolling Stones, and the Beatles,
agree to determine the order of performance based
on a random selection. Each band’s name is
written on one of five cards. The cards are placed
in a hat and then five cards are drawn out, one at a
time. The order in which the cards are drawn
determines the order in which the bands perform.
What is the probability of the Rolling Stones
performing fourth and the Beatles last?
Solution
We begin by applying the definition of
probability to this situation.
P(Rolling Stones fourth, Beatles last) =
(permutations with Rolling Stones fourth, Beatles last)
(total number of possible permutations)
We can use the Fundamental Counting Principle to find
the total number of possible permutations.
5 4 3 2 1 = 120
Solution cont.
We can also use the Fundamental Counting Principle
to find the number of permutations with the
Rolling Stones performing fourth and the Beatles
performing last. You can choose any one of the
three groups as the opening act. This leaves two
choices for the second group to perform, and only
one choice for the third group to perform. Then
we have one choice for fourth and last.
32111=6
There are six lineups with Rolling Stones fourth and
Beatles last.
Solution cont.
Now we can return to our probability fraction.
P(Rolling Stones fourth, Beatles last) =
(permutations with Rolling Stones fourth, Beatles last)
(total number of possible permutations)
= 6/120 = 1/20
The probability of the Rolling Stones
performing fourth and the Beatles last is 1/20.
Example Probability and Combinations:
Winning the Lottery
Florida’s lottery game, LOTTO, is set up so
that each player chooses six numbers from 1
to 53. If the six numbers chosen match the
six numbers drawn randomly twice weekly,
the player wins (or shares) the top cash
prize. With one LOTTO ticket, what is the
probability of winning this prize?
Solution
Because the order of the six numbers does
not matter, this is a situation involving
combinations. We begin with the formula
for probability.
P(winning) = number of ways of winning
total number of possible combinations
We can use the combinations formula to find
the total number of combinations.
Solution cont.
53!
53!
53 C6
(53 6)!6! 47!6!
53 52 51 50 49 48 47!
22,957,480
47!6 5 4 3 2 1
There are nearly 23 million number combinations
possible in LOTTO, and only one is a winning
combination.
P(winning) = number of ways of winning
total number of possible combinations
= 1/22,957,480
Example 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:
a. three men.
b. one man and two women.
Solution
We begin with the probability of selecting
three men.
P( 3 men)=number of ways of selecting 3 men
total number of possible combinations
12C3 = 12!/((12-3)!3!) = 220
5C3 = 5!/((5-3)!(3!)) = 10
P(3 men) = 10/220 = 1/22
Solution part b
We set up the fraction for the probability that
the selected group consists of one man and
two women. P(1 man and 2 women) =
number of ways of selecting 1 man and 2 women
total number of possible combinations
We know the denominator is 12C3 = 220.
Next we move to the numerator of the
probability fraction.
Solution part b cont.
The number of ways of selecting r = 1 man
from n = 5 men is
5C1 = 5!/(((5-1)!1!) = 5
The number of ways of selecting r = 2 women
from n=7 women is
7C2 = 7!/((7-2)!2!) = 21
Solution part b cont.
By the Fundamental Counting Principle, the number
of ways of selecting 1 man and 2 women is
5C1
7C2 = 5 21 = 105
Now we can fill in the numbers in our probability
fraction. P(1 man and 2 women) =
number of ways of selecting 1 man and 2 women
total number of possible combinations
= 105/220 = 21/44
Thinking
Mathematically
Probability with the Fundamental
Counting Principle, Permutations,
and Combinations