§8.3 Probability Applications of Counting Principles
In this section, we will see how we can apply the counting principles from the previous two sections in solving probability problems. Many of the probability problems involving dependent events that were solved earlier by using tree diagrams can also be solved by using permutations and combinations. Permutations and combinations are especially helpful when the numbers involved are large.
To compare the method of using permutations and combinations with the method of tree diagrams used in Section 7.5, the first example repeats Example 8 from that section. (It is a good idea to review that example from section 7.5 before/after going through the following example.)
☼ Example 1. The Environmental Protection Agency is considering inspecting 6 plants for environmental compliance: 3 in Chicago, 2 in Los Angeles, and 1 in New York. Due to a lack of inspectors, they decide to inspect two plants selected at random, one this month and one next month, with each plant equally likely to be selected, but no plant selected twice.
What is the probability that 1 Chicago plant and 1 Los Angeles plant are selected?
· Solution. We note that, although the plants are selected one at a time, with one labeled as the first plant and the other as the second, the probability that 1 Chicago plant and 1 Los Angeles plant are selected should not depend upon the order in which the plants are selected (the outcomes, Chicago followed by Los Angeles, and Los Angeles followed by Chicago are both contained in the required event). So, we may use combinations. The number of ways to select 1 Chicago plant out of 3 Chicago plants and 1 Los Angeles plant out of 2 Los Angeles plants is
3 1
×2 1
= 3 × 2 = 6.
The number of ways to select any 2 plants out of 6 is
6 2
= 6!
4! × 2! = 6 × 5 × 4!
4! × 2 × 1 = 6 × 5 2 = 15.
Thus, the probability that 1 Chicago plant and 1 Los Angeles plant are selected is 6
15 = 2 5 . This agrees with the answer found earlier.
☼ Example 2. From a group of 22 nurses, 4 are to be selected to present a list of grievances to management.
(a) In how many ways can this be done?
· Solution. 4 nurses can be selected from a group of 22 in 22 4
ways. (We use combinations, since the group of 4 is an unordered set.)
22 4
= 22!
18! × 4! = 22 × 21 × 20 × 19 × 18!
18! × 4 × 3 × 2 × 1 = 22 × 21 × 20 × 19
4 × 3 × 2 = 7315 . There are 7315 ways to choose 4 nurses from 22.
(b) One of the nurses is Julie Davis. Find the probability that Davis will be among the 4 selected.
· Solution. The probability that Davis will be selected is given by n(E)
n(S), where E is the event that the chosen group includes Davis, and S is the sample space for the experiment of choosing a group of 4 nurses.
There is only 1 1
= 1 way to choose Davis. The number of ways that the other 3 nurses can be chosen from the remaining 21 nurses is
21 3
= 21!
18! × 3! = 21 × 20 × 19
3 × 2 × 1 = 1330.
Therefore, the probability that Davis will be one of the 4 chosen is
P (Davis is chosen) = P (E) = n(E) n(S) =
1 1
×
21 3
22 4
= 1330
7315 ≈ 0.1818 .
Notice that the two numbers in red in the numerator, 1 and 21, add up to the number in red in the denominator, 22. This indicates that 22 nurses have been split into two groups, one of size 1 (Davis) and the other of size 21 (the other nurses). Similarly, the green numbers indicate that the 4 nurses chosen consist of two groups of size 1 (Davis) and size 3 (the other nurses chosen).
(c) Find the probability that Davis will not be selected.
· Solution. The probability that Davis will not be chosen is 1 − P (E) = 1 − 0.1818 = 0.8182 .
Remark. The problems in the previous two sections – 8.1 and 8.2 – of this chapter asked how many ways a certain operation can be done. The problems in this section ask what is the probability that a certain event occurs; the solution involves answering questions about how many ways the event and the operation can be done.
• If a problem asks how many ways something can be done, the answer must be a nonnegative integer.
• If a problem asks for a probability, the answer must be a number between 0 and 1.
☼ Example 3. When shipping diesel engines abroad, it is common to pack 12 engines in one container that is then loaded on a rail car and sent to a port. Suppose that a company has received complaints from its customers that many of the engines arrive in nonworking condition. To help solve this problem, the company decides to make a spot check of containers after loading. The company will test 3 engines from the container at random; if any of the 3 are nonworking, the container will not be shipped until each engine in it is checked. Suppose that a given container has 2 nonworking engines. Find the probability that container will not be shipped.
· Solution. The container will not be shipped if the sample of 3 engines contains at least 1 defective engine, that is, 1 or 2 defective engines (note that the container contains only 2 defective engines). If P (1 defective) represents that probability of exactly 1 defective engine in the sample, then
P (not shipping) = P (1 defective) + P (2 defective).
There are 12 3
ways to choose the 3 engines for testing:
12 3
= 12!
9! × 3! = 220.
There are2 1
ways of choosing 1 defective engine from the 2 in the container, and for each of these ways there are 10
2
ways of choosing 2 good engines from among the 10 good engines in the container. By the multiplication principle, there are
2 1
×10 2
= 2!
1! × 1! × 10!
8! × 2! = 2 × 45 = 90 ways to choose a sample of 3 engines containing 1 defective engine. Thus,
P (1 defective) = 90 220 = 9
22. There are 2
2
ways of choosing 2 defective engines from the 2 defective engines in the container, and10
1
ways of choosing 1 good engine from among the 10 good engines. Thus there are
2 2
×10 1
= 1 × 10 = 10
ways of choosing a sample of 3 engines containing 2 defective engines. So, P (2 defective) = 10
220 = 1 22. Finally,
P (not shipping) = P (1 defective) + P (2 defective)
=
2 1
×10 2
12 3
+
2 2
×10 1
12 3
= 9
22 + 1 22 = 10
22 ≈ 0.4545 .
Remark. Observe that in Example 3, the complement of finding 1 or 2 defective engines is finding 0 defective engines. Then instead of finding the sum P (1 defective) + P (2 defective), the result in Example 3 could be found as 1 − P (0 defective).
P (not shipping) = 1 − P (0 defective)
= 1 −
2 0
×10 3
12 3
= 1 − 1 × 20 220
= 1 − 120
220 = 100
220 ≈ 0.4545 .
☼ Example 4. In a common form of the card game poker, a hand of 5 cards is dealt to each player from a deck of 52 cards. There are a total of
52 5
= 52!
47! × 5! = 2, 598, 960
such hands possible. Find the probability of getting each of the following hands.
(a) A hand containing only hearts, called a heart flush.
· Solution. There are 13 hearts, and 52 − 13 = 39 other cards in a deck. Thus there
are 13
5
×39 0
= 13!
8! × 5! × 1 = 13 × 12 × 11 × 10 × 9
5 × 4 × 3 × 2 × 1 = 1287
different hands containing only hearts. Hence, the probability of a heart flush is
P (heart flush) =
13 5
×39 0
52 5
= 1287
2, 598, 960 ≈ 0.0004952 .
(b) A flush of any suit (5 cards of the same suit).
· Solution. There are 4 suits in a deck, so
P (flush) = 4 × P (heart flush) = 4 × 0.0004952 ≈ 0.001981 . (c) A full house of aces and eights (3 aces and 2 eights).
· Solution. There are 4 3
ways to choose 3 aces from among the 4 in the deck, and
4 2
ways to choose 2 eights out of the 4 in the deck. Thus,
P (3 aces, 2 eights) =
4 3
×4 2
×44 0
52 3
= 4 × 6 × 1
2, 598, 960 ≈ 0.000009234 .
(d) Any full house (3 cards of one value, 2 of another).
· Solution. There are 13 values in a deck — King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3, 2, Ace, and 4 cards of each value. Thus, there are 13 choices for the first value, and
4 3
ways to choose 3 cards from among the 4 cards that have that value. This leaves 12 choices for the second value (order is important here, since a full house of 3 aces, and 2 eights is not the same as a full house of 3 eights, and 2 aces). From the 4 cards that have the second value, 2 cards can be chosen in 4
2
ways. The probability of any full house is then
P (full house) =
13 ×4 3
× 12 ×4 2
2, 598, 960 ≈ 0.001441 .
☼ Example 5. A music teacher has 3 violin pupils, Fred, Carl, and Helen. For a recital, the teacher selects a first violinist and a second violinist. The third pupil will play with the others, but not solo. If the teacher selects randomly, what is the probability that Helen is first violinist, Carl is second violinist, and Fred does not solo?
· Solution. We use permutations to find the number of arrangements in the sample space.
P (3, 3) = 3! = 6
(We can think of this as filling the positions of the first violin, second violin, and no solo.) The 6 arrangements are equally likely, since the teacher will select randomly. Now there is only one arrangement where Helen is first violinist, Card is second violinist, and Fred does not solo. Thus, the required probability is 1
6 .
☼ Example 6. Suppose a group of n people is in a room. Find the probability that at least 2 of the people have the same birthday.
· Solution. “Same birthday” refers to the month and the day, not necessarily the same year. Also, we ignore leap years, and assume that each day in the year is equally likely as a birthday. To see how to proceed, we look at the case in which n = 5 and find the probability
that no 2 people from among the 5 people have the same birthday. There are 365 different birthdays possible for the first of the 5 people, 364 for the second (so that the people have different birthdays), 363 for the third, and so on. The number of ways that 5 people can have different birthdays is thus the number of permutations of 365 days taken 5 at a time or
P (365, 5) = 365 × 364 × 363 × 362 × 361.
The number of ways that 5 people can have the same or different birthdays is 365 × 365 × 365 × 365 × 365 = (365)5.
Finally, the probability that none of the 5 people have the same birthday is P (365, 5)
3655 = 365 × 364 × 363 × 362 × 361
365 × 365 × 365 × 365 × 365 ≈ 0.9729.
Thus, the probability that at least 2 of the 5 people do have the same birthday is 1 − 0.9729 = 0.0271 .
Now this result can be extended to more than 5 people. Generalizing, the probability that no 2 people among n people have the same birthday is
P (365, n) 365n .
Therefore, the probability that at least 2 of the n people do have the same birthday is 1 − P (365, n)
365n .
The following table shows this probability for various values of n.
Number of People, Probability That Two
n Have the Same Birthday
5 0.0271
10 0.1169
15 0.2529
20 0.4114
22 0.4757
23 0.5073
25 0.5687
30 0.7063
35 0.8144
40 0.8912
50 0.9704
> 365 1
The probability that 2 people among 23 have the same birthday is 0.5073, a little over half.
This is quite surprising!
☼ Example 7. Ray and Nate are arranging a row of fruit at random on a table. They have 5 apples, 6 oranges, and 7 lemons. What is the probability that all fruit of the same kind are together?
· Solution.
Method 1:
Ray can’t tell individual pieces of fruit of the same kind apart. All apples look the same to him, as do all oranges and all lemons. So, in the denominator of the probability, he calculates the number of distinguishable ways to arrange the 5 + 6 + 7 = 18 pieces of fruit, given that all apples are indistinguishable, as are all oranges and all lemons.
18!
5! × 6! × 7! = 14, 702, 688
As for the numerator, the only choice is how to arrange the 3 kinds of fruit, for which there are 3! = 6 ways. Thus
P (all fruit of the same kind are together) = 6
14, 702, 688 = 0.4081 × 10−7 .
Method 2:
Nate has better eyesight than Ray and can tell the individual pieces of fruit apart. So in the denominator of the probability, he calculates the number of ways to arrange the 18 pieces of fruit, which is
18! = 6.4024 × 1015.
For the numerator he must choose how to arrange the 3 kinds of fruit, for which there are 3! ways. Then there are 5! ways to arrange the apples, 6! ways to arrange the oranges, and 7! ways to arrange the lemons, for a total number of possibilities of
3! × 5! × 6! × 7! = 2, 612, 736, 000.
Therefore,
P (all fruit of the same kind are together) = 2, 612, 736, 000
6.4024 × 1015 = 0.4081 × 10−7 .
The results for Method 1 and Method 2 are the same. The probability does not depend on whether a person can distinguish individual pieces of the same kind of fruit.
