Page 145 - Probability Demystified

134 CHAPTER 8 Other Probability Distributions

5 3 2
3. n ¼ 5, x 1 ¼ 2, x 2 ¼ 2, x 3 ¼ 1, and p ¼ 10 , p ¼ 10 , and p ¼ 10
2
3
1

5! 5 2 3 2 2 1
The probability is ¼ 30ð0:0045Þ¼ 0:135
2!2!1! 10 10 10
4. n ¼ 10, x 1 ¼ 5, x 2 ¼ 3, x 3 ¼ 2, and p 1 ¼ 0.5, p 2 ¼ 0.3, and p 3 ¼ 0.2

10!
3
5
2
The probability is ð0:5Þ ð0:3Þ ð0:2Þ ¼ 2520 ð0:00003375Þ¼
5!3!2!
0:08505
9 3 3
5. n ¼ 8, x 1 ¼ 3, x 2 ¼ 2, x 3 ¼ 2, x 4 ¼ 1, and p ¼ , p ¼ , p ¼ ,
1
2
3
1 16 16 16
and p ¼
4
16
1
2
3
2
8! 9 3 3 1
The probability is ¼1680ð0:000013748Þ¼
3!2!2!1! 16 16 16 16
0:0231
The Hypergeometric Distribution
When a probability experiment has two outcomes and the items are selected
without replacement, the hypergeometric distribution can be used to compute
the probabilities. When there are two groups of items such that there are
a items in the first group and b items in the second group, so that the total
number of items is a þ b, the probability of selecting x items from the first
group and n x items from the second group is
C C
a x b n x
C
aþb n
where n is the total number of items selected without replacement.

EXAMPLE: A committee of 4 people is selected at random without replace-
ment from a group of 6 men and 4 women. Find the probability that the
committee consists of 2 men and 2 women.

SOLUTION:
Since there are 6 men and 2 women, a ¼ 6, b ¼ 4 and n ¼ 6 þ 4 or 10. Since the
committee consists of 2 men and 2 women, x ¼ 2, and n x ¼ 4 2 ¼ 2.