A school site committee is to be chosen from 6 men
and 5 women. If the committee consists of 4 members, what is the probability that 2 of them
are men? How many men do you expect to be on the committee?
Let XX = the number of men on the committee of 4. The men are the group of interest (first
group).
XX takes on the values 0, 1, 2, 3, 4, where r=6r=6, b=5b=5 , and n=4n=4.
X~H(6, 5, 4)X~H(6, 5, 4)
Find
P
(X=2
)P
(X2
).
P
(X=2
)=0.4545
P
(X2
)
0.4545
(calculator or computer)
Currently, the TI-83+ and TI-84 do not have hypergeometric probability
functions. There are a number of computer packages, including Microsoft Excel, that
do.
The probability that there are 2 men on the committee is about 0.45.
The graph of X~H(6, 5, 4)X~H(6, 5, 4) is:

The yy-axis contains the
probability of XX, where XX =
the number of men on the
committee.
You would expect m=2.18m=2.18(about 2) men on the committee.
The formula for the mean is
μ=
n⋅r
r+b
=
4⋅6
6+5
=2.18
μ
n⋅r
r+b
4⋅6
6+5
2.18
The formula for the variance is fairly complex. You will find it in the Summary of the Discrete Probability Functions Chapter [link pending].