Question

A committee has seven men and four women. If four people are selected to go to a conference, what is the chance that the group is two men and two women?

Answer 1

Given:

A committee has seven men and four women.

four people are selected to go to a conference

We will use the combinations as follows

the number of ways to choose the four people are:

$7C4+7C3+7C2+7C1+7C0$

The rule of combinations is:

$\text{nCr}=\frac{n!}{(n-r)!\cdot r!}$

so, the number of ways will be:

$35+35+21+7+1=99$

The number of ways to make a group of two men and two women will be:

$7C2=21$

So, the chance that the group is two men and two women =

$\frac{21}{99}=\frac{7}{33}$

so, the answer will be 7/33