Question

5 women and 7 men are eligible to serve on a committee. If a committee of 4 is randomly selected, what is the probability that it will consist of 2 women and 2 men?

Answer 1

Answer:

The required  probability is $\frac{14}{33}$.

Step-by-step explanation:

Probability:

The ratio of the favorable outcomes to the all possible outcomes.

Given that, 5 women and 7 men are eligible to serve on a committee.

A committee of 4 is randomly selected.

Choosing 2 women out of 5 women is

$=^5C_2$

$=\frac{5!}{2!(5-2)!}$

$=\frac{5!}{2!3!}$

=10

Choosing 2 men out of 7 men is

$=^7C_2$

$=\frac{7!}{2!(7-2)!}$

$=\frac{7!}{2!5!}$

=21

Total number of eligible person is = (5+7)=12

Choosing 4 member out of 12 member is

=$^{12}C_4$

$=\frac{12!}{4!(12-4)!}$

=495

The required  probability is

$=\frac{10\times 21}{495}$

$=\frac{14}{33}$