Question

A committee of three has to be chosen from a group of $4$ men and $5$ women. If the selection is made at random, what is the probability that exactly two members are men?

• A. $\frac{5}{14}$
• B. $\frac{1}{21}$
• C. $\frac{3}{14}$
• D. $\frac{8}{21}$

Answer 1

The correct option is A $\frac{5}{14}$

Total number of members $=4+5=9$

A committee of $3$ has to be chosen.

$\therefore n\left(S\right)=9C_3=84$

Let $E$ be the event where exactly two members are men.

$\therefore n\left(E\right)=4C_2\times 5C_1=6\times 5=30$

Therefore $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{30}{84}=\frac{5}{14}$