Question

A committee of $5$ persons is to be randomly selected from a group of $5$ men and $4$ women and a chairperson will be randomly selected from the committee. The probability that the committee will have exactly $2$ women and $3$ men and the chairperson will be a man is $p,$ then $7p$ is equal to

Answer 1

Answer: 2

Total cases $={}^9{C}_5\times {}^5{C}_1$ ( $\therefore$ there are 5 ways to choose a chairperson) Favourable cases $=3$ men can be chosen in ${}^5{C}_3$ ways, 2 women can be chosen in ${}^4{C}_2$ ways and there are 3 ways to choose the chairperson
Total number of favourable ways $={}^5{C}_3\times {}^4{C}_2\times 3$
$\begin{array}{l}\text{ Required probability }=\frac{{}^5{C}_3\times {}^4{C}_2\times 3}{{}^9{C}_5\times 5}\\ =\frac{10\times 6\times 3}{126\times 5}=\frac{2}{7}\end{array}$