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

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} $