JEE Main 2017 — Probability Question with Solution

The number of ways to form a committee having at least one woman is

$$={}^5{C}_1\times {}^{10}{C}_3+{}^5{C}_2\times {}^{10}{C}_2+{}^5{C}_3\times {}^{10}{C}_1+{}^5{C}_4$$

$$=\frac{5!}{4!}\times \frac{10!}{7!\times 3!}+\frac{5!}{2!3!}\times \frac{10!}{8!2!}+\frac{5!}{3!2!}\times \frac{10!}{9!1!}+\frac{5!}{4!}$$

$$=5\times \frac{10\times 9\times 8}{3\times 2}+\frac{5\times 4}{2\times 1}\times \frac{10\times 9}{2}+\frac{5\times 4}{2\times 1}\times 10+5$$

$$=600+450+100+5=1155$$

The number of ways to form a committee having more women than men is

$${}^5{C}_2\times {}^{10}{C}_2+{}^5{C}_4=\frac{5!}{6!2!}\times \frac{10!}{9!}+\frac{5!}{4!}$$

$$=10\times 10+5=105$$

Therefore, the probability for the committees to have more women than men is

$$\frac{105}{1155}=\frac{1}{11}$$