Solution

$$=5_{C_1}\times 10_{C_3}+5_{C_2}\times 10_{C_2}+5_{C_3}\times 10_{C_7}+5_{C_4}$$
$$=\displaystyle\frac{5!}{4!}\times \frac{10!}{7!3!}+\frac{5!}{2!3!}\times \frac{10!}{8!2!}+\frac{5!}{3!2!}\times  \frac{10!}{9!}+\frac{5!}{4!}$$
$$=\displaystyle 5\times \frac{10 \times 9\times 8}{3\times 2}+\frac{5\times 4}{2\times 1}\times \frac{10\times 9}{2}+\frac{5\times 4}{2\times 1}\times 10+5$$
$$=5\times 5\times 3\times 8+5\times 2\times 5\times 9+10\times 10+5$$
$$=600+450+100+5$$
$$=1155$$
$$\Rightarrow$$ Total no. of ways.
No. of ways to form a committee having more women than men.
$$\displaystyle =5_{C_3}\times 10_{C_1}+5_{C_4}=\displaystyle\frac{5!}{3!2!}\times \frac{10!}{9!}+\frac{5!}{4!}$$
$$=10\times 10+5$$
$$=105$$
Probability$$=\displaystyle\frac{105}{1155}=\frac{1}{11}$$.