QJEEM17Ma94b6: JEE Main 2017 Mathematics - Class 12 JEEM17Ma94b6 Solution: From a group of 10 men and 5 women, four member committees are…

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

$= {}^5{C_1} \times {}^{10}{C_3} + {}^5{C_2} \times {}^{10}{C_2} + {}^5{C_3} \times {}^{10}{C_1} + {}^5{C_4}$

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

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

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

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

${}^5{C_2} \times {}^{10}{C_2} + {}^5{C_4} = {{5!} \over {6!2!}} \times {{10!} \over {9!}} + {{5!} \over {4!}}$

$= 10 \times 10 + 5 = 105$

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

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