Question

From a group of $10$ men and $5$ women, four member committees are to be formed each of which must contain at least one women. Then the probability for these committees to have more women than men, is.

• A. $\frac{3}{11}$
• B. $\frac{2}{23}$
• C. $\frac{1}{11}$
• D. $\frac{21}{220}$

Answer 1

Answer: (C)

$\frac{1}{11}$

Total number of ways ${=}^5{C}_1{\times }^{10}{C}_3{+}^5{C}_2{\times }^{10}{C}_2{+}^5{C}_3{\times }^{10}{C}_1{+}^5{C}_4$

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

$=5\times \frac{10\times 9\times 8}{3\times 2}+\frac{5\times 4}{2\times 1}\times \frac{10\times 9}{2}+\frac{5\times 4}{2\times 1}\times 10+5$
$=5\times 5\times 3\times 8+5\times 2\times 5\times 9+10\times 10+5$
$=600+450+100+5$

$=1155$
Total no. of ways=$1155$
No. of ways to form a committee having more women than men.

$=5_{C_3}\times {10}_{C_1}+5_{C_4}=\frac{5!}{3!2!}\times \frac{10!}{9!}+\frac{5!}{4!}$
$=10\times 10+5$

$=105$

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