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

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$\frac{2}{23}$

$\frac{3}{11}$

$\frac{21}{220}$

$\frac{1}{11}$

answer is C.

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Detailed Solution

a group of 10m, 5 women 4 members are Selected at least on women

10 M	5 W	no of ways
3	1	${}^{10}{c_{3}} \times {}^{5}{c_{1}}$
2	2	${}^{10}{c_{2}} \times {}^{5}{c_{2}}$
1	3	${}^{10}{c_{1}} \times {}^{5}{c_{3}}$
0	4	${}^{10}{c_{0}} \times {}^{5}{c_{4}}$

$\begin{array}{rcl} n (S) & = & 120 \times 5 + 45 \times 10 + 10 \times 10 + 5 \\ = & 600 + 450 + 100 + 5 = 1155 \end{array}$

E=These Committees to have more women than men

$\begin{array}{rcl} n (E) & = & 100 + 5 = 105 \\ P (E) & = & \frac{105}{1155} = \frac{1}{11} \end{array}$

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