From a group of 10 men and 5 women,a four-mem | vedclass

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(C) Total ways to choose a $4$-member committee from $15$ people is $^{15}C_4 = \frac{15 	imes 14 	imes 13 	imes 12}{4 	imes 3 	imes 2 	imes 1}

Vedclass · Accepted Answer

$\frac{1}{11}$

Vedclass · Answer

(C) Total ways to choose a $4$-member committee from $15$ people is $^{15}C_4 = \frac{15 	imes 14 	imes 13 	imes 12}{4 	imes 3 	imes 2 	imes 1} = 1365$. Let $S$ be the set of committees with at least one woman. The number of committees with no women is $^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$. So,the number of committees with at least one woman is $n(S) = 1365 - 210 = 1155$. We want the probability that the committee has more women than men. This happens if the committee has $3$ women and $1$ man,or $4$ women and $0$ men. Number of ways for $3$ women and $1$ man: $^{5}C_3 	imes ^{10}C_1 = 10 	imes 10 = 100$. Number of ways for $4$ women and $0$ men: $^{5}C_4 	imes ^{10}C_0 = 5 	imes 1 = 5$. Total favorable ways = $100 + 5 = 105$. The required probability is $\frac{105}{1155} = \frac{1}{11}$.

From a group of $10$ men and $5$ women,a four-member committee is to be formed such that it must contain at least one woman. The probability that the committee has more women than men is:

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