From a group of 8 boys and 3 girls,a committe | vedclass

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(B) Total number of ways to form a committee of $5$ members from $8$ boys and $3$ girls is given by $C(11, 5) = \frac{11!}{5!6!} = \frac{11 	imes 10

Vedclass · Accepted Answer

$\frac{2}{11}$

Vedclass · Answer

(B) Total number of ways to form a committee of $5$ members from $8$ boys and $3$ girls is given by $C(11, 5) = \frac{11!}{5!6!} = \frac{11 	imes 10 	imes 9 	imes 8 	imes 7}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 462$. If $2$ particular girls must be included,we need to select the remaining $3$ members from the remaining $9$ people ($8$ boys and $1$ girl). The number of ways to select the remaining $3$ members is $C(9, 3) = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$. Therefore,the required probability is $\frac{84}{462} = \frac{2}{11}$.

From a group of $8$ boys and $3$ girls,a committee of $5$ members is to be formed. Find the probability that $2$ particular girls are included in the committee.

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