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Question

(A) Total number of persons $= 3 + 2 + 4 = 9$.We need to choose $4$ persons out of $9$,which can be done in ${}^9C_4$ ways.${}^9C_4 = \frac{9 	imes 8

Vedclass · Accepted Answer

$\frac{10}{21}$

Vedclass · Answer

(A) Total number of persons $= 3 + 2 + 4 = 9$.We need to choose $4$ persons out of $9$,which can be done in ${}^9C_4$ ways.${}^9C_4 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.We want exactly $2$ children out of $4$ children,which can be chosen in ${}^4C_2$ ways.${}^4C_2 = \frac{4 	imes 3}{2 	imes 1} = 6$.The remaining $2$ persons must be chosen from the $5$ adults ($3$ men $+ 2$ women),which can be done in ${}^5C_2$ ways.${}^5C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$.The number of favorable outcomes $= {}^4C_2 	imes {}^5C_2 = 6 	imes 10 = 60$.The required probability $= \frac{60}{126} = \frac{10}{21}$.

If four persons are chosen at random from a group of $3$ men,$2$ women,and $4$ children,then the probability that exactly two of them are children is:

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