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(D) The total number of persons is $3 + 2 + 4 = 9$.The total number of ways to select $4$ persons from $9$ is given by ${}^{9}C_{4} = \frac{9 	imes 8

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$\frac{10}{21}$

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(D) The total number of persons is $3 + 2 + 4 = 9$.The total number of ways to select $4$ persons from $9$ is given by ${}^{9}C_{4} = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$.We want to select exactly $2$ children from $4$ children,which can be done in ${}^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$ ways.To complete the selection of $4$ persons,the remaining $2$ persons must be selected from the $5$ adults ($3$ men + $2$ women). This can be done in ${}^{5}C_{2} = \frac{5 	imes 4}{2 	imes 1} = 10$ ways.The number of favorable outcomes is ${}^{4}C_{2} 	imes {}^{5}C_{2} = 6 	imes 10 = 60$.The probability $P(E)$ is $\frac{60}{126} = \frac{10}{21}$.

Four persons are selected at random from a group of $3$ men,$2$ women and $4$ children. The chance that exactly two of them are children is

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