Question

Four persons are chosen at random from a group containing $3$ men, $2$ women and $4$ children. The probability that exactly two of them will be children is

• A. $\frac{5}{21}$
• B. $\frac{6}{21}$
• C. $\frac{10}{21}$
• D. $\frac{12}{21}$

Answer 1

Answer: (C)

$\frac{10}{21}$

Since we want exactly $2$ of them to be children, we choose the $2$ children first in ${}^4{C}_2=6$ ways
Choose the other $2$ people from the men and women in ${}^5{C}_2=10$ ways
The universal set had ${}^9{C}_4=126$ number of ways
Hence, probability is $\frac{10\times 6}{126}=\frac{10}{21}$