Question

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

• A. $\frac{10}{21}$
• B. $\frac{11}{13}$
• C. $\frac{13}{25}$
• D. $\frac{21}{32}$

Answer 1

The correct option is A $\frac{10}{21}$

4 person chosen -

2 of them are children ${\Rightarrow }^4{C}_2.$

also, choosing rest 2 from $\left[\right(3+2+4-4)=5]$ person will be ${=}^5{C}_2$

(only from men and woman)

$\therefore$ Prob $\frac{{}^4{C}_2{.}^5{C}_2}{{}^8{C}_4}$ (${}^9{C}_4\to$ selection of 4 from 9 person)

$\Rightarrow P=\frac{10}{21}$