Q. 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) $\frac{21}{32}$
Explanation
(a)
Let $E =$Number of ways of choosing 2 children out of 4 and 2 persons out (3+2)persons
$\therefore\ n(E) = (\ ^{4}C_{2} \times \ ^{5}C_{2})$
$\therefore$Required probability$= \frac{\ ^{4}C_{2}{\ \times}^{5}C_{2}}{\ ^{9}C_{4}} = \frac{60}{126} = \frac{10}{21}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.