Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
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)$\frac{1}{9}$
(b)$\frac{1}{5}$
(c)$\frac{1}{12}$
(d)$\frac{10}{21}$
(e)None of these
Answer
Answer (as printed): D
Explanation
$n(S)=$ number of ways of choosing 4 persons out of 9 $={ }^{9} C_{4}=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}=126$. $n(E)=$ Number of ways of choosing 2 children out of 4 and 2 persons out of $(3+2)$ persons $=\left({ }^{4} C_{2} \times{ }^{5} C_{2}\right)=\left(\frac{4 \times 3}{2 \times 1} \times \frac{5 \times 4}{2 \times 1}\right)=60$. $\therefore P(E)=\frac{n(E)}{n(S)}=\frac{60}{126}=\frac{10}{21}$.
Explanation as extracted from the printed page; notation may be imperfect.