Question

Four persons are selected at random out of 3 men, 2 women and 4 children. The probability that there are exactly 2 children in the selection is
(a) 11/21
(b) 9/21
(c) 10/21
(d) none of these

Answer 1

(c) 10/21

There are nine persons (three men, two women and four children) out of which four persons can be selected in ⁹C₄ = 126 ways.
∴ Total number of elementary events = 126
Exactly two children means selecting two children and two other people from three men and two women.
This can be done in ⁴C₂ × ⁵C₂ ways.
∴ Favourable number of elementary events = ⁴C₂ × ⁵C₂ = 60
So, required probability = $\frac{60}{126}=\frac{10}{21}$