**Question **: A father has 3 children with at least one boy. The probability that he has 2 boys and 1 girl is

(A) 1/4

(B) 1/3

(C) 2/3

(D) none of these

**Answer** : option (B)

**Solution** :

Consider the following events:

A: Father has at least one boy

B: father has 2 boys and one girl

Then, A = one boy and 2 girls, 2 boys and one girl, 3 boys and no girl

AnB = 2 boys and one girl

Now, the required probability is

P(A/B) = P(AnB)/P(B)

=1/3