Explanation

(c)
Total cases are $BB.BG,GB,\ GG = 4$
Favourable cases are $BB,\ BG,\ GB = 3$
Let $P(A) =$Probability of a boy in two children
$= \frac{3}{4}$
Let $P(B) =$The probability that the second child is also a boy
$= \frac{1}{4}$
Here, $P(A \cap B) = \frac{1}{4}$
$\therefore P\left( \frac{B}{A} \right) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$