$\begin {array} {1 1} (A)\;\large\frac{1}{2} & \quad (B)\;\large\frac{1}{4} \\ (C)\;\large\frac{1}{3} & \quad (D)\;None\: of \: these \end {array}$

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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

$ = \large\frac{3}{4}$

The probability that the second child is also a boy is = $ \large\frac{1}{4}$

We have to find $P(B/A) = \large\frac{P(A \cap B}{P(A)}$

$ = \Large\frac{\Large\frac{1}{4}}{\Large\frac{3}{4}}$

$ \large\frac{1}{\not 4} \times \large\frac{ \not 4}{3}$

$ = \large\frac{1}{3}$

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