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# A person has two children. If one of them is boy, then the probability that other is also a boy is

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

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 $P(B/A) = \large\frac{P(A \cap B}{P(A)}$
$= \Large\frac{\Large\frac{1}{4}}{\Large\frac{3}{4}}$
$\large\frac{1}{\not 4} \times \large\frac{ \not 4}{3}$
$= \large\frac{1}{3}$