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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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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

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


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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}$
answered Jan 16, 2014 by thanvigandhi_1

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