A family has two children. Given that one of the children is a boy, what is the probability that both children are boys?

Question

I was doing this question using conditional probability formula.... i.e P(2/1)=? Probability of 2nd child to be boy given that 1st child is a boy. By formula, P(2/1)=P(2,1)/P(1) which is equal to P(2).P(1)/P(1)...since second child to be boy doesn't depend on first child and vice versa. So independent..Please provide the detailed solution and correct me if I am wrong.

P(2,1)= probability of 2nd and 1st child both to be boys.

Answer 1

Two children :-

Possibilities :-

B B

G G

B G

G B

P(One Boy Child) = $\frac{3}{4}$

P(Both Boy / One Boy Child) = (P(Both Boy) and P(One Boy child) )/ P(One Boy Child) =

$\frac{(\frac{1}{4}) } {\frac{3}{4}} = \frac{1}{4}*\frac{4}{3} = \frac{1}{3}$

 

This can also be seen more intuitively like :-

It has been said that one of the children is a boy.

So our sample space reduces to only 3 possibilities:-

B B

B G

G B

Out of the three , only one has both as boy.

Thus required P = $\frac{1}{3}$

Comments

Thank you for your response. But can you help me why you have applied bias formula rather than conditional probability rule.

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Thanks for pointing out , I wrote it incorrectly , updated :)