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

I was doing this question using conditional probability formula.

Suppose, (1) is the event, that the first child is a boy, and (2) is the event that the second child is a boy.

Then the probability of the second child to be boy given that first child is a boys by formula, $P((2)|(1))=\frac{P((2) \cap (1))}{P((1))}=\frac{P((2))P((1))}{P((1))} = P((2))$ ...since second child to be boy doesn't depend on first child and vice versa. Please provide the detailed solution and correct me if I am wrong.

2 Answers 2

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Let's write the sample space:

BB, BG, GB, GG

But we know that one child is a boy, so that means that GG isn't a possibility. Thus, the sample space is reduced to: BB, BG, GB.

Therefore, the probability of both children being boys given that one is a boy is 1/3.

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The flaw in your solution is to write $$P(2,1)=P(2).P(1)$$ What would be the logic behind this? Think again, the event $(2,1)$ means "there are $2$ boys and there is at least $1$ boy". Therefore, $(2,1)$ is equal to the event $(2):$ "there are two boys".

Can you fix your solution yourself now?