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The question below is a part of a more complex question in probability theory. I can't get my head around why is my solution incorrect...

So here it goes. Distribution of number of boys in a family with $n$ children is binomial with parameters $n$, $\frac{1}{2}$ (my guess is that it means $\mathbb{P}(X=k)=\binom{n}{k}(\frac{1}{2})^k(\frac{1}{2})^{n-k}=\binom{n}{k}(\frac{1}{2})^n$). What is $\mathbb{P}(X=2)$?

I thought it would be $$\sum_{n=2}^{\infty}\binom{n}{2}\left(\frac{1}{2}\right)^n=\frac{1}{2}\sum_{n=2}^{\infty}n(n-1)\left(\frac{1}{2}\right)^n=\frac{1}{8}\left(\sum_{n=2}^{\infty}\left(\frac{1}{2}\right)^n\right)''=\frac{1}{8}\left(\frac{2}{(1-\frac{1}{2})^3}\right)$$

But it definitely is not true, since the above-mentioned sum is $2$...

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  • $\begingroup$ $\mathbb{P}(X=2|N=n)=\binom{n}{k}(\frac{1}{2})^n$ as you say. You also need a distribution for $N$ to find $\displaystyle \mathbb{P}(X=2)=\sum_{n=2}^{\infty} \mathbb{P}(X=2|N=n)\mathbb{P}(N=n)$. $\endgroup$ – Henry Sep 1 '14 at 19:00
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There is no reason to sum. For families with $n$ children, the probability, as you observed, is $\binom{n}{2}\cdot\frac{1}{2^n}$.

In order to find the probability that a randomly chosen family has exactly $2$ boys, you would have to know the distribution of family sizes.

If the probability that a family has $n$ children is $p_n$, then the probability a randomly chosen family has $2$ boys is $$\sum_0^\infty p_n \binom{n}{2}\frac{1}{2^n}.\tag{1}$$

Without knowing the $p_n$, we cannot go beyond (1).

Remark: We have used some assumptions that are in fact false. The probability of a boy is not exactly $\frac{1}{2}$. And the implicit assumption of independence is quite dubious.

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  • $\begingroup$ Dang it, you're totally right, I think I need more sleep. Thank you. $\endgroup$ – Jules Sep 1 '14 at 19:02
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    $\begingroup$ You are welcome. Yes, sleep (or coffee) are quite useful. $\endgroup$ – André Nicolas Sep 1 '14 at 19:04

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