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I've reviewed a similar proposed question, however the help given wasn't exactly what I was looking for (unfortunately). So if the probability that a family will have exactly n children are equally likely is $2^{-n}$ for n=1,2,..., and if all $2^{n}$ permutations are equally likely, I understand this probability to be defined as $\sum_{n=1}^{\infty}$ ${n \choose k}$$\alpha$($\frac{p}{2}$)$^{n}$. What I do not understand how the solution results in a solution of $\frac{4}{3^{k+1}}$ using conditional probability.

So let $A$ be the event that a family has $k$ boys and let $B_{n}$ be the event that it has exactly $n$ children. Then we know that $P(B_{n})=2^{-n}$. However, I am not sure exactly what $P(A|B_{n})$ is because I need it to calculate $P(A)=\sum_{n=1}^{\infty}P(A|B_{n})P(B_{n})$

I would really appreciate some clarity on this problem. Probability is not my strong suit at all. Thanks!

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  • $\begingroup$ might be just me but I don't understand one heck of what you are trying to calculate. $\endgroup$
    – user126540
    Oct 1, 2014 at 1:53
  • $\begingroup$ Sorry for not being clear. I will try to edit my post. $\endgroup$
    – user42864
    Oct 1, 2014 at 1:55
  • $\begingroup$ If you are asking about $P(A_k | B_n)$, it is binomially distributed $\endgroup$
    – user126540
    Oct 1, 2014 at 2:19
  • $\begingroup$ Is there a way that I can present that using conditional probability? That is where I am confused. $\endgroup$
    – user42864
    Oct 1, 2014 at 2:22

1 Answer 1

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If there are $n$ children, there are $2^n$ possible outcomes; of these, exactly $\binom{n}{k}$ have $k$ boys. So $P(A|B_n) = \frac{\binom{n}{k}}{2^n}$. If you wish to assume that $P(B_n) = 2^{-n}$, then the sum is $$P(A) = \sum_{n=1}^\infty P(A|B_n)P(B_n) = \sum_{n=1}^\infty \frac{\binom{n}{k}}{2^{2n}}.$$ But I'm not clear as to why that is your assumption. (In particular, for example, this means that the probability of having no children must be zero.)

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  • $\begingroup$ So how does this equate to $\frac{4}{3^{k+1}}$? Geometric? $\endgroup$
    – user42864
    Oct 1, 2014 at 2:27
  • $\begingroup$ Note: $\mathsf P(A) =\sum\limits_{n=\color{blue}{k}}^\infty {n\choose k} 4^{-n}$ because $\forall n: (n < k) \to \mathsf P(A\mid B_n) = 0$ $\endgroup$
    – Graham Kemp
    Oct 1, 2014 at 3:40
  • $\begingroup$ I can see that if take the sum of $4^{-n}$ to infinity it equals $\frac{4}{3}$, but I am not seeing how the solution is supposed to be $\frac{4}{3^{k+1}}$. $\endgroup$
    – user42864
    Oct 1, 2014 at 12:13

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