The problem comes from Feller's introduction to probability. The problem states:
"Let the probability $p_n$ that a family has exactly $n$ children be $ap^n$ when $n \geq 1$, and $p_0 = 1 - ap(1+p+p^2+...)$. Suppose that all sex distributions of $n$ children have the same probability. Show that for $k \geq1$ the probability that a family has exactly $k$ boys is $\frac{2ap^k}{(2-p)^{k+1}}$"
My problem right now is that i think that the answer may have a typo (lots of answers on the book have them) because if you take the events:
$A = $ The family has $n$ children
$B = $ The family has exactly $k$ boys
Then the required probability is given by $Pr(A\bigcap B) = Pr(B|A)Pr(A)$
By the problem statement: $Pr(A) = ap^n$ and then the conditional probability $Pr(A\bigcap B)$ should be given by $\frac{n\choose k}{2^n}$ (because all sex distributions of $n$ children have the same probability and then probability of exactly $k$ boys will depend on how many ways you can choose those elements given that the family has exactly $n$ children).
But then we get:
$Pr(A\bigcap B) = \frac{{n\choose k}ap^n}{2^n}$
Which is clearly different than the expected answer, so where is my mistake so far?