Probability that a family with $n$ children has exactly $k$ boys Let the probability $p_n$ that a family has exactly $n$ children be $\alpha p^n$ when $n\geq1$, and $$p_0=1-\alpha p(1+p+p^2+\cdots).$$ Suppose that all the sex distributions have the same probability. Show that for $k\geq1$ the probability that a family has exactly $k$ boys is $2\alpha p^k/(2-p)^{k+1}$.
 A: Extended hint:  We sketch an argument that uses only basic notions.
Note that the probability $b_k$ of $k$ boys is, by a conditional probability argument, given by
$$b_k=\sum_{n=1}^\infty \alpha p^n \binom{n}{k} \left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{n-k}.$$
This simplifies to 
$$b_k=\sum_{n=1}^\infty \alpha \binom{n}{k}\left(\frac{p}{2}\right)^n.\qquad\qquad(\ast)$$
(We define $\binom{n}{k}$ to be $0$ if $n<k$.)
There are many tools for evaluating $(\ast)$. We do it using not much machinery. 
Recall the combinatorial identity
$$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}.$$
Substitute for $\binom{n}{k}$ in $(\ast)$. 
We obtain
$$b_k=\sum_{n=1}^\infty \alpha \binom{n-1}{k}\left(\frac{p}{2}\right)^n +\sum_{n=1}^\infty \alpha \binom{n-1}{k-1}\left(\frac{p}{2}\right)^n.\qquad\qquad(\ast\ast)$$
The first term in $(\ast\ast)$ is just $\dfrac{p}{2}b_k$. The second term is $\dfrac{p}{2}b_{k-1}$.
So we have derived the recurrence
$$b_k=\frac{p}{2}b_k+\frac{p}{2}b_{k-1}$$
or equivalently 
$$b_k=\frac{p}{2-p}b_{k-1}.$$
This almost settles things: each time we increment $k$ by $1$, the probability gets multiplied by $\dfrac{p}{2-p}$.  To get the process started, we need $b_1$.  We have
 $$b_1=\alpha\sum_{n=1}^\infty n \left(\frac{p}{2}\right)^n.$$
There is a  trick for finding $\sum_{n=1}^\infty n x^{n}$. Using the fact that for $|x|<1$,
$$1+x+x^2+x^3+ x^4+ \cdots=\frac{1}{1-x},$$
we find, by differentiating, that
$$\frac{1}{(1-x)^2}=1+2x+3x^2+ 4x^3+\cdots.$$
It follows that 
$$x+2x^2+3x^3+4x^4+\cdots =\frac{x}{(1-x)^2}.$$
Comment: Alternately, one could evaluate the sum $(\ast)$ by a repeated differentiation argument that generalizes the method we used for $b_1$. 
The recurrence $b_k=\dfrac{p}{2}b_k+\dfrac{p}{2}b_{k-1}$ can also be obtained directly, bypassing the series manipulation. 
A: Required probability=$\alpha \Sigma [(p/2)^n *nCk]$
[Summation on  “n” running from k to infinity]
The above expression=$\alpha \Sigma [(k+r) Ck *(p/2)^{k+r} ]$
[Summation on r running from zero to infinity]
Expression
     =$\alpha (p/2)^k \Sigma (p/2)^r (k+r)Ck$  -------(1)
Now,
$\Sigma (p/2)^r (k+r)Ck$
=The coefficient  of $x^k$ in the sum given below:
$(p/2+x)^0+(p/2+x)^1+(p/2+1)^2 ……..$ ------- ---- (2)
[We have an infinite number of terms in the above series].
We choose x such that:
$p<p/2+x<1$    --------[inequality A]   
$=>0<p/(2-p)<2x/(2-p)<1$   --------[Inequality B]
The series given by (2)  evaluates to
$1/(1-x-p/2) = 2/(2-2x-p)$
$=2(2-2x-p)^{-1}$
$=[2/(2-p)]  [1-2x/(2-p)]^{-1}$
Inequality A ensures the convergence of the series expressed by (2).
Coefficient  of $x^k$ in the above expression is:
$2/(2-p) 2^k/(2-p)^k$  ----(3)
Inequality B ensures the workability of the binomial expansion to obtain (3)
Using expression (3) in (1) we obtain the probability=
$\alpha (p/2)^k*2/(2-p) 2^k/(2-p)^k$
$=2\alpha p^k/(2-p)^{k+1}$
