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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 $2\alpha p^k/(2-p)^{k+1}$.

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    I have made the correction as suggested by B. M scott2011-11-26

2 Answers 2

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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.) 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.

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    @Anamitra Palit: Thank you for noticing the typo, and telling me about it. Fixed.2011-12-03
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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

--------[inequality A]

$=>0

--------[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=

$\alpha (p/2)^k*2/(2-p) 2^k/(2-p)^k$

$=2\alpha p^k/(2-p)^{k+1}$

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    @Mathematics:We are considering the probability of exactly 'k' boys out of 'n' children[n>=k].Probability of 'n' children${\eq\alpha\p^n}$. The stated probability for n children tends to zero for n tending to infinity, p being a fraction.This is apparent from the series defining p_0: the series can converge only if p is a fraction.Now 'n' could be any integral value greater than or equal to k. So we have considered the summation from n=k to n=infinity2017-11-24