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 $2\alpha p^k/(2-p)^{k+1}$.
Probability that a family with $n$ children has exactly $k$ boys
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5Please don't write your questions as if you are giving us a homework assignment! Show us what you already tried and where you got stuck. – 2011-11-26
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0Assuming the probability of a boy being born=probability of a girl being born=1/2, I am getting Required probability$=\Sigma\alpha p^n nCk*(1/2^n)$.,summation over n extending from 1 to infinity.Is it the correct interpretation? – 2011-11-26
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0Shouldn’t $p_0$ be $1-\alpha(p+p^2+\dots)$? – 2011-11-26
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0Summation over n should be from k to infinity in my last comment. – 2011-11-26
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0I have made the correction as suggested by B. M scott – 2011-11-26
2 Answers
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
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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0There is a minor typo:It should be $b_1=\alpha\Sigma_1^{\infty}np^n$. The coefficient $\alpha$ has been left out. This does not reduce the elegance of the proof – 2011-12-03
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0There is a minor typo:It should be $b_1=\alpha\Sigma_1^{\infty}n(p/2)^n$. The coefficient $\alpha$ has been left out. This does not reduce the elegance of the proof – 2011-12-03
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0@Anamitra Palit: Thank you for noticing the typo, and telling me about it. Fixed. – 2011-12-03
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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1Your writing style and presentation leave much to be desired. – 2011-11-26
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0I have put out my calculations over here just to assert that I have been working on the problem – 2011-12-03
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0@Anamitra Palit: Nice work. The convergence issues you spent some time on turn out not to matter, since yours is essentially a generating function argument. – 2011-12-03
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0@AnamitraPalit Why is sumation running from $k$ to $\infty$ ? – 2017-11-22
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0@Mathematics :we are considering the probability of exactly 'k' boys out of 'n' children[n>=k].Probability of 'n' children=alpha*p^n. The stated probability for n childeren 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=infinity – 2017-11-24
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0@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=infinity – 2017-11-24