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The probability that a randomly chosen family having exactly k children is $p_k = \alpha p^k$ for $k=1,2,...$ and $p_0=1-\frac{\alpha p}{1-p}$. Suppose all gender distributions of k children are equally likely. Find the probability that a family has exactly r boys, $r\ge1$.

So far, I have $P(r)=\sum_{k=1}^\infty \alpha p^k$$k\choose r$$(\frac12)^r(\frac12)^{k-r}=$$\sum_{k=1}^\infty \alpha $$k\choose r$$(\frac{p}2)^k$

Not sure how to proceed from here, or if I need to proceed at all.

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Also, a follow up question is: Find the conditional probability that a family has at least 2 boys, given that it has at least 1 boy?

My thought for this is $P(r\ge2|r\ge1)=1-P(r\lt2|r\ge1)=1-P(r=1)$. Would I just plug $r=1$ into whatever I get for the 1st part?

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  • $\begingroup$ Are you familiar with negative binomial expansion? $\endgroup$ – StubbornAtom Sep 19 '17 at 10:37
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I don't know if there is a general formula for $\sum\alpha{k \choose r}(\frac{p}{2})^k$, but you can calculate particular cases for $r=0$ and $r=1$, it is: $$P_{r=0}=p_0+\sum\alpha{k \choose 0}(\frac{p}{2})^k=p_0+\alpha\frac p {2-p}$$ and $$P_{r=1}=\sum\alpha{k \choose 1}(\frac{p}{2})^k=2\alpha p(\frac 1 {2-p})^2$$ Probability you are looking for will be $P=\frac{1-P_{r=0}-P_{r=1}}{1-P_{r=0}}=\frac{\frac{\alpha p}{1-p}-\frac{\alpha p}{2-p}-\frac{2\alpha p}{(2-p)^2}}{\frac{\alpha p}{1-p}-\frac{\alpha p}{2-p}}=\frac{p}{2-p}$

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Let $B_k$ be the event denoting that a family has $k$ children and $A_r$ be the event that it has $r$ boys.

Then, $\qquad\qquad P(A_r)=\displaystyle\sum_{k=0}^\infty P(A_r|B_k)P(B_k)$

$$=\sum_{k=r}^\infty (\alpha p^k)\binom{k}{r}\frac{1}{2^k}=\alpha\left(\frac{p}{2}\right)^r\sum_{t=0}^\infty \binom{t+r}{t}\left(\frac{p}{2}\right)^t\quad, t=k-r$$

$$=\alpha\left(\frac{p}{2}\right)^r\left(1-\frac{p}{2}\right)^{-(r+1)}$$

For the 2nd part the required probability is

$$P\left(\bigcup_{r=2}^\infty A_r|\bigcup_{r=1}^\infty A_r\right)=\dfrac{\sum _{r=2}^\infty P(A_r)}{\sum _{r=1}^\infty P(A_r)}\quad(\text{as the events are disjoint})$$

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