The probability of a family chosen at random having exactly k children is $\alpha p^k$, where $0\lt p \lt 1$. The probability that any child has blue eyes is b, where $0 \lt b \lt 1$ (independently of other children). What is the probability that a family chosen at random has exactly $r \ge 0$ children with blue eyes? (AN INTRODUCTION TO PROBABILITY AND STATISTICS,ROHATGI and EHSANES SALEH).
My approach:
This is equivalent to the probability that the family has $k \ge r$ children (event A) and that r of those k children are blue-yed (event B).
$$P(A \cap B) = P(A)P(B|A)$$
Let $A_n$ denote the pobability that the family has n children. Then $P(A)=P(A_k)=1-\sum_{i=0}^{r-1}P(A_i) = 1- \sum_{i=0}^{r-1}\alpha p^i$
Now $P(B) = {r\choose k}b^r(1-b)^{k-r}$ as it is a binomial distribution.
Putting it together we get:$$P(A \cap B)= (1- \sum_{i=0}^{r-1}\alpha p^i)({r\choose k}b^r(1-b)^{k-r})$$.
The answer seems overly complicated, i would like to verify that the method is correct.
Any help would be greatly appreciated.
Best Answer
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Let $B$ count the number of Blue eyed children in the family, and $A$ the number of children.
We are provided $\mathsf P(A{=}k)=\alpha p^k\mathbf 1_{k\in \Bbb N}$ and $\mathsf P(B{=}r\mid A{=}k)=\tbinom{k}{r}b^r(1-b)^{k-r}\mathbf 1_{r\in[0..k]\cap\Bbb N}$
So, since $0\leq r\leq k$ is a given (since you cannot have more blue eyed children than you have children). $$\require{enclose}\begin{align}\mathsf P(B{=}r) &= \color{red}{\enclose{circle}{\color{black}{~\sum\limits_{k=r}^\infty~}}}\mathsf P(B{=}r\mid A{=}k)\mathsf P(A{=}k)\\[1ex] &=\mathbf 1_{r\in\Bbb N}\sum\limits_{k=r}^\infty \tbinom krb^r(1-b)^{k-r}\alpha p^k\\[1ex]&=\alpha b^r(1-b)^{-r}\mathbf 1_{r\in\Bbb N}\sum_{k=r}^\infty\tbinom kr ((1-b)p)^k\end{align}$$
Now use the negative binomial identity (when $\lvert x\rvert<1$):
$$\sum_{k=r}^\infty\tbinom kr x^k = x^r(1-x)^{-(1+r)}$$
Similarly you should be able to determine $\alpha$ from the identity $\sum\limits_{k=0}^\infty x^k=(1-x)^{-1}$ when $\lvert x\rvert<1$.