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i am faced with the following problem and here is my attempt at solving it. Let me know what you think about it!

So here is the problem:

Assume the probability that a family has exactly n children is given by $P_n = cp^n$ where $c \gt 0$ and $0\lt p \lt 1$ and $P_0 = 1 - \sum_{i=1}^\infty P_n$ . We also assume the sex distributions of children to be equiprobable.

a) Compute the probability that a family of n children has exactly k boys for $k = 0,1,...,n$

b) Compute the probability that a family has no boys

c) Compute the expected number of boys

I know the problem is similar to that of Probability that a family with $n$ children has exactly $k$ boys However the questions are different.

Here is what i tried:

QUESTION ONE

Let $N$ the random variable counting the number of children and $X$ the random variable counting the number of boys.

We compute $$ P[ X=k ¦ N = n] = \binom{n}{k}\left(\frac{1}{2}\right)^n$$

QUESTION TWO

$$P[X=0] = \sum_{n=0}^\infty P[X=0 ¦ N = n]P[N=n]$$ $$P[X=0] = P_0 + \sum_{n=1}^\infty cp^n\binom{n}{0}\left(\frac{1}{2}\right)^n$$ $$P[X=0] = 1 - \sum_{n=1}^{\infty} cp^n + \sum_{n=1}^{\infty} c\left(\frac p2\right)^n$$ $$P[X=0] = 1 - c\left(\frac{p}{1-p} - \frac{p/2}{1-p/2}\right) = 1 - c\left(\frac{p}{(1-p)(2-p)}\right)$$

QUESTION THREE

We have $$E[X ¦ N=n] = n\left(\frac{1}{2}\right)$$

Hence $$E[X] = E[E[X¦N]]= E[N/2] = \frac{1}{2}E[N]$$

However $$E[N] = \sum_{n=1}^\infty ncp^n$$

We use the power series $\frac{x}{(1-x)^2} = \sum_{i=1}^\infty nx^n$

Thus $$E[X] = \frac{1}{2}E[N] = \frac{1}{2}c\frac{p}{(1-p)^2}$$

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Problem solved, got the answers checked and they are correct.

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