Geometric and binomial distribution problem Let $X \sim Bi(n,p)$, and $Y \sim \mathcal{G}(p)$. 
(a) Show that $P(X=0)=P(Y>n)$.
(b) Find the number of kids a marriage should have so as the probability of having at least one boy is $\geq \frac{8}{9}$.
For (a), $P(X=0)$ is the probability of having $0$ successes in $n$ trials with $p$ the probability of success. So $P(X=0)=\binom{n}{0}(1-p)^n=(1-p)^n$. I am a bit confused with the calculation of $P(Y>n)$. Maybe it is easier to calculate $$P(Y>n)=1-P(Y\leq n).$$
I've calculated that expression but I am not at all positive it is correct. $P(Y\leq n)$ would be the sum of the probabilities of the first success happening at the i-th trial for $1\leq i \leq n$. So I thought $$P(Y \leq n)=\sum_{i=1}^n (1-p)^{i-1}p.$$
Bu, in general, $(1-p)^n \neq 1-\sum_{i=1}^n (1-p)^{i-1}p$, so I am clearly doing something wrong.
For (b) I have no idea what to do, I would really appreciate suggestions and corrections on my work in (a).
 A: (a) Given the expressions of the pmfs for the binomial and geometric distributions:
$$
\mathbb{P}\{X=0\} = \binom{n}{0} p^0(1-p)^n = (1-p)^n
$$
while indeed
\begin{align}
\mathbb{P}\{Y > n\} &= 1 - \mathbb{P}\{Y \leq n\} = 1 - \sum_{k=1}^n (1-p)^{k-1}p \\
&= 1 - p\sum_{k=0}^{n-1} (1-p)^{k} = 1-p\frac{1-(1-p)^n}{1-(1-p)}\\
&= 1-p\frac{1-(1-p)^n}{p} = (1-p)^n = \mathbb{P}\{X=0\}.
\end{align}
(using the closed form expression of a geometric series).
For (b), let $Z$ be the number of boys amongst $n$ children, assuming that each chldren is a girl or a boy independently, with probability $1/2$. Then you are looking for the smallest $n$ for which $\mathbb{P}\{Z=0\} \leq 1-8/9=1/9$. What is the distribution of $Z$? (Hint: you should use (a) to find $\mathbb{P}\{Z=0\}$ as a function of $n$ only). 
A: The easiest way for a) is to note that the probability that $Y\gt n$ is the probability of $n$ failures in a row, which is $(1-p)^n$.
However, the path you started on works also. For $\sum_1^{n-1}p(1-p)^{k}$ is the sum of a finite geometric series with first term $p$ and common ratio $1-p$. By the usual formula, this sum is 
$$p\frac{1-(1-p)^n}{1-(1-p)},$$
which simplifies to $1-(1-p)^n$.
It would have been a little faster, if you were going to sum a series, to find 
$$\sum_n^{\infty} p(1-p)^k.$$
This is an infinite geometric series with first term $p(1-p)^n$ and common ratio $1-p$. The sum is $(1-p)^n$.
A little faster, but no series is faster still.   
For the second problem, the probability of no boys in $n$ trials is $\frac{1}{2^n}$. We want this to be $\le \frac{1}{9}$. Three kids won't quite do it, they will have to go for $4$. Then the probability of no boys will be $\frac{1}{16}$.
The thing I can't figure out is why they want at least one boy. 
