Problem source

A coin has probability $p$ ($0 < p < 1$) of showing a head when tossed. Give a careful argument to show that the $k$th head in a series of consecutive tosses is achieved after \textit{exactly} $n$ tosses with probability 
\[
\binom{n-1}{k-1}p^{k}(1-p)^{n-k}\qquad(n\geqslant k).
\]
Given that it took an even number of tosses to achieve exactly $k-1$ heads, find the probability that exactly $k$ heads are achieved after an even number of tosses. 
If this coin is tossed until exactly 3 heads are obtained, what is the probability that \textit{exactly }2 of the heads occur on even-numbered
tosses?

Solution source

We must have a sequence consisting of $\underbrace{HTT\cdots TH}_{k-1\text{ heads and }n-k\text{ tails}}\underbrace{H}_{k\text{th head}}$. There are $\binom{n-1}{k-1}$ ways to chose how to place the $k-1$ heads in the first $n-1$ flips, and each sequence has probability $p^{k-1}(1-p)^{n-k}p$ which gives a probability of $\displaystyle \binom{n-1}{k-1} p^k (1-p)^{n-k}$.

Given that it took an even number of tosses to achieve $k-1$ heads, this is equivalent to the problem of what is the probability that the first head occurs on an even flip, ie

\begin{align*}
\mathbb{P}(\text{even flip}) &= \mathbb{P}(2\text{nd flip}) +\mathbb{P}(4\text{th flip}) +\mathbb{P}(6\text{th flip}) + \cdots \\
&= (1-p)p + (1-p)^3p + (1-p)^5p + \cdots \\
&= (1-p)p \left ( \sum_{r=0}^\infty (1-p)^{2r}\right) \\
&= \frac{p(1-p)}{1-(1-p)^2} \\
&= \frac{p(1-p)}{2p-p^2} \\
&= \frac{1-p}{2-p}
\end{align*}

The ways to achieve $2$ heads on even tosses are $EEO$, $EOE$, $OEE$. The probability of going from $O$ to $E$ is the same as the initial probability of an $O$ flip, etc.

Therefore

\begin{align*}
\mathbb{P}(EEO) &=\left( \frac{1-p}{2-p} \right)^2 \left ( 1- \frac{1-p}{2-p} \right) \\
&=  \left( \frac{1-p}{2-p} \right)^2 \left ( \frac{1}{2-p} \right) \\
\mathbb{P}(EOE) &= \left( \frac{1-p}{2-p} \right) \left ( \frac{1}{2-p} \right)^2 \\
\mathbb{P}(OEE) &=  \left ( \frac{1}{2-p} \right)^2 \left( \frac{1-p}{2-p} \right)\\
\mathbb{P}(2 \text{ heads on even tosses}) &= \frac{(1-p)^2 + 2(1-p)}{(2-p)^3} \\
&= \frac{(1-p)(2-p)}{(2-p)^3} \\
&= \frac{1-p}{(2-p)^2}
\end{align*}