A coin when flipped shows heads with probability $p$ and tails with probability $q = 1 − p$. It is flipped repeatedly. Assume that the outcome of different flips is independent. Let $X$ be the length of the initial run and $Y$ the length of the second run.
Evaluate $P(X=2)$ and $P(Y=3)$ and $P(X=2\:$ and $\:Y=3)$
Since for the first run of length 2 is either $HHT$ or $TTH$, I have $P(X=2)=p^2q+q^2p\:$, but I'm not sure how to evaluate the second and third part of the question. I'm thinking that the second run is either $...THHHT$ or $...HTTTH$ so $P(Y=3)=q^2p^3+p^2q^3$.
But the notes my lecturer asked us to refer to show that $P(Y=3)=q^2p^{3-1}+p^2q^{3-1}=q^2p^2+p^2q^2$, which I dont understand. The notes also show the probability is derived from an infinite sum, which we are specified NOT to do, but to do it by conditioning on the first toss instead.
let the probability of tossing a tail be $q=1-p$
$P(X=2)$ is the probability of tossing $HHT$ or $TTH$, $$\therefore P(X=2)=p^2q+q^2p$$ $P(Y=3)$ is the probability of tossing $H...HTTTH$ or $T...THHHT$. Thus we need to account for all possible values of the first run
$$ \begin{equation} \begin{split} \therefore P(Y=3)&=(p+p^2+p^3+...)(q^3p)+(q+q^2+q^3+...)(p^3q) \\ &=p(\sum_{n=0}^{+\infty} p^n)q^3p+q(\sum_{n=0}^{+\infty} q^n)p^3q \\ &=p^2q^3\frac{1}{1-p}+q^2p^3\frac{1}{1-q}\\ &=p^2q^2+p^2q^2 \\ &=2p^2q^2 \end{split} \end{equation} $$ $P(X=2\; \text{and} \; Y=3)$is the probability of tossing $HHTTTH$ or $TTHHHT$, $$ \begin{equation} \begin{split} \therefore P(X=2\; \text{and} \; Y=3)&=p^2q^3p+q^2p^3q \\ &=2p^3q^3 \end{split} \end{equation} $$