Answer

Define \(X\) as the following random variable

\[ \begin{alignat}{1} X = \begin{cases} 1, p = \frac{1}{4} &\mbox{$HHH$ or $TTT$}\newline 0, p = \frac{3}{4} &\mbox{otherwise} \end{cases}\newline \end{alignat} \]

1. \(K\) is simply a binomial distribution, where we want the \(2^{nd}\) success to happen at the \(K+1\)th trial.

\[ \begin{aligned} p_{K}(k) = \binom{k}{1}\frac{1}{4}^{2}\frac{3}{4}^{k-1} \quad\text{since the last trial is success} \end{aligned} \]

2. \(M\) = number of tails before first success. Let the success be at \(N+1\). Defin \(Y\) as

\[ \begin{alignat}{2} Y &= \begin{cases} 1 \quad p=\frac{1}{2} &\mbox{$HHT$, $HTH$, or $THH$}\newline 2 \quad p=\frac{1}{2} &\mbox{$HTT$, $THT$, or $TTH$} \end{cases}\newline E[Y] &= 1 \times \frac{1}{2} + 2 \times \frac{1}{2}\newline Var(Y) &= (1 - \frac{3}{2})^{2} \times \frac{1}{2} + (2 - \frac{3}{2})^{2} \times \frac{1}{2}\newline E[N+1] &= \frac{1}{p} = 4\newline Var(N+1) &= Var(N) = \frac{1-p}{p^{2}} = \frac{1 - \frac{1}{4}}{\frac{1}{4}^{2}}\newline M &= Y_{1} + Y_{2} + \cdots Y_{N}\newline E[M] &= E[Y_{1} + Y_{2} + \cdots Y_{N}]\newline Var(M) &= Var(Y_{1} + Y_{2} + \cdots Y_{N})\newline \end{alignat} \]

Note that both \(Y\) and \(N\) are random variables here. Using the formulae for random number of random variables,

\[ \begin{aligned} E[M] &= E[E[M|N]] = E[NE[Y]] = E[N]E[Y] = (4-1) \times \frac{3}{2} = \frac{9}{2}\newline Var(M) &= Var(E[M|N]) + E[Var(M|N)] = Var(NE[Y]) + E[NVar(Y)]\newline &= E[Y]^{2}Var(N) + E[N]Var(Y) = \frac{9}{4} \times 12 + 3 \times \frac{1}{4} = \frac{111}{4} \end{aligned} \]