Problem source

An unbiased twelve-sided die has its faces marked $A,A,A,B,B,B,B,B,B,B,B,B.$
In a series of throws of the die the first $M$ throws show $A,$ the next $N$ throws show $B$ and the $(M+N+1)$th throw shows $A$.
Write down the probability that $M=m$ and $N=n$, where $m\geqslant0$ and $n\geqslant1.$ Find
\begin{questionparts}
\item the marginal distributions of $M$ and $N$, 
\item the mean values of $M$ and $N$. 
\end{questionparts}
Investigate whether $M$ and $N$ are independent. 
Find the probability that $N$ is greater than a given integer $k$, where $k\geqslant1,$ and find $\mathrm{P}(N > M).$ Find also $\mathrm{P}(N=M)$
and show that $\mathrm{P}(N < M)=\frac{1}{52}.$

Solution source

\begin{align*}
\mathbb{P}(M = m, N = n) &= \left ( \frac{3}{12} \right)^m \left ( \frac{9}{12} \right)^n \frac{3}{12} \\
&= \frac{3^n}{4^{m+n+1}} 
\end{align*}

\begin{questionparts}
\item
\begin{align*}
\mathbb{P}(M = m) &= \sum_{n = 1}^{\infty} \mathbb{P}(M=m,N=n) \\
&= \sum_{n = 1}^{\infty} \frac{3^n}{4^{m+n+1}} \\
&= \frac{1}{4^{m+1}} \sum_{n = 1}^{\infty} \left ( \frac34\right)^n \\
&= \frac{1}{4^{m+1}} \frac{3/4}{1/4} \\
&= \frac{3}{4^{m+1}} \\
\\
\mathbb{P}(N = n) &= \sum_{m = 0}^{\infty} \mathbb{P}(M=m,N=n) \\
&= \sum_{m = 0}^{\infty} \frac{3^n}{4^{m+n+1}} \\
&= \frac{3^n}{4^{n+1}} \sum_{m = 0}^{\infty} \left ( \frac14\right)^n \\
&= \frac{3^n}{4^{n+1}} \frac{1}{3/4} \\
&= \frac{3^{n-1}}{4^{n}} \\
\end{align*}

\item $M+1 \sim Geo(\frac34) \Rightarrow \mathbb{E}(M) = \frac43 -1 = \frac13$

$N \sim Geo(\frac14) \Rightarrow \mathbb{E}(N) = 4$
\end{questionparts}

$M,N$ are independent since $\mathbb{P}(M = m, N =n ) = \mathbb{P}(M=m)\mathbb{P}(N=n)$

\begin{align*}
\mathbb{P}(N > k) &= \sum_{n=k+1}^{\infty} \mathbb{P}(N = n) \\
&= \sum_{n=k+1}^{\infty} \frac{3^{n-1}}{4^{n}} \\
&= \frac{3^k}{4^{k+1}} \sum_{n = 0}^{\infty} \left ( \frac34\right)^n \\
&= \frac{3^k}{4^{k+1}} \frac{1}{1/4} \\
&= \frac{3^k}{4^k}
\end{align*}

\begin{align*}
\mathbb{P}(N > M) &= \sum_{m=0}^{\infty} \mathbb{P}(N > m) \mathbb{P}(M = m) \\
&= \sum_{m=0}^{\infty} \left (\frac34 \right)^m \frac{3}{4^{m+1}}\\ 
&=\sum_{m=0}^{\infty} \frac{3^{m+1}}{4^{2m+1}}\\
&= \frac{3}{4} \frac{1}{13/16} \\
&= \frac{12}{13} \\
\\
\mathbb{P}(N=M) &= \sum_{m=1}^{\infty} \mathbb{P}(N=m, M=m) \\
&=  \sum_{m=1}^{\infty} \frac{3^m}{4^{2m+1}} \\
&= \frac{3}{64} \sum_{m=0}^{\infty} \left ( \frac{3}{16} \right)^m \\
&= \frac{3}{64} \frac{1}{13/16} \\
&= \frac{3}{52}\\
\\
\mathbb{P}(N < M) &= 1 - \frac34 - \frac3{52} \\
&= 1 - \frac{48}{52} - \frac{3}{52} \\
&= 1 - \frac{51}{52} \\
&= \frac{1}{52}
\end{align*}