= Solution
Let $(m,n)$ denote the clock position and the queue population, including any customer in service. By the <memoryless property> of the <exponential distribution>, the nonzero transition rates are
$$
(m,n)\longrightarrow
\begin{cases}
(m+1,n)&\text{at rate }\nu,\\
(m-1,n+1)&\text{at rate }\lambda_m,\\
(m,n-1)&\text{at rate }\mu\quad(n>0).
\end{cases}
$$
Set $\rho=\nu/\mu<1$. We claim that
$$
\boxed{\pi(m,n)=p_m(1-\rho)\rho^n,\qquad n\geq0,}
$$
where $p_m$ is the clock's <stationary distribution> from part (a). To check <global balance for a continuous-time Markov chain>, divide the incoming probability flux by $\pi(m,n)$. The clockwise contribution is $\nu p_{m-1}/p_m=\lambda_m$. The arrival contribution, present only for $n>0$, is $\lambda_{m+1}(p_{m+1}/p_m)\rho^{-1}=\mu$. The service contribution is $\mu\rho=\nu$. Their sum is the outgoing rate $\nu+\lambda_m+\mu\mathbf1_{\{n>0\}}$.
The proposed <probability distribution> is normalized, so it is the equilibrium law. In particular, the clock position and queue population are <independent random variables> at any fixed equilibrium time, and the population has the <geometric distribution> of a stable <M M 1 queue>. Independence of these two random variables does not assert independence of the two entire processes.
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