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Put $\mu ={\mu }_1+{\mu }_2$. In this heterogeneous two-server queue, the total population alone is not a Markov process: when only one customer is present, its departure rate depends on which server is occupied. Use states $0,a,b,2,3,\dots$, where $a$ and $b$ distinguish the two single-customer possibilities, and $n\ge 2$ records the total population. The memoryless property of the exponential distributions gives these nonzero transition rates:and $q_{n,n+1}=\nu$ for $n\ge 2$, $q_{n,n-1}=\mu$ for $n\ge 3$. In particular, a departure from state $2$ leaves the other server occupied.

The detailed balance equations suggestFor example, both ${\pi }_a\nu ={\pi }_2{\mu }_2$ and ${\pi }_b\nu ={\pi }_2{\mu }_1$ hold; these are the two transitions that would be lost by using only population as the state. All other adjacent-state detailed balance equations hold as well. Since $\nu /\mu <1$, the geometric series converges, and normalization givesThus the stationary distribution exists, and the requested probability isWith equal server rates ${\mu }_1={\mu }_2=m$, this reduces to ${\nu }^2/[m(2m+\nu )]$, as for an M-M-s queue with two servers.

To prove the output claim, put the continuous-time Markov chain in its stationary distribution. The preceding detailed balance equations make it a reversible Markov chain: time reversal of a continuous-time Markov chain has the same transition rates. A forward departure becomes a reversed arrival. In every reversed state, the total rate of such arrivals is $\nu$: from $0$ the rates $\nu /2$ add, from $a$ or $b$ the upward rate is $\nu$, and from every $n\ge 2$ it is also $\nu$. The reversed process can therefore be constructed with an independent rate-$\nu$ Poisson process of arrival epochs, choosing the idle server by a fair coin when appropriate. Consequently reversed arrival counts on disjoint time intervals are independent Poisson random variables with the corresponding means. Reversing those intervals proves that the stationary departure stream is a rate-$\nu$ Poisson process, rather than merely a stream with mean rate $\nu$.

Stationarity matters here. If the queue starts empty, a departure by time $h$ requires both an arrival and a completed service time, so its probability is $O(h^2)$; a rate-$\nu$ Poisson process would have probability $\nu h+O(h^2)$. Thus the output assertion is the equilibrium assertion in the context of the preceding stationary calculation.