= Solution
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 $0,a,b,2,3,\ldots$, where $a$ and $b$ distinguish the two single-customer possibilities, and $n\geq2$ records the total population. The <memoryless property> of the <exponential distributions> gives these nonzero transition rates:
$$
q_{0a}=q_{0b}=\nu/2,\quad q_{a0}=\mu_1,\quad q_{b0}=\mu_2,\quad q_{a2}=q_{b2}=\nu,\quad q_{2a}=\mu_2,\quad q_{2b}=\mu_1,
$$
and $q_{n,n+1}=\nu$ for $n\geq2$, $q_{n,n-1}=\mu$ for $n\geq3$. In particular, a departure from state $2$ leaves the other server occupied.
The <detailed balance equations> suggest
$$
\pi_a=\frac{\nu}{2\mu_1}\pi_0,\qquad
\pi_b=\frac{\nu}{2\mu_2}\pi_0,\qquad
\pi_n=\frac{\nu^2}{2\mu_1\mu_2}\left(\frac{\nu}{\mu}\right)^{n-2}\pi_0\quad(n\geq2).
$$
For 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 gives
$$
\pi_0^{-1}=1+\frac{\nu\mu}{2\mu_1\mu_2}+\frac{\nu^2\mu}{2\mu_1\mu_2(\mu-\nu)}
=1+\frac{\nu\mu^2}{2\mu_1\mu_2(\mu-\nu)}.
$$
Thus the <stationary distribution> exists, and the requested probability is
$$
\boxed{\Pr\{\text{both busy}\}=\sum_{n\geq2}\pi_n
=\frac{\nu^2\mu}{2\mu_1\mu_2(\mu-\nu)+\nu\mu^2}.}
$$
With 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 $n\geq2$ 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 \b[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.
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