= Solution
For $0\leq k\leq n$, the event $\{X=k,X+Y=n\}$ is $\{X=k,Y=n-k\}$. Use <independence> in its numerator and the <Poisson distribution> of the sum in its denominator:
$$
\begin{aligned}
\mathbb P(X=k\mid X+Y=n)
&=\frac{e^{-\lambda}\lambda^k/k!\;e^{-\mu}\mu^{n-k}/(n-k)!}{e^{-(\lambda+\mu)}(\lambda+\mu)^n/n!}\\
&=\boxed{\binom nk\left(\frac\lambda{\lambda+\mu}\right)^k\left(\frac\mu{\lambda+\mu}\right)^{n-k}}.
\end{aligned}
$$
Outside $0\leq k\leq n$ the conditional probability is zero. Consequently
$$
\boxed{X\mid(X+Y=n)\sim\operatorname{Bin}\!\left(n,\frac\lambda{\lambda+\mu}\right)}.
$$
This <binomial distribution> includes $n=0$, when it is concentrated at zero. Conditioning on the total converts the independent counts into an allocation of that total between the two rates.
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