Let $X$~$\mathsf{Poisson}$$(rate=\lambda_1)$ and $Y$~$\mathsf{Poisson}$$(rate=\lambda_2)$. Let $Z=X+Y$. Find the distribution of $X|Z=n$ and name it.
Here's what I tried:
$$\begin{align}\mathsf{M_{X|X+Y=n}}(t) &\stackrel{?}{=} \mathsf{E}[e^{tX}|X+Y=n]\tag 1 \\[1ex]&= \mathsf{E}[e^{tX}|X=n-Y]\tag 2 \\[1ex]&\stackrel{?}{=} \mathsf{E}[e^{t(n-Y)}] \tag 3\\[1ex]&= \mathsf{E}[e^{tn}e^{-tY}] \tag 4\\[1ex]&\stackrel{?}{=} e^{tn}\mathsf{E}[e^{(-t)Y}]\tag 5 \\[1ex]&= e^{tn}\mathsf{M_Y}(-t)\tag 6 \\[1ex]&= e^{tn} \cdot e^{\lambda_2(e^{-t}-1)}\tag 7\end{align}$$
And this is where I am stuck. I don't see how I can work the $tn$ in with the other exponent in such a way that I have a moment generating function that I recognize as the MGF of a distribution I know. I am also uncertain about some of the steps I took to get this MGF, so I put question marks over those equals signs. Lastly, my homework does not include the information that $X$ and $Y$ are independent, but if this information is necessary to make the problem workable then I intend to assume that my instructor forgot to include it. Thanks for the help.
There is no need to use mgfs. Assume that $X$ and $Y$ are independent. Note that (for $0\leq k\leq n)$ $$ P(X=k\mid X+Y=n)=\frac{P(X=k, Y=n-k)}{P(X+Y=n)}=\frac{P(X=k) P(Y=n-k)}{P(X+Y=n)}.\tag{0} $$ where we use independence in the final step. Now since $X,Y$ are independent $X+Y$ is Poisson with mean $\lambda_1+\lambda_2$. After some simplification (which I leave to you) we wind up with $$ P(X=k\mid X+Y=n)=\binom{n}{k}\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^k \left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{n-k} $$ which is the pmf of a binomial distribution.