Let $X$ and $Y$ be independent Poisson variables with respective parameters $\lambda$ and $\mu$. Show that:
(a) $X + Y$ is Poisson, with parameter $\lambda + \mu$
(b) the conditional distribution of $X$, given $X + Y = n$, is binomial, and find its parameters.
MY ATTEMPT
(a) According to the convolution theorem, we have \begin{align*} p_{Z}(Z) & = \sum_{x=0}^{z}f_{X}(x)f_{Y}(z-x) = \sum_{x=0}^{z}\frac{e^{-\lambda}\lambda^{x}}{x!}\times\frac{e^{-\mu}\mu^{z-x}}{(z-x)!}\\\\ & = \frac{e^{-\lambda-\mu}\mu^{z}}{z!}\sum_{x=0}^{z}\frac{z!}{x!(z-x)!}\left(\frac{\lambda}{\mu}\right)^{x} = \frac{e^{-\lambda-\mu}\mu^{z}}{z!}\sum_{x=0}^{z}{z\choose x}\left(\frac{\lambda} {\mu}\right)^{x}\\\\ & = \frac{e^{-(\lambda+\mu)}\mu^{z}}{z!}\left(1+\frac{\lambda}{\mu}\right)^{z} = \frac{e^{-(\lambda + \mu)}(\mu + \lambda)^{z}}{z!} \end{align*}
Therefore $Z\sim\text{Poisson}(\lambda+\mu)$.
(b) According to the definition of conditional distribution, we have \begin{align*} \textbf{P}(X = x | X + Y = n) & = \frac{\textbf{P}(X = x, X + Y = n)}{\textbf{P}(X + Y = n)} = \frac{\textbf{P}(X = x, Y = n - x)}{\textbf{P}(X + Y = n)}\\\\ & = \frac{\textbf{P}(X = x)\textbf{P}(Y = n - x)}{\textbf{P}(X + Y = n)} = \frac{\displaystyle\frac{e^{-\lambda}\lambda^{x}}{x!}\times\frac{e^{-\mu}\mu^{n-x}}{(n-x)!}}{\displaystyle\frac{e^{-(\lambda+\mu)}(\lambda+\mu)^{n}}{n!}}\\\\ & = \frac{n!}{x!(n-x)!}\frac{\lambda^{x}\mu^{n-x}}{(\lambda+\mu)^{n}} = {n\choose x}\left(\frac{\lambda}{\mu}\right)^{x}\left(\frac{\mu}{\mu+\lambda}\right)^{n} \end{align*}
This is as far as I can get. Could someone tell me how to proceed from here? Thanks!
From your last step, write $$\binom{n}{x} \left(\frac{\lambda}{\lambda + \mu}\right)^x \left(\frac{\mu}{\lambda + \mu}\right)^{n-x}.$$