Suppose $X$ and $Y$ are independent Poisson random variables. Find the conditional probability mass function $P(X=k\mid X+Y=m)$ Suppose $X$ and $Y$ are independent Poisson random variables with parameters $\lambda$ and $\mu$, respectively. Find the conditional probability mass function $P(X=k\mid X+Y=n)$.

Don't know how to approach this. Additionally, the question asks me to find $E[X\mid X+Y=n]$. What is the distinction? Does it have a different distribution?
 A: You need to provide more context for your question; in particular, even if you say you don't know how to approach the problem, you need to be able to demonstrate what you DO know.
The one question I will answer, however, is that $$\Pr[X = k \mid X + Y = n]$$ is a conditional probability.  Specifically, it is the probability that, given you observe the sum of the Poisson variables to be equal to $n$, that you also observed $X = k$.  So, obviously, this probability cannot be less than zero, nor greater than one.
By contrast, $$\operatorname{E}[X \mid X + Y = n]$$ is a conditional expectation.  It is, in some sense, the "average value" of $X$ given that the sum $X+Y$ was observed to be $n$.  This is a number that need not be between zero and one; in fact, if $n$ is very large, say $1000$, we would also intuitively reason that the expected value of $X$ should be quite a bit larger than one.
A: First you need to find the distribution of $X+Y$, there are several ways to do this,you can use the moment generating function to find the distribution of $X+Y$ 
$E[e^{t(X+Y)}]=E[e^{tX}]E[e^{tY}]=e^{\lambda(e^t-1)}e^{\mu(e^t-1)}=e^{(\lambda+\mu)(e^t-1)}$~Poisson($\lambda+\mu)$.
After you can computate the probability making
$$P(X=k|X+Y=n)=\frac{P(X=k)P(Y=x-n)}{P(X+Y=n)}$$
