Binomial dependent on a Poisson I have been working on a problem with a binomial rv dependent on a poisson rv and have worked through to this point: 
$P(X=x) = \sum_{n=x}^{\infty} \dfrac{n!}{x!(n-x)!} p^x(1−p)^{n−x} \dfrac{\lambda^n e^{-\lambda}}{n!}$
Might someone guide me through a next step?
 A: What you need is a simple transformation: Let $j=n−x$. Then
$P(X=x) = \sum_{j=0}^{\infty} p^x(1−p)^j \dfrac{\lambda^{x+j} e^{-\lambda}}{x!j!}$
And proceed. Hint: pull out the "x" terms. i remember similar from somewhere...
A: See the comment of André (at the end $\frac{\lambda^{n}e^{-\lambda}}{n!}$ instead of $\frac{\lambda^{x}e^{-\lambda}}{x!}$).
Let $N,X_{1},X_{2},\dots$ be independent rv's with $N\sim$ Poisson$\left(\lambda\right)$
and $X_{i}\sim$ Bernouilli$\left(p\right)$. 
Now define $X:=X_{1}+\cdots+X_{N}$.
Note that $X\sim$ Binom$\left(n,p\right)$ under condition $N=n$.
If $q:=1-p$ then:
$$P\left\{ X=x\right\} =\sum_{n=x}^{\infty}P\left\{ X=x\mid N=n\right\} P\left\{ N=n\right\} =\sum_{n=x}^{\infty}\binom{n}{x}p^{x}q^{n-x}e^{-\lambda}\frac{\lambda^{n}}{n!}=$$
$$e^{-\lambda}\frac{p^{x}\lambda^{x}}{x!}\sum_{n=x}^{\infty}\frac{q^{n-x}\lambda^{n-x}}{\left(n-x\right)!}=e^{-p\lambda}\frac{\left(p\lambda\right)^{x}}{x!}$$
(Last equality because: $\sum_{n=x}^{\infty}\frac{q^{n-x}\lambda^{n-x}}{\left(n-x\right)!}=\sum_{n=0}^{\infty}\frac{q^{n}\lambda^{n}}{n!}=e^{q\lambda}$)
So $X\sim$ Poisson$\left(\lambda p\right)$.
