Summation of binomial number of poisson random variables Z is summation of K random variables that each has Poisson distribution with different means. But, K is a Binomial random with parameters of n and p.
I was wondering what is the distribution of Z?
 A: I doubt $Z$ has a simple distribution.  But it might be possible to describe its moments.
For example if you have a random variable $X_i$ with a Poisson distribution with mean $\lambda_i$ then its variance is also $\lambda_i$ and its second moment is $\lambda_i+\lambda_i^2$.  So if you have a random variable $Y_i$ which is $X_i$ with probability $p$ and $0$ with probability $1-p$, the mean of $Y_i$ is $p\lambda_i$, its second moment is $p\lambda_i+p\lambda_i^2$ and its variance is $p\lambda_i+p\lambda_i^2 - p^2\lambda_i^2$.
So taking $Z$ to be the sum of $n$ such independent $Y_i$s, the mean of $Z$ is $$E[Z]=\sum_{i=1}^n p\lambda_i$$ and the variance of $Z$ is $$\text{Var}(Z) = \sum_{i=1}^n \left(p\lambda_i+p\lambda_i^2 - p^2\lambda_i^2\right)$$ so the second moment is $$E[X^2] = \sum_{i=1}^n \left(p\lambda_i+p\lambda_i^2 - p^2\lambda_i^2\right) + \left(\sum_{i=1}^n p\lambda_i\right)^2$$  and you might with some effort extend this to higher moments.  Since the variance is larger than the mean, this does not have a Poisson distribution.
You can also calculate the probability that 
$$\Pr(Z=0)=\prod_{i=1}^n \left(1-p+pe^{-\lambda_i}\right)$$ and $$\Pr(Z=1)=p \left(\sum_{j=1}^n  \frac{ \lambda_j e^{-\lambda_j}}{1-p+pe^{-\lambda_j}}\right) \left(\prod_{i=1}^n \left(1-p+pe^{-\lambda_i}\right)\right)$$ but I suspect higher values of $Z$ are harder.
