The problem I am trying to solve is formulated as following:
Random number $\mathit{N}$ of unsorted elements in a set is Poisson distributed with $\lambda>0$. These elements are found independently (and independently of $\mathit{N}$) with a probability of $\mathit{p}\in(0,1)$ and placed in the set on their right places. What is a number of still unsorted elements after above process?
My approach:
After the process, the number of unsorted elements is $\mathit{K}$. It is also $\mathcal{B}_{(n, 1-p)}$-distributed, in other words $P(K=k) \mid N=n) = \mathcal{B}_{n, 1-p} \{k\}$. Then to be found is probability mass function: $\mathit{P(K=k)}$. After using the law of total probability I obtained:
$$P(K=k)=\sum_{n=k}^{\infty} \binom{n}{k} (1-p)^k p^{n-k}\frac{\lambda^n}{n!}e^{-\lambda}$$
Here's where I got stuck and do not on how to proceed further. I know that in that formula, there is a possibility of simplifying it due to $\frac{\lambda^n}{n!}$ and $\binom{n}{k} (1-p)^k p^{n-k}$. Any help would be appreciated.