probability of exactly k success among n trials when the number of trial follows a Poisson distribution Let $n$ denotes the number of ideas, which follows a Poisson distribution.
Given $n=0, 1, 2, ...$, the successful innovation out of $n$ ideas follows a binomial distribution $(n, p)$. That is, probability $p$, each idea becomes a success.
Then, how can I compute the probability that exactly $k$ innovations are realized? For example, $k=1$. Then, it is the sum of probability that one innovation out of one idea, one innovation out of two ideas, and so on.  I know the expression of this, but I need to simply it as much as possible. 
More specifically, $p=1-e^{-j}$ where $j$ is each innovator's investment level.
That is, I need
$$P(X=k) = \sum_{n=0}^\infty P(X=k\mid n) P(n)=\sum_{n=k}^\infty {n \choose k}p^k (1-p)^{n-k} e^{-\lambda} \frac{\lambda^n}{n!}$$
where  $p=1-e^{-j}$. 
The article says that it is given by 
$$ \tilde{\lambda}^k e^{-\tilde{\lambda}} / k!$$
where $\tilde{\lambda}=\lambda(1-e^{-j})$. Unfortunately, I cannot understand this. I need to understand this because I need to present this paper in the class. Please help me out. 
 A: You know that 
$$P(X=k\mid n)= {n \choose k} p^k (1-p)^{n-k}$$
You want
$$P(X=k) = \sum_{n=0}^\infty P(X=k\mid n) P(n)=\sum_{n=k}^\infty {n \choose k}p^k (1-p)^{n-k} e^{-\lambda} \frac{\lambda^n}{n!}\\= e^{-\lambda} \frac{p^k}{(1-p)^k k!} \sum_{n=k}^\infty \frac{s^n}{(n-k)!} $$
with $s=\lambda (1-p)$
Can you go on from here?
A: HINT I assume when you say you know the expression, you mean you're using the law of total probability and writing $$P(K=k) = \sum_{n=0}^\infty p(K=k|N=n)P(N=n) $$ where $N$ and $K$ denote the number of ideas and the number of successful inventions respectively. $P(N=n)$ is just the Poisson distribution and $P(K=k|N=n)$ is related to the binomial distribution.
When you plug everything in, the sum can be done in closed form (it winds up just involving the power series for an exponential). The answer is unsurprising if you consider the ideas to not only be Poisson-distributed, but as the result of a Poisson process. If only some of the ideas wind up being viable (and they are chosen with a given probability as in this problem), you can think of the viable ideas as also being generated by a Poisson process with a smaller rate. 
