$Y\sim Poi(\lambda)$ trials are conducted with a probability of success $p$. Find the distribution of the number of successful trials. 
For a given experiment, $Y\sim Poi(\lambda)$ independent trials are conducted, with each trial having a probability of success $p$ and a probability of failure $1-p$. Find the distribution of the number of successful trials $X$."

First of all, $Y$ Bernouilli trials are conducted, thus
$$X\sim\mathit{Bin}(Y, p)$$
Therefore
$$P(X=k)=\binom{Y}{k}p^k(1-p)^{Y-k}$$
From $Y\sim Poi(\lambda)\Longrightarrow P(Y=t)=\frac{\lambda^t}{t!}e^{-\lambda}$
Am I on the right path? How can I continue now?
 A: You can condition on number of independent trials $Y$ and then integrate the conditional distribution against the distribution of $Y$.
So $P(X=k|Y=y) = {y \choose k}p^k (1-p)^{y-k}$
$P(X=k) = \sum_{y=0}^{\infty} P(X=k|Y=y) P(Y=y)$, using the law of total probability.
$P(X=k) = \sum_{y=0}^{\infty} {y \choose k}p^k (1-p)^{y-k} {\lambda^y \over y!} e^{-\lambda} = e^{-\lambda}({p \over 1-p})^k\sum_{y=k}^{\infty} {y \choose k} {\lambda^y \over y!}(1-p)^y$
A: In addition to the previous answer:
$P(X=k) = \sum_{y=k}^{\infty} {y \choose k}p^k (1-p)^{y-k} {\lambda^y \over y!} e^{-\lambda} $ hence
$$P(X=k) = \sum_{y=k}^{\infty} {y \choose k}p^k(1-p)^{y-k} {\lambda^y \over y!} e^{-\lambda} = e^{-\lambda} p^k \sum_{y=k}^{\infty} {y \choose k} {\lambda^y \over y!} (1-p)^{y-k}=$$
$$ = e^{-\lambda}p^k \sum_{y=k}^{\infty} \frac{y!}{k! (y-k)!} {\lambda^y \over y!} (1-p)^{y-k}= \frac{e^{-\lambda} \lambda^k}{k!} p^k \sum_{y=k}^{\infty} \frac{\lambda^{y-k}}{(y-k)!} (1-p)^{y-k} = $$
$$ = \frac{e^{-\lambda} \lambda^k}{k!} p^k \sum_{z = 0}^{\infty} \frac{(\lambda(1-p))^{z}}{(z)!}  = \frac{e^{-\lambda} \lambda^k}{k!} p^k e^{-(1-p)\lambda}= \frac{(p\lambda)^k e^{-p\lambda}}{k!}.$$
Thus $X \sim Pois(p\lambda)$.
A: Another way is through the moment-generating function. We have $$ E( e^{tX}) =  \sum_{n=0}^\infty P(Y=n) E( e^{tX}\mid Y=n ) = \sum_{n=0}^\infty e^{-\lambda}\frac{\lambda^n}{n!} ((1-p) + pe^t)^n \\= e^{-\lambda} e^{\lambda((1-p)+pe^t)} = e^{p\lambda(e^t-1)}, $$ where we used the fact that the MGF for a binomial(n,p) is $((1-p) + pe^t)^n.$ Then, note that $e^{p\lambda(e^t-1)}$ is the MGF for a Poisson distribution with mean $\lambda p.$
