Let $X$ denotes the number of eggs; poisson-distributed with parameter $λ$. Each egg hatches independently of the other eggs with probability $p$. Find the mean number of descendants.
We know that $X\sim Po(\lambda)$. I think that we have to find $E(X|Y=1)$ where $Y$ is an event that describes whether or not it hatch.
Then we have $$ E(X|Y)=\frac{1}{P(Y)} \int_{Y=1} X(\omega)dP $$ since $P(Y)=p$ we have $$E(X|Y)= \frac{1}{p} \int_{Y=1} X(\omega)dP $$
and I am stuck here. I guess i should use the probability density function for Poisson distribution but I am not sure about integration limits. However I am not sure if my idea is correct.
Which "it"? There are $X$ eggs, each hatching independently with probability $p$.
Rather, let $Y$ be that which you wish to count: the descendants. This is clearly the count for eggs which are lain and hatch.
The count for successes among a number of independent trials with identical success rate $p$ is, by definition, a Binomial Distributed Random Variable.
$\qquad Y\mid X\sim \mathcal{Bin}(X, p)$
So, begin here:
$\qquad\begin{align}\mathsf E(Y) &=\mathsf E(\mathsf E(Y\mid X))\\[1ex] &=\sum_{x=0}^\infty\sum_{y=0}^x y\,\mathsf P(X=x)\,\mathsf P(Y=y\mid X=x)\\[1ex]&= \sum_{x=0}^\infty\sum_{y=0}^x y\cdot\dfrac{\lambda^x\mathrm e^{-\lambda}}{x!}\cdot\dfrac{x!}{y!~(x-y)!}\,p^y\,(1-p)^{x-y}\\[1ex]&=\sum_{y=0}^\infty\sum_{x=y}^\infty\dfrac{y\,\lambda^{y+x-y}\,p^y\,(1-p)^{x-y}~\mathrm e^{-\lambda}}{y!~(x-y)!}\\&~~\vdots\end{align}$