The number of eggs laid by an insect in one generation is X where X is a Poisson distribution of parameter $\lambda$. The probability of an insect developing from each egg is $\alpha$. Assuming insects develop from eggs independently, compute the probability that an insect has k descendants (hatched eggs) in one generation.
Let $E_k$ be the event that the insect has k descendants in one generation. We have: $$P(E_k) = \sum_{i=k}^{\infty}P(E_k|X=i)P(X=i) = \sum_{i=k}^\infty {i \choose k}\alpha^k(1-\alpha)^{i-k}e^{-\lambda}\frac{\lambda^i}{i!}$$
Wondering if there is any more meaningful way to simplify/express this probability?
We can remove the constants (in terms of $i$) from $$\sum_{i=k}^\infty {i \choose k}\alpha^k(1-\alpha)^{i-k}e^{-\lambda}\frac{\lambda^i}{i!}$$to get $$\frac{\alpha^k e^{-\lambda}}{k!} \sum_{i=k}^\infty \frac{1}{(i-k)!}(1-\alpha)^{i-k}\lambda^i.$$ Taking out a factor of $\lambda^k$ and reindexing the sum with $j=i-k$ we have $$\frac{(\alpha\lambda)^{k} e^{-\lambda}}{k!} \sum_{j=0}^\infty \frac{(\lambda(1-\alpha))^{j}}{j!}.$$ The sum is now $e^{\lambda(1 - \alpha)}$, so we have $$P(E_k)=\frac{(\alpha\lambda)^{k} e^{-\lambda\alpha}}{k!},$$ so it's equivalent to observing $k$ from a Poisson$(\alpha\lambda)$ random variable.