# A question on probability involving Binomial distribution

First I will state the problem.

A chicken lays $n$ eggs. Each egg independently does or doesn’t hatch, with probability $p$ of hatching. For each egg that hatches, the chick does or doesn’t survive (independently of the other eggs), with probability $s$ of survival. Let $N\sim \text{Bin}(n,p)$ be the number of eggs which hatch, $X$ be the number of chicks which survive, and $Y$ be the number of chicks which hatch but don’t survive (so $X+Y =N$). Find the marginal PMF of $X$.

Intuitively, the probability that a egg hatches and chick survives is $ps$. We can consider each egg as a Bernoulli trial each with a success (hatching and surviving) probability $ps$. There are $n$ independent trials, so $X\sim \text{Bin}(n,ps)$.

But I am trying to prove this more rigorously.

For any $1\le i\le n$ we have $$P(X=i)=\sum_{j=0}^nP(X=i|N=j)P(N=j)\\=\sum_{j=i}^nP(X=i|N=j)P(N=j)\\=\sum_{j=i}^n\binom{j}{i}s^i (1-s)^{j-i}\binom{n}{j}p^j(1-p)^{n-j}$$

How do I show that the above sum collapses to $\binom{n}{i}(ps)^i(1-ps)^{n-i}$?

\begin{aligned} \mathbb{P}(X = i) &= \sum_{j=i}^n {j \choose i} s^i (1-s)^{j-i} {n \choose j} p^j (1-p)^{n-j} \\[8pt] &= \sum_{j=i}^n \frac{j!}{i! (j-i)!} \frac{n!}{j! (n-j)!} s^i (1-s)^{j-i} p^j (1-p)^{n-j} \\[8pt] &= \sum_{j=i}^n \frac{n!}{i! (j-i)! (n-j)!} s^i (1-s)^{j-i} p^j (1-p)^{n-j} \\[8pt] &= \sum_{r=0}^{n-i} \frac{n!}{i! r! (n-i-r)!} s^i (1-s)^r p^{r+i} (1-p)^{n-i-r} \\[8pt] &= \frac{n!}{i! (n-i)!} (ps)^i (1-p)^{n-i} \sum_{r=0}^{n-i} \frac{(n-i)!}{r! (n-i-r)!} (1-s)^r p^r (1-p)^{-r} \\[8pt] &= {n \choose i} (ps)^i (1-p)^{n-i} \sum_{r=0}^{n-i} {n-i \choose r} \bigg( \frac{(1-s) p}{1-p} \bigg)^r \\[8pt] &= {n \choose i} (ps)^i (1-p)^{n-i} \bigg( 1 + \frac{(1-s) p}{1-p} \bigg)^{n-i} \\[8pt] &= {n \choose i} (ps)^i (1-p)^{n-i} \bigg( \frac{1-p + p-ps}{1-p} \bigg)^{n-i} \\[8pt] &= {n \choose i} (ps)^i (1-p)^{n-i} \bigg( \frac{1-ps}{1-p} \bigg)^{n-i} \\[8pt] &= {n \choose i} (ps)^i (1-ps)^{n-i} \\[8pt] &= \text{Bin}(i | n, ps). \\[8pt] \end{aligned}