If $X \sim \text{Bin}(n,p)$ and $Y|X =k \sim \text{Bin}(k,r)$, can we say that $Y \sim \text{Bin}(n,pr)$?

Question

Question:

There are $n$ flower buds in a garden, each of which opens (independently) with a probability $p$ and each of the blooms survive (independently) with a probability $r$. What's the distribution of the number of buds that survive as a flower?

Attempt:

Let $X$ be the number of buds that open and $Y$ be the number of flowers that survive. Then $X \sim \text{Bin}(n,p)$ and $Y|X =k \sim \text{Bin}(k,r)$. I think eventually we will get $Y \sim \text{Bin}(n,pr)$, but I don't know how to get there.

$$\begin{align}P(Y = y) &= \sum_{k=y}^{n} P(Y = y | X = k)P(X=k) \\&= \sum_{k=y}^{n} {k \choose y}r^y(1-r)^{k-y}{n \choose k}p^k (1-p)^{n-k}\\ &= ?\end{align}$$

Kindly don't redirect me to other similar questions (unless the approach there is exactly similar to mine). This is (possibly) a well-known question, but I am trying to solve it in this exact manner or know if and why my approach is incorrect.

Answer 1

The result is obvious probabilistically, but can also be derived by manipulating binomial coefficients:

Observe that $${k \choose y} {n \choose k}={n \choose y}{n-y \choose k-y}\,.$$ Therefore, writing $\ell=k-y$ and recognizing a binomial expansion, $$ \sum_{k=y}^{n} {k \choose y} (1-r)^{k-y}{n \choose k}p^{k-y} (1-p)^{n-k} = $$ $${n \choose y} \sum_{\ell=0}^{n-y} {n-y \choose \ell} (1-r)^{\ell} p^{\ell}(1-p)^{n-y-\ell}=$$ $${n \choose y}\bigl((1-r)p+(1-p)\bigr)^{n-y} ={n \choose y} (1-rp)^{n-y} \,, $$ where we have used the binomial theorem in the second step.

Multiplying both sides of the last display by $r^yp^y$ gives

$$P(Y = y) = \sum_{k=y}^{n} {k \choose y} r^y(1-r)^{k-y}{n \choose k}p^k (1-p)^{n-k} = {n \choose y}(rp)^y (1-rp)^{n-y} \,,$$ as required.

Answer 2

Well, the key is that you know where you want to go. So the next step is to distribute out what you want to keep. Next show that what remains in the series will close to $1$.

Recall that $\dbinom ab=\dfrac{a!}{b!~(a-b)!}$, so...

$~$

$\begin{align}P(Y = y) &= \sum_{k=y}^{n} P(Y = y | X = k)P(X=k) \\&= \sum_{k=y}^{n} {k! \over y!~(k-y)!}r^y(1-r)^{k-y}{n! \over k!~(n-k)!}p^k (1-p)^{n-k}\\ &=\dfrac{n!}{y!~(n-y)!}(rp)^y(1-rp)^{n-y}\sum_{k=y}^n\dfrac{(n-y)!}{(k-y)!~(n-k)!}\dfrac{(1-r)^{k-y}p^{k-y}~(1-p)^{n-k}}{(1-rp)^{n-y}}\\&~~\vdots \\[4ex] &=\binom n y(rp)^y(1-rp)^{n-y} \end{align}$