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

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.

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

• Before someone points out to this similar problem, I am looking to continue with my approach which is different from the one there. Nov 11, 2022 at 11:16
• Yes, this is described on Wikipedia here: en.m.wikipedia.org/wiki/… and probably already in a question here. Nov 11, 2022 at 12:49
• Your problem is the same as in the question that you linked to. Just compare the following 'flower buds opening = eggs hatching' and 'blooms surviving = chick survives'. Nov 11, 2022 at 13:00