Suppose that in a coffeehouse there are $\text{Poi}(10)$ guests, each one of the guests drinks $\text{Poi(}1)$ cups of coffee, and raises his hand, independently, $\text{Poi}(1)$ amout of times.
The number of guests, the amount of coffee that each guest drink, and the number of hand raises each guest perform - are all independent.
Suppose that in the same day, no cups of coffee have been sold. Given that information, calculate the expected value of the total hand raises.
Well, my attempt was to let $X\sim \text{Poi}(10)$ be the number of guests at the coffeehouse, let $Y$ be the number of cups of coffee and $Z$ be the number of hand raises.
Since the hand raises are independent where each one has $\text{Poi}(1)$ distribution , and there are $X$ guests, we can conclude that $Z\sim \text{Poi}(X)$. Similarly, $Y\sim\text{Poi}(X)$. So if I understand correctly, I need to find the value of $E[Z|Y=0]$
How can I proceed from here? How the fact the $Y=0$ affects $Z$?
Using Bayes' theorem we get that:
$$P(X=k|Y=0)=\frac{e^{-\frac{10}{e}}(\frac{10}{e})^k}{k!}$$
Therefore, $Z\sim\text{Poi}(\text{Poi}(\frac{10}{e}))$, so $E[Z]=\frac{10}{e}$