A certain toy store sells toys. During a particular hour, customers arrive at the store according to a Poisson Process with mean $10$ per hour. Let $X$ be the number of customers who arrive during this hour. For each customer, there is a $\frac{1}{5}$ chance that the customer will buy exactly one toy and a $\frac{4}{5}$ chance that the customer will not buy any toy, independent of other customers. Let $Y$ be the number of toys sold in the one hour period. By conditioning $Y$ on $X$, use the law of iterated expectation to determine $\mathbb{E}(Y)$ and then use the law of total variance to determine $\mathrm{Var}(Y)$.
I have been trying to solve this problem for hours but without success. Any assistance you can provide is much appreciated.
Since customer arrival is a Poisson process with an intensity of $10$, the Number of customers in an hour will have distribution Poisson$(10)$, and each customer buys a toy with probability $1/5$ therefore expected number of toys sold given $x$ many customers arrived ($\mathbb{E}(Y|X=x)$) is $x/5$. Thus
$$ \mathbb{E}(Y) = \sum_{x=0}^{\infty}\mathbb{E}(Y|X=x)\mathbb{P}(X=x) = \sum_{x=0}^{\infty}\frac{x}{5}\frac{e^{-10}10^{x}}{x!} = \frac{1}{5}\sum_{x=0}^{\infty}\frac{x}{5}\frac{e^{-10}10^{x}}{x!} =2, $$ since $\sum_{x=0}^{\infty}{x}\frac{e^{-10}10^{x}}{x!}$ is the expectation of Poisson$(10)$, which is $10$.
You can write $$ Var(Y) = Var(\mathbb{E}(Y|X)) +\mathbb{E}(Var(Y|X)). $$ you can use this formula and compute the two terms separately. Try it!
Hint 1: $Y|X \sim$ Binomial$(X,\frac{1}{5})$.
Hint 2: $X=$ the Number of customers in an hour. It has distribution Poisson$(10)$