In a mall, a survey found that the number of people who pass by JCPenney between 4:00 and 5:00 pm is a Poisson random variable with parameter λ = 100. Assume that each person may enter the store, independently of the other person, with a given probability p = 0.15. What is the expected number of people who enter the store during the given period? (Answer key 15)
My biggest problem is trying to figure out how to set up the problem. It is in the textbook section on conditional expected value. My thought process is that on average we have a 100 people enter the store between 4:00 and 5:00 pm therefore $$\begin{align*} E(X|Y=100)&=\sum_ {x=0}^{100} xP(X=x|Y=100)\\&=\sum_{x=0}^{100}\frac{x(0.15)100^x}{e^{100} x!}\\&=\frac{0.15}{e^{100}}\sum_{x=0}^{100}\frac{x100^x}{x!}\\&=100(0.15)=15\end{align*}$$ where $Y$ is poisson random variable and $X$ is the number of people that enter the store. I am not very confident at word problems so I would like to know if I set it up correctly or just got lucky.
Here's the easy way:
People are passing by the store with a constant average rate of $100$ per that interval, and each one has an independent probability of entering the store of $0.15$. Therefore people are entering the store at a constant average rate of $15$ per that interval.
The hard way (what you were trying): $$\begin{align}\mathsf E(X) =&~ \mathsf E(\mathsf E(X\mid Y)) \\[1ex] =&~ \sum_{y=0}^{\infty}\dfrac{100^y\,\mathsf e^{-100}}{y!}\sum_{x=0}^y \dfrac{x~y!~0.15^x 0.85^{y-x}}{x!(y-x)!}\\[1ex]\vdots&~ \\[1ex]=&~ 15\end{align}$$
You may find the closed forms from first principles, but they are well known, so quickly: