Poisson process in a restaurant

Question

While preparing for a midterm, I came across this question

Suppose a restaurant is visited by 10 clients per hour on average, and clients follow a homogeneous Poisson Process. Independantly of other client, each client has a 20% chance to eat here and 80% to take away. In average, how many clients should be expected before one eats here ? 

Proposed answer :

• 8
• 4
• 2

For me the correct answer is 4, but many of friends have answered "2" because they've decomposed the poisson into two poisson process one of parameter 0.210 and another one with parameter 0.810.

Who is right? The question is really tricky, isn't it?

Thanks for you help !

Answer 1

The rate at which clients arrive does not matter. All that matters is that each client has $0.2$ chance to eat here, so the expectation of the geometric distribution is $5$ clients before one eats here. As the question is written, you are asked about how many not counting the one that eats here, so your $4$ is correct.

Answer 2

Suppose $X$ clients arrive before one eats at the restaurant, and take $T$ as the arrival time of the first client to eat at the restaurant. Then $T\sim \exp(2)$ and $X|T\sim \text{Poisson}(8T)$. We get for $x=0,1,2,...$ that $$\begin{eqnarray*}P(X=x)&=&\int_0^{\infty}P(X=x|T=t)f_{T}(t)\mathrm{d}t \\ &=& \int_0^{\infty}{e^{-8t}(8t)^x \over x!}\cdot2e^{-2t}\mathrm{d}t \\ &=& (0.8)^x\cdot 0.2\end{eqnarray*}$$ So, $$\mathbb{E}(X)=\sum_{x=0}^{\infty}x(0.8)^x(0.2)=4$$