Poisson process in a restaurant

Solution 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.

Solution 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, $$\mathbb{E}(X)=\sum_{x=0}^{\infty}x(0.8)^x(0.2)=4$$