Let $A(t)$ denote how often a specific event happens during $[0, t]$., $A(0) = 0$. With iid constant arrival rates, we could write
$$ \mathbf P[A(t+\delta t) - A(t) = 1] = \lambda \delta t + o(\delta t)$$
Working on this formulation then gives rise to the Poisson distribution.
I'm wondering what happens if events take positive time. Say $A(t)$ denotes the number of customers searching through your store. However, once a customer is in your store, it takes him $\gamma$ time to leave your store (where $\gamma$ is another poisson rate). While he is in your store, he blocks any other customer from entering.
In other words, if $B(t) \in \{0, 1\}$ denotes whether a customer is blocking your store,
$$ \mathbf P[A(t+\delta t) - A(t) = 1] = \begin{cases}\lambda \delta t + o(\delta t) & B(t) = 0\\ 0 & B(t) = 1\end{cases}$$
$$ \mathbf P[B(t+\delta t) = 1 \mid B(t) = 1] = \gamma \delta t \\ \mathbf P[B(t+\delta t) = 1 \mid B(t) = 0] = \lambda \delta t $$
However, I wouldn't know what to do with this. Is this correctly set up? Is there a smarter way to set up the system? And how would I approach solving the arrival rate of customers conditional on $B(t) = 0$?