In a shop, the average customers per 5 minutes is 3. What is the probability that the shopkeeper has to wait at least 6 minutes before the second customer walks in.
I don't know which distribution I have to use and how to solve this question. I was thinking about Poisson, negative binomial, exponential maybe, but I honestly don't know how to solve this.
On
The number of customers per unit time is Poisson. Equivalently, the time between consecutive customers being exponentially distributed, the two variables' means having product $1$. The sum of two exponential iids has a $k=2$ Erlang distribution. I'll leave the rest to you.
If the question means, right now, before any customers have walked in, what is the probability that we will need to wait at least 6 minutes before the second customer walks in? then we might interpret it as what is the probability that either no customers or only one customer arrives in the next 6 minutes (in which case the second customer could only arrive after 6 minutes)?
In that case we could remodel the situation as $X\sim Po(\lambda_6)$ where the $\lambda_6=\frac{18}{5}$ is the average number of customers per 6 minutes (based on 3 customers per 5 minute period). Then, $P(X=0\lor X=1)\approx 0.126$.