People getting to a builduing in time gaps that distribute exponential, in 3 people per minute rate.
let X be the rate of people that entering the building. X~exponential(3).
What is the probability for the 3rd person to get to the building in less than 3 minutes after the 1st one?
Let $\theta$ be the exponential parameter, which is 3 in this example.
The first person doesn't matter; this is equivalent to asking for the probability that the second person gets into the building within 3 minutes of it opening.
Assuming that people enter independently, if $X$ and $Y$ are both $\mathrm{Exp}(\theta)$ then $X + Y = \mathrm{Erlang}(2, \theta)$. See the Erlang Distribution for more details.
So we need $Pr(X+Y \leq 3)$, which will reduce in this case to solving
$$ \int_0^3 \theta^2xe^{-\theta x} dx $$
which I calculated as around 99.88% from Wolfram Alpha.
