There is a bus, whose departure at a stop is distributed as poisson(mu). People arriving to the same stop is distributed as poisson(lambda). Find the PMF of the number, N, of people on any given bus.
P_N (n) = ??
I could not come up with a way to approach other than, i know i should condition on the inter arrival time of the bus.
conditioning I meant The number of people on a bus can be seen as the number of failures (people arriving) until the first success (bus departure).
Assuming both Poisson processes are independent and all buses have infinite capacity which is able to let all people waiting at the stop to enter the bus.
Let $T_k$ be the arrival time of the $k$-th bus and $X(t)$ be the cumulative number of arrivals of the people at the stop. Then $T_k \sim \text{Gamma}(k, \mu)$ and $X(t) \sim \text{Poisson}(\lambda t)$, and the number of people enter the $k$-th bus at this stop is $X(T_k) - X(T_{k-1})$.
But actually the situation is much simpler, as by the assumption we know that all the inter-arrival times of the buses are i.i.d. $\text{Exp}(\mu)$, and the number of people waiting at the stop will be zero right after each bus arrival. So actually the number of people enter in each bus has the same distribution (that's why the question use $N$ to represent without subscript to specify which bus).
So WLOG we can just consider the number of people entering the first bus as $N$ has the same distribution as $X(T_1)$
$$ \begin{align} \Pr\{X(T_1) = n\} &= \int_0^{\infty} \Pr\{X(t) = n\} f_{T_1}(t)dt \\ & = \int_0^{\infty} e^{-\lambda t} \frac {(\lambda t)^n} {n!} \mu e^{-\mu t}dt \\ & = \frac {\mu\lambda^n} {n!} \int_0^{\infty} t^n e^{-(\lambda + \mu)t}dt \\ & = \frac {\mu\lambda^n} {n!} \frac {\Gamma(n+1)} {(\lambda + \mu)^{n+1}} \\ & = \frac {\mu} {\lambda + \mu} \left(\frac {\lambda} {\lambda +\mu}\right)^n, n = 0, 1, 2, \ldots \end{align}$$
So $\displaystyle N \sim \text{Geometric}\left(\frac {\lambda} {\lambda +\mu}\right)$ (with support $\{0, 1, 2, \ldots\}$)