Buses leave a bus station according to a Poisson process $\{B_t\}$ of rate $\lambda$, while travellers reach the same station according to (an independent) Poisson process $\{T_t\}$ of rate $\mu$. Travellers get on the bus and leave the station as soon as a bus departures.
$(1)$ Let $N$ be the random variable, expressing the number of passengers on a departuring bus. Find $P(N=k),~k=0,1,2,\ldots$
$(2)$ Let $\Lambda$ be the random variable, expressing the number of departuring buses till the time the $1000$th traveller has reached the station. Find $P(\Lambda=k),~k=0,1,2,\ldots$
Attempt.
(1) In order for $N$ to be $k$ there must be $k$ travellers reaching the bus station and no bus leaving in between. The time needed for k travellers to reach the station is $W\sim Gamma(k,\mu)$, so $$P(N=k)=P(B_W=0)=\int_{0}^{+\infty}P(B_w=0)f_W(w)dw=\int_{0}^{+\infty}e^{-\lambda w} \frac{\mu^k}{(k-1)!}w^{k-1}e^{-\mu w}dw.$$
(2) The time needed for the $1000$th traveller to reach the station is $W\sim Gamma(1000,\mu)$, so $$P(\Lambda=k)=P(B_W=k)=\int_{0}^{+\infty}P(B_w=k)f_W(w)dw=\int_{0}^{+\infty}e^{-\lambda w}\frac{(\lambda w)^k}{k!} \frac{\mu^{1000}}{999!}w^{999}e^{-\mu w}dw.$$
Am I on the right path? Thank you!
For sake of completeness I post my answer, based on the useful remarks made by @Furrer:
(1) $$P(N\geq k)=P(B_W=0)=\int_{0}^{+\infty}e^{-\lambda w} \frac{\mu^k}{(k-1)!}e^{-\mu w}dw=\Big(\frac{\mu}{\lambda+\mu}\Big)^k$$ and $$P(N=k)=P(N\geq k)-P(N\geq k+1)=\frac{\lambda}{\lambda+\mu}~\Big(1-\frac{\lambda}{\lambda+\mu}\Big)^k$$ (geometric distribution)
(2) $$P(\Lambda= k)=P(B_W=k)=\int_{0}^{+\infty}e^{-\lambda w}\frac{(\lambda w)^k}{k!} \frac{\mu^{1000}}{999!}w^{999}e^{-\mu w}dw=\binom{999+k}{k}\Big(\frac{\lambda}{\mu+\lambda}\Big)^k\Big(1-\frac{\lambda}{\mu+\lambda}\Big)^{1000}.$$
Question: if I have to plug in numbers for $\lambda,~\mu$, they have to be on the same time unit, for example both rates expressed in days, or both in months?