Poisson process: Probability that $n$ busses of bus route A arrive before the first bus of route B 
There are two bus routes that stop at a bus stop. Busses arrive according to a poisson process with rate $\lambda_1t$ buses per minute for the first route and $\lambda_2t$ for the second route.
Find the probability that at least $n$ buses of route 1 will pass by berfor the first bus of route 2 arrives.
Hint: You may use the fact that if $X_1 \sim \operatorname{Expo}(\lambda_1)$ and $X_2 \sim \operatorname{Expo}(\lambda_2)$, then $P(X_1 < X_2) = \frac{\lambda_1}{\lambda_1 + \lambda_2}$.

I didn't know how to make use of the hint. I thought the following way:
Let $Z_1(t) \sim \operatorname{Pois}(\lambda_1t)$ be the number of buses of route 1 that stop in a time interval of size $t$, and $Z_2(t) \sim \operatorname{Pois}(\lambda_2t)$ the number of buses of route 2. Then we are looking for $P(Z_1 \geq n \cap Z_2=0)$:
\begin{align}
P(Z_1 \geq n \cap Z_2=0) &= \int_{t=0}^{t=\infty} P(Z_1 \geq n)(t) \cdot P(Z_2 = 0)(t) dt
\end{align}
And there I realize that I'm probably off the track.
 A: You may already know that the time it takes for the first bus of route 1 to arrive is exponentially distributed with parameter $\lambda_1$. Likewise, the arrival time of the other bus is exponentially distributed with parameter $\lambda_2$. So, the hint tells you that the probability of the event "first bus to reach the bus stop is of route 1" is $\frac{\lambda_1}{\lambda_1+\lambda_2}$. (Do you see why?)
Using other properties of Poisson processes, you can show that the probability of the event "the second bus to reach the bus stop is of route 1" is also $\frac{\lambda_1}{\lambda_1+\lambda_2}$, and the similarly for the $n$th bus. Even more, these events are all independent! [You will need to justify these facts.] With this information, you can easily find the answer to your question.
The "merging" of two Poisson processes is called "superposition," in case you want to learn more about it.
A: Assuming that the Poisson processes are independent, the successive routes are i.i.d. and route 1 has probability $p=\lambda_1/(\lambda_1+\lambda_2)$. Hence the $n$ first routes are all route 1 with probability $p^n$.
