Consider a taxi station where taxis and customers arrive in accordance with Poisson process with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. I'm trying to find the probability that precisely one taxi arrives between 11:00 and 11:02 and it's still waiting at 11:02.
I've been stuck for a while and not exactly sure how to approach this problem.
Best Answer
A few things need to happen:
Alternatively, these conditions can be equivalently expressed as
Note that by Poisson superposition the combined process of arrivals (of either type) has a rate of 3 arrivals per minute. So the probability that "there are $k$ arrivals in two minutes" is $e^{-6} 6^k/k!$, and conditioned on this, the probability that "the $k$ arrivals are $k-1$ customers followed by a single taxi" is $(2/3)^{k-1}(1/3)$. Thus, the final probability is $$\sum_{k=1}^\infty e^{-6} \frac{6^k}{k!} \frac{2^{k-1}}{3^k} = \frac{e^{-6}}{2}\sum_{k=1}^\infty \frac{4^k}{k!} = \frac{e^{-6}}{2} (e^4-1).$$