Probability that only one taxi arrives in a 2-minute interval and it's still waiting after 2 minutes

Solution 1:

A few things need to happen:

• Exactly one taxi arrives between 11:00 and 11:02; let its arrival time be $T$.
• In the interval between $T$ and 11:02, no customers arrive. (However, any number of customers may arrive before $T$.)

Alternatively, these conditions can be equivalently expressed as

• The arrivals (either taxis of customers) between 11:00 and 11:02 are some [possibly zero] number of customers followed by a single taxi.

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 $$\sum_{k=1}^\infty e^{-6} \frac{6^k}{k!} \frac{2^{k-1}}{3^k} = \frac{e^{-6}}{2}\sum_{k=1}^\infty \frac{4^k}{k!} = \frac{e^{-6}}{2} (e^4-1).$$