Expected waiting time for bus Suppose that the inter-arrival time between consecutive buses is 15 minutes with probability 0.5 and 30 minutes with probability 0.5. You just arrived at a bus stop. What is your expected waiting time?

My attempt: my gut tells me (since no real information about the type of distribution is provided) that it should just be 22.5 by splitting them down the middle. Is this correct?
 A: The expected waiting time = 11.25 minutes, in my opinion.
Expected time if the inter-arrival time is 15 minutes with probability 1 = 7.5 mins
Expected time if the inter-arrival time is 30 minutes with probability 1 = 15 mins
Now, expected waiting time = $0.5 * (7.5 + 15)=11.25$
A: Because the bus interval arrival time is 30-minute or 15-minute with equal probability,
the probability your arrival falls into a 30-minute interval is $\frac{30}{30+15}=\frac{2}{3}$
the probability your arrival falls into a 15-minute interval is $\frac{15}{30+15}=\frac{1}{3}$
if your arrival falls into a 30-minute interval, the expected waiting time is 15 minutes
if your arrival falls into a 15-minute interval, the expected waiting time is 7.5 minutes
So, the overall expected waiting time is $15\times\frac{2}{3}+7.5\times\frac{1}{3}=12.5$ minutes
A: If the inter-arrival time $T$ is 15 then the waiting time $W$ is bounded by 15, if not its bounded by 30. Now you have that
$$
\mathrm{E}[W]=\mathrm{E}[W|T=15]\Pr [T=15]+\mathrm{E}[W|T=30]\Pr [T=30]
$$
Now we can assume that $W|T\sim U([0,T])$, therefore we have that $\mathrm{E}[W]=\frac{15}{2}\cdot \frac1{2}+\frac{30}{2}\cdot \frac1{2}=\frac{45}{4}=11.25$.
