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?
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 = $0.5 * (7.5 + 15)=11.25$
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
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 $\mathrm{E}[W]=\frac{15}{2}\cdot \frac1{2}+\frac{30}{2}\cdot \frac1{2}=\frac{45}{4}=11.25$.
