The distribution of an event whose waiting time is exactly k

Question

I am trying to solve the following problem

Assume busses arrive with the interval exactly 10 min one after another. At the end of a working day, I take a bus home. Let α be my expected waiting time for the train. Find α.

So, my approach here is as follows, first I need to model the following distribution, then integrate the PDF of the corresponding distribution to find the expected value. The problem is, I am not sure whether the distribution is Exponential with lambda 1/10. I am given the waiting time between two successive events, but it is not the average waiting time, and that confuses me. Can you help me understand the distribution??

Answer 1

I see the problem is in modelling, so let me propose two models:

1. Buses arrive every 10 minutes all day, starting at midnight 0:00. I finish my work somewhere between 5:00 and 5:10 pm and I wish to take the first bus. Let's say the time I finish work is uniformly distributed.
2. I finish my work at 5:00 pm sharp. Buses arrive every 10 minutes, but their starting time is random. Let's say the first bus arrives somewhere between 6:00 and 7:00 am, and again with uniform distribution.

In all possible models, my waiting time $X$ is a random variable with values in $[0,10]$ (measured in minutes). In both cases above, $X$ is actually uniformly distributed, which gives us the estimates time of $5$ minutes.

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The modelling part is not strictly mathematical. The author of the problem probably wants you to assume (or realize, in part) that the waiting time is uniformly distributed in $[0,10]$. However, one can come up with different models:

3. The first bus is at midnight 0:00 and I always finish work at 5:03, in which case I always wait 7 minutes.
4. I always finish at 5:00 and the first bus arrives between 6:00 and 6:58 (with uniform distribution), in which case the estimated time is slightly smaller than 5 minutes.
5. I'm taking a bus, but I'm actually waiting for a train, so I missed my chance entirely.