I have a question in a Probability theory and think that it could be solved by exponential distribution. However I'm not confident for dealing with this. Hope to get some helps. Thanks in advance.
A working schedule of buses at a bus stop as follows: the first bus of day comes in at 7am and buses come in every of 15 minutes. Suppose a passenger arrives at the bus stop between 7 and 7:30. What is the probability that the passenger will have to wait
a) less than 10 minutes.
b) at least 12 minutes.
Best Answer
Edit: The wording has changed in a way that makes the problem entirely different. We keep the answer to the original problem below, and show how to solve the new problem at the end of this post.
Solution to old problem: Using an exponential distribution model seems unreasonable, but you are probably expected to do so. Then use the fact that the exponential distribution is memoryless. The waiting time, from the time you arrive, until the first bus has exponential distribution with mean $15$ minutes.
Calculation: Our exponential has mean $15$, and therefore density function $\frac{1}{15}e^{-x/15}$ (for $x\gt 0$). Thus if $X$ is the waiting time, then $$\Pr(X\lt 10)=\int_0^{10} \frac{1}{15}e^{-x/15}=1-e^{-10/15}.$$
The new problem: We assume that we arrive at the stop at a time uniformly distributed between 7 and 7:30 (unreasonable!). Then our waiting time is less than $10$ minutes if we arrive between 7:05 and 7:15, or between 7:20 and 7:30. That has total length $20$ minutes, so the probability our arrival time is in this region is $\frac{20}{30}$.