Waiting time in a bus stop 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.
 A: 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 $\frac{20}{30}$. 
A: If we can state that there is a 1 in 15 probability of a bus arriving any given minute, then the probability of a no bus arriving for a certain length of time is the integral
$$\int_0^t{e^{-t/\tau}dt}$$
where $\tau$ is the probability of a bus arriving in unit time (1/900 if we are working in seconds). This is an easy integral to evaluate.
Of course anyone who has ever waited for buses understands that, especially in the morning, there is a high correlation in arrival times. The "first" bus is slow because it has to pick up lots of passengers; the one behind will catch up, and soon you get the situation where you wait a long time, only to find a full bus followed immediately by an empty one.
update
on the other hand, if the buses arrive every 15 minutes (not "on average" but actually at 7:00, 7:15 etc) then a different scheme is needed.  You can look at every possible minute that the person could arrive (7:01, 7:02 etc) and see whether the wait time is longer or shorter than a given limit.  Thus
Arriving after 7:05 or after 7:20, there will be less than 10 minute wait. This is a probability of 2 in 3.
Arriving between 7:00 and 7:03, and 7:15 and 7:18, the wait will be more than 12 minutes. Probability 6 in 30 or 1 in 5.
This is conveniently ignoring the "running after the bus", time that the bus is actually at the stop, etc... 
