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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.

user
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  • Do we believe that the buses arrive every 15 minutes? What does "average frequency" mean?... Does it mean there is a 1/15 probability of a bus arriving any given minute? – Floris May 18 '14 at 16:34

2 Answers2

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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 $\frac{20}{30}$.

André Nicolas
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  • I think we are saying the same thing. – Floris May 18 '14 at 16:40
  • @AndréNicolas. Thank you. I just know from your hint, $\lambda =15$. If the exponential distribution is expected to use so I need something for (a) like $P(X<10)=1-e^{-15 \times 10}$. But the assumption concerning with 7 and 7:30 hasn't used yet. – user May 18 '14 at 16:44
  • If the buses arrive truly independently, then the fact that a bus arrived at 7 am ought to be completely irrelevant (and so the information is a red herring). But if the buses arrive every 15 minutes, then you need to use a different method. – Floris May 18 '14 at 16:57
  • A lot of the information is there to confuse you. Your computation is not correct, I will add something. – André Nicolas May 18 '14 at 17:09
  • @Floris. You're right. The buses arrive every 15 minutes. I edited it. – user May 18 '14 at 17:11
  • @AndréNicolas. I didn't see the assumption the passenger arriving at the bus stop between 7 and 7:30. Why we don't need this? – user May 18 '14 at 17:23
  • The problem has changed, now the buses arrive according to a rigid schedule. Before, we were using an exponential model for bus arrival times, and the memorylessness of the exponential made the passenger's time of arrival irrelevant. – André Nicolas May 18 '14 at 17:32
  • I am a little bit miffed that this incomplete answer (only answering the first of two questions) got marked as accepted. Especially since I had prodded OP to actually get the wording right, and I had given the solution to both a and b for the "correct" scenario quite a while ago. – Floris May 18 '14 at 17:50
  • @AndréNicolas. Thank you for your answer solving both versions. Actually, I mean that the first bus of day comes in at 7am and after any 15 minutes there is other comimg. In the case my attemping correction is still not clear. Can we do a "trick" that whenever problems refer an assumption like "7 and 7:30", the problems is automatically considered over uniform distribution? – user May 18 '14 at 17:53
  • I imagine that @user meant to accept yours, and hope the acceptance of mine will be corrected. – André Nicolas May 18 '14 at 17:55
  • @user: The problem should have specified the distribution, particularly since in this case the assumption of uniformity is implausible. But it is true that in "exercises" uniformity is often a default assumption. – André Nicolas May 18 '14 at 18:01
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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...

Floris
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  • Thank you for your detailed solution solving the problem over uniformly distributed version. – user May 18 '14 at 18:07