I've been given the following question which was part of a past exam paper.

Q3) Trains arrive at a station at the rate of 25 per hour.

a) What is the probability that the time between trains arriving is 6 minutes or less?

I have attempted this question again, following some of the formulae which I found in some of the links given to me. These are my workings.
I apologise if this isn't the typical way of demonstrating workings, I will reformat if needs be.

b) The train company wants to state that waiting times (in minutes) between trains have been targeted for improvement. It fails to meet its target 40% of the time. What target has it set itself?

For this part, this is what I've got so far: P(Xt<x)=0.6, since they are succeeding in having a train arrive within this time 60% of the time, which means 0.6=1-e^-0.4167x but I think I've gone wrong again as this is giving me the answer that x=-1.23 which can't be right because . Not sure where to go from here.

c) A train enthusiast notes that a goods train arrives, on average, once an hour. What is the probability that two goods trains will arrive in one hour?

Not sure how to approach this one.

d) If the enthusiast wishes to stay at the station for an hour, what is the probability that he will see at least one goods train in that time?

Not sure how to approach this one.

I've been in correspondence with my lecturer and she says this is a pretty straightforward problem. I know you're not meant to provide exact solutions to problems and just guide me to the answers myself, but I can't help but think I'm missing something really obvious which is making this hard for me.

It's worth noting that we've not been given anything about exponential probability distribution functions in our textbooks etc so I don't know if there's some other way of getting the answers I need.

Any help appreciated.

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    $\begingroup$ Welcome to Cross Validated. Since this is a homework question, please add the self-study tag and read its wiki, which includes guidelines for asking self-study questions. One such guideline is to show that you have made a "good faith" attempt to solve the problem yourself and show us where you are stuck. Questions that do not show a good faith attempt may be subject to closure. $\endgroup$ Commented Apr 11, 2016 at 21:34
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    $\begingroup$ yes, please do that $\endgroup$ Commented Apr 11, 2016 at 21:40
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    $\begingroup$ @Havesspeas wrote "On average, the time interval is 2.4 minutes between trains, so [I think] intuitively, I got the answer that P(X<=6) = 1." Your intuition is wrong, and is obviously not attuned to probability calculations, so don't rely on it (or use it at all until better developed). Systematically work through the mathematics of the probability calculations. Do you know what a Poisson process is, and how it relates to the probability distribution of time between events? $\endgroup$ Commented Apr 11, 2016 at 21:47
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    $\begingroup$ @Havesspeas, that's fine that you don't have faith in them being correct. If you did, you wouldn't be on this site :). Getting feedback/criticism on your work is a normal part of learning. Be sure to show your attempts at answering in the original question (not in the comments) by editing it $\endgroup$ Commented Apr 11, 2016 at 21:55
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    $\begingroup$ Important hints on Poisson processes: The probability that an event will occur has nothing to do with whether some other event has already occurred. You thus can start measuring time intervals from any particular starting time you wish. Sometimes it helps to start that measurement from the time that some other event occurred. $\endgroup$
    – EdM
    Commented Apr 11, 2016 at 22:00

1 Answer 1


To develop intuition for the Poisson distribution, it helps to consider the special case where there is an average of exactly 1 event per unit time. Use the Poisson distribution directly rather than thinking about exponential distributions of inter-arrival times. If you plug in to the formula, you will find that the probability of no event in 1 unit of time is $1/e\approx 0.37$. So the probability of at least 1 event in 1 unit of time is approximately 1-0.37, or 0.63.

In this special case, the probability of having exactly 1 event in 1 unit of time is also $1/e$. The probability of having exactly 2 events in 1 unit of time is one half of the probability of having exactly 1 event, the probability of having exactly 3 events in 1 unit of time is one-third of the probability of having exactly 2 events, and so on.

Some people find it useful to re-define time units so that there is exactly 1 event on average per re-defined time unit. So for part (a), you could define a "train-time unit" (TTU) as the 2.4 minutes that correspond to 25 trains per hour. Then the question about 6 minutes becomes one about 2.5 TTU.

For part (a), the probability of no train in any 1 TTU is about 0.37. Events and time periods are independent, so the probability of no trains at all in 2 TTU is about $0.37^2$, and $0.37^3$ for no trains in 3 TTU, etc. You will see that this approach with 2.5 TTU gives an answer like the one you got for part (a) with the exponential distribution, except perhaps for some rounding difference.

For part (b), check your calculations. It's easy to miss a minus sign, as I think you have done somewhere in your calculation. For a close estimate to check your work, note that a 40% chance of no train is very close to the $1/e$ chance of no event in 1 time unit if there is on average 1 event per time unit. (My guess is that this approximation was how your instructor expected you to approach this problem, although the exponential distribution of inter-arrival times also works.)

For parts (c) and (d), you are now working with a situation where there is on average 1 event per hour. The hints in the first 2 paragraphs here should suffice.

For looking forward to tests or other applications, note that instead of using events per unit time you can use the same approach for events per unit length on a line, per unit area (as famously represented in the novel Gravity's Rainbow), or, as my laboratory is presently doing, events per unit volume, where dispensing cells from a liquid suspension into multiple wells, at an average of one cell per well, means about 37% empty wells, 37% wells with exactly 1 cell, and 26% of wells with more than one.

Also, don't forget that all this is based on knowing that the distributions really are Poisson. In the real world, if trains are trying to meet a fixed schedule, for example, then they won't be.

  • $\begingroup$ Thanks very much for your help, I managed to get part c) and d) done quite easily, I think in my haste and frustration I'd not really looked at them properly to see they were much more straightforward, at least for me. I expected as much with the second question, I have reviewed my workings but am yet to see any glaringly obvious errors, though I'm sure it's there somewhere. I appreciate your help very much and the link you suggested, it cleared up at least the first part for me and has improved my overall understanding. $\endgroup$
    – user111815
    Commented Apr 12, 2016 at 15:20

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