Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Customers line up to be serviced according to a Poisson process at an average rate of five per hour. If the time it takes to serve one customer is a continuous uniform random variable on $[0,4]$, independent of arrival times, what is the probability that the second customer who arrives will have to wait to be served?

I'm really rather lost I think. I'm saying the time $T_n$ that the $n$th customer arrives is a sum of exponential variables with parameter 5, which ends up being a Gamma distribution with parameters $n, 5$. The time that it takes to serve the $n$th customer is $S\sim U(0,4)$. I figure that the second customer waits whenever $T_1+S>T_2$, so that's the probability that I'm looking for. The problem is, I don't know how to find that probability. I tried taking the convolution of $S$ and $T_1$ to get the distribution function of $T_1 +S$ but something seems to be going wrong with that:

$$F_{T_1+S}(a)=\int_{-\infty}^{\infty}F_{T_{1}}(a-y)f_{S}(y)dy=\frac{1}{4}\int_{0}^{4}(1-e^{-5(a-y)})dy \\ =\frac{1}{4}(4-\frac{1}{5}e^{-5a}[e^{20}-1])$$ which is clearly not a CDF (it's negative between $0$ and $4$), so I must have done that wrong. But even if I figure out what I did wrong there, I wouldn't know what to do next.

share|cite|improve this question
Hint: $T_2=T_1+R$ with $R$ independent of $T_1$ and of known distribution hence $[T_1+S\gt T_2]=[S\gt R]$. – Did Aug 13 '12 at 7:52
@did ahh, R is just exp(5), yes? So I'm looking for P(R>exp(5)) – crf Aug 13 '12 at 7:58
up vote 1 down vote accepted

Hint: $T_2=T_1+R$ with $R$ independent of $T_1$ and $S$, and $[T_1+S\gt T_2]=[S\gt R]$.

Sub-hint: $\mathrm P(R\gt S)=\int\limits_0^4\mathrm P(R\gt s)\,\frac{\mathrm ds}4$ and $\mathrm P(R\gt s)=\mathrm e^{-5s}$ for every $s\geqslant0$.

share|cite|improve this answer
I don't know why this particular problem has been causing me so much trouble but I think I've got it now. I'm thinking of this as $P(S>R)=P(0<R<S)$ and then if this were a discrete case, that would be something like $\sum_0^4 P(S=s)P(R<s)$ so $\int_0^4 f_S(s)F_R(s)ds$ ought to get me the right thing, which is $\int_{0}^{4}\frac{1}{4}(1-e^{-5s})ds$. The only thing is this reasoning isn't totally right because it's not discrete and I'm not looking at $P(S=s)$. Am I making a good analogy and how can I make this rigorous? Is this better suited to a different question? – crf Aug 13 '12 at 19:23
As soon as $S$ is continuous with density $f_S$ and $R$ is independent of $S$, the formula $P(R\lt S)=\int f_S(s)F_R(s)ds$ is entirely rigorous. – Did Aug 13 '12 at 20:37
:) thank you. I'll have to go over proving that, but it makes sense to me. – crf Aug 13 '12 at 20:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.