# Find the probability that the second customer to arrive has to wait to be served if arrival time is exponential and serving time is uniform

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.

• 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
Commented Aug 13, 2012 at 7:52
• @did ahh, R is just exp(5), yes? So I'm looking for P(R>exp(5))
– crf
Commented Aug 13, 2012 at 7:58

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 $s\geqslant0$.

• 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
Commented Aug 13, 2012 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
Commented Aug 13, 2012 at 20:37
• :) thank you. I'll have to go over proving that, but it makes sense to me.
– crf
Commented Aug 13, 2012 at 20:40