The exact question is not quite the same (it was hard to put in a concise title) but has the same principle:
The time between customers has an exponential distribution. What is the probability that $T_1$, the time between opening the shop and the first customer, is larger than $k$ times $T_2$, the time between the first and second customer?
In other words, what is $P(T_1>kT_2)$?
This is what I have so far:
- $T_1 > kT_2 \iff T_1+T_2 > (k+1)T_2$
- $T_1$ and $T_2$ have the exponential distribution $f_{T_i}(t)=\lambda e^{-\lambda t}$
- $T := T_1+T_2$ has gamma distribution $f_{T}(t) = \frac{\lambda^2}{2}te^{-\lambda t}$.
I don't know how to proceed now. I don't know how to convert $P(T_1>kT_2)$ to something where I can use the distribution functions.
Thanks in advance.
By definition of the exponential distribution, $$ \mathbb P(T_1>t)=\mathbb P(T_2>t)=e^{-\lambda t},\qquad t\geq 0. $$ Thus, $$ \mathbb P(T_1>kT_2)=\mathbb E\bigl[\mathbb P(T_1>kT_2\mid T_2)\bigr]=\mathbb Ee^{-\lambda k T_2}=\int_0^{\infty}e^{-\lambda k x}\cdot \lambda e^{-\lambda x} dx, $$ where in the last line we are integrating against the distribution function. Performing the integral yields $$ \mathbb P(T_1>kT_2)=\left[\frac{e^{-\lambda(k+1)x}}{k+1}\right]_0^{\infty}=\frac{1}{k+1}, $$ as desired.