The number of people queuing at a supermarket check-out during some time interval of length $t$ minutes follows a Poisson process with mean $\displaystyle\frac{t}{2}$. The variable $Y$ denotes the time in minutes from when the supermarket opens until the first customer arrives at the check-out.
(a) Find the p.d.f. of the variable $Y$.
(b) On a certain day the person running the check-out arrives 5 minutes late (5 minutes after opening), but no customers have yet arrived at the check-out. Find the distribution of the time that the check-out attendant will have to wait for the first customer to arrive at the check-out.
I found the p.d.f. in part (a) to be, in terms of $t$
\begin{array}{@{}c@{\quad}l@{}} 0.5e^{-0.5t} & \text{if $t$ is positive}\\ 0 & \text{otherwise}\\ \end{array} So my question is whether the answer to (b) is
\begin{array}{@{}c@{\quad}l@{}} 0.5e^{-0.5(5-t)} & \text{if $t>5$}\\ 0 & \text{otherwise}\\ \end{array}
However, I'm not sure whether this is correct. If this is an incorrect answer, could someone please explain why?
Any help much appreciated, as always.
Let's do some conditional probabilities!
We want to compute the conditional probability $$ P(Y-5 = t | Y > 5)=\frac{P(Y=t+5)}{P(Y>5)} = \frac{\frac{e^{-\frac{t+5}{2}}}{2}}{1-\int_0^5\frac{e^{-\frac{t}{2}}}{2}dt} = \frac{\frac{e^{-\frac{t+5}{2}}}{2}}{e^{5/2}} = \frac{e^{-\frac{t}{2}}}{2} $$
So the final distribution is exactly the same!
This is because the distribution was an exponential distribution, and exponential distributions are memoryless.