Customers arrive at a neighborhood grocery store in a Poisson process

Question

Customers arrive at a neighborhood grocery store in a Poisson process with a rate of 5 customers per hour, starting at 8:00 a.m. Upon arrival, a customer remains for Exponentially distributed time with a parameter $\lambda$ = 3 in hours until he finishes his business, regardless of time , his arrival, or the other customers and the number of customers in the grocery store. Given that the grocery store closes at 17:00 and only one customer arrived after 16:00. What The probability that he will finish his business before the grocery store closes?

now here's what I did , I said let Y be the time that it takes him to arrive after 4 o'clock, and T is the time that takes him to leave the store, and we have to calculate p($Y+T \leq 1$) p($Y \leq 1-T$) = $\int f(Y)dt$ , I set the bounds for T is $1\geq T\geq 0 $

f(Y) = $\lambda$$e^{-x\lambda}$= $3e^{-3(1-t)}$ so $\int_0^1 f(t)dt$ = 1-$e^{-3}$

is this correct ?

Answer 1

Conditioned on there being one arrival in the last hour, that arrival is uniformly distributed over that time period. So Y∼Unif(0,1) and T∼Exp(3), and P(Y+T⩽1)=P(T⩽1−Y)=∫10∫1−y03e−3t dt dy=13(2+e−3). (Note that 1−Y∼Unif(0,1) as well.) Your computation is the probability that the customer leaves the store given that he arrives exactly one hour before closing, which is not what was asked.

Answer 2

Conditioned on there being one arrival in the last hour, that arrival is uniformly distributed over that time period. So $Y\sim\mathsf{Unif}(0,1)$ and $T\sim\mathsf{Exp}(3)$, and $$ \mathbb P(Y+T\leqslant 1) =\mathbb P(T\leqslant 1-Y)= \int_0^1\int_0^{1-y}3e^{-3t}\ \mathsf dt\ \mathsf dy = \frac13(2+e^{-3}). $$ (Note that $1-Y\sim\mathsf{Unif}(0,1)$ as well.) Your computation is the probability that the customer leaves the store given that he arrives exactly one hour before closing, which is not what was asked.