# Customers arrive at a neighborhood grocery store in a Poisson process

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

is this correct ?

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.