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 ?
Best Answer
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.