How Much Time Must the Shopkeeper Wait? -- Exponential Distribution In a store, the distance between customer arrivals follows an exponential distribution with a parameter of 8 minutes. The second seller starts his shift at 10:30 while the last customer entered the store at 10:28. How much time should the second seller wait for the first customer to enter in his shift?
Can Anyone Help Me ?
 A: The waiting time after the second seller's arrival until his/her first customer arrives averages 8 minutes. So judged as of 10:30, one would expect the average arrival time of that next customer to be at 10:38.
The fact that a previous customer arrived before the second seller's arrival is irrelevant because of the no-memory property of the exponential distribution. Additionally:

*

*At the time the previous customer arrived, the expected waiting
time for the next arriving customer would have been 8 min., but that is
not the question being asked here; the waiting time starts afresh when
the second seller arrives.


*Moreover, if it happens that the no customer arrives
within 10 min. after the second seller's arrival, then as of 10:40, the expected arrival time for the next customer is again 8 min. later--at 10:48 on average.
The exponential no-memory property may seem counter-intuitive. One can argue whether an exponential model is the correct one to use for customer arrivals. However, once you say to use the exponential model, you're
automatically agreeing to its no-memory property. In practice, exponential models have proved useful for modeling customer arrivals in many settings.
Addendum. Maybe this simulation helps.
set.seed(2021)
x = rexp(10^6, 1/8)  # a million exponential waiting times

mean(x)
[1] 7.997945         # aprx E(X) = 8
mean(x[x > 2]-2)
[1] 7.997422         # aprx E(X - 2| X > 2) = 8

                     # add'l time given no event by time 2
mean(x > 8)        
[1] 0.367389         # aprx P(X > 8) = 0.3678794
1 - pexp(8, 1/8)   
[1] 0.3678794        # exact
mean(x[x > 2] > 10)
[1] 0.3667555        # aprx P(X > 10 | X > 2) = 0.3678794

Proof:  Let $W\sim\mathsf{Expo}(\mu).$ Then
$$P(W > t+s\, |\,W > s) = \frac{P(W > t+s, W > s)}{P(W>s)}$$
$$= \frac{P(W > t+s)}{P(W>s)} = \frac{e^{-(t+s)/\mu}}{ e^{-s/\mu}}
= e^{-t/\mu} = P(W > t).$$
