Arrival of customers at shop ~ Poisson. What is the probability that a single customer arrive in 1/4 of the time?

Question

problem statement: Cutomers arrive at a store according to a Poisson distribution with $\lambda = 1$. Suppose we observe customer arrivals from time t = 0 to t = T and find only 1 customer arrived during this period. What is the probability that this single customer arrived between t = 0 and t = T/4?

The possible answers are:

Since it is just a single customer that arrives in that time period, and assuming this person arrives uniformly, isn't the answer simply 1/4?

How does $\lambda$ come in to play here? And what does $\lambda = 1$ mean? 1 person per minute, 1 per per hour, 1 person per T?

Answer 1

Since it is just a single customer that arrives in that time period, and assuming this person arrives uniformly, isn't the answer simply 1/4?

Yes, however, the property that determines this is the independence of arrivals, not that they arrive uniformly; they don’t or at least the don’t necessarily.

How does λ come in to play here?

It doesn’t.

And what does λ=1 mean? 1 person per minute, 1 per per hour, 1 person per T?

1 person per unit period in whatever period λ is expressed in. The question doesn’t tell us.

Answer 2

$\lambda=1$ means you expect one customer in the time period of interest. $T$ could be a different period, so you could measure per hour and expect one per hour, then observe for two hours and see only one customer in that span. There is no linkage in the question between $T$ and $\frac 1\lambda$, but they shouldn't be far different.

Yes, by symmetry the chance that customer came in the first quarter is $\frac 14$. Divide the period $T$ into quarters. You have a customer arriving in one of them and no customers in the other three. There is no reason to prefer one quarter over another.