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Question: Your friend owns a hardware store. On Saturdays, the store is open from noon to 7:00 p.m. During these hours, customers arrive according to a Poisson process at a rate of λ customers per hour. One Saturday, after your friend closes his store, you ask him how many customers he had that day. “Just one!” your friend replies. Without asking him when the customer came, how is that customer’s arrival time distributed under the information you now have?

I know that a Poisson process has a pmf given by: $P_X(k)=\frac{e^{-\mu}\mu^k}{k!}$. I find the wording of the question a bit confusing, so I'm not sure how to tackle this.

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Let the arrival time of the customer be $T$ hours after noon. Then $T$ will be uniformly distributed on the half-open interval $(0, 7]$. This follows from calculating the probability that $T \leq t$, given the information that $X = 1$, where $X$ is the number of customers in the given $7$ hour period. $$P(T \leq t \mid X = 1) = \frac{P(T \leq t \cap X = 1)}{P(X = 1)}$$ The probability $P(T \leq t \cap X = 1)$ is the probability that $1$ customer comes in the first $t$ hours, and that no customers come in the next $7 - t$ hours. Using properties of the Poisson process, $$P(T \leq t \mid X = 1) = \frac{e^{-\lambda t} (\lambda t) \cdot e^{-\lambda(7 - t)}}{7 \lambda e^{-7 \lambda}}$$ $$= \frac{t}{7}.$$ This is independent of the value of the value of $\lambda$.

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