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Problem:

A store owner observes that there are $3$ (in average) customers visiting the store per hour. He wants to find the probability that there are at least $1$ customer visiting his store in $10$ minute, using Poisson distribution.

As I read in a probability book, the solution is to use the Poisson distribution with rate $\lambda = 3/6 = 1/2$ per $10$ minutes (since there are $3$ customers visiting the store per hour). But why don't we use the Poisson distribution with rate $\lambda = 3$ per hour and re-state the question as "What is the probability that there are at least $6$ customers visiting the store within an hour".

I've tried both ways and they gave different answers. Some may argue that having at least $1$ customer visiting the store within $10$ minutes isn't equivalent to having at least $6$ customers visiting the store within an hour. But I think having $3$ customers visiting the store, in average, per hour isn't equivalent to having $1/2$ customer visiting the store, in average, per $10$ minutes.

Can anyone explain the solution of the book for me ?

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The correct way to state the problem while using the $\lambda = 3\ \text{hr}^{-1}$ rate is to ask what is the probability that at least $1$ customer arrives within $t = \frac{1}{6}\ \text{hr}$.

Because of the form of the probability distribution for the Poisson process $\left(\frac{(\lambda t)^n e^{-\lambda t}}{n!}\right)$, as long as the $\lambda t$ term and $n$ terms don't change, you'll have the same probability distribution. But trying to change $n$ while keeping $n/t$ constant will clearly change the probability distribution.

As a freebie, the probability that there is at least $1$ customer in the shop within ten minutes is:

$$\mathbb{P}[N(1/3)\geq1] = 1 - \mathbb{P}[N(1/3)=0] = 1 - e^{-1/2}$$

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