# How to solve this problem with Poisson distribution

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 ?

• $6$ customers in $1$ hour $\not \equiv$ $1$ customer in $10$ minutes. Commented Mar 21 at 1:35

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