Suppose that the times at which customers enter a store can be modelled as a Poisson process. Over a 10-minute period, 8 customers entered the store.

  1. How many customers would you expect to see enter the store in a two hours period?

  2. What is the probability that no customers enter the store in a two minutes period?

  3. What is the probability of waiting an additional 3 minutes for a customer to enter the store, if the last customer entering the store came n minutes ago?

I don't think I understand the Poisson process properly. Would the intensity, $\lambda = 0.8$, since we expect (theoretically) 0.8 customers every minute?

From that, I assume that the answer to (1) is $\lambda t = (0.8)(120) = 96$.

For (2), I think want $P(N_2 = 0)$, where $N_2 \sim PP(1.6)$. So I get $P(N_2 = 0) = \frac{1.6^0 e^{-1.6}}{0!} = 0.201$. Am I on the right track?

I am completely lost on (3). Could anyone perhaps provide me a tip to get started? Thanks!

  • $\begingroup$ For (3) you have memorylessness $\endgroup$
    – Henry
    Nov 17, 2014 at 8:46

1 Answer 1


To find the intensity, recall as you write, that the average number of customers in a time interval $[t_1,t_2]$ is $\lambda(t_2-t_1)$. Thus, if $t$ is given in minutes, $$ 10\lambda=8, $$ from which you deduce your answer $\lambda=0.8\,\mathrm{min}^{-1}$.

  1. Your answer is correct.

  2. Your answer is correct, since $N_2\sim\mathrm{Poiss}(2\lambda)$ (I'm not sure what your notation $PP(\cdot)$ is, it should reference the Poisson distribution, not the Poisson process).

  3. Expanding on Henry's tip, recall that the increments of the Poisson process are independent and exponentially distributed. So it is asked to compute $$ \mathbb P\left(X\ge n+3\mid X\ge n\right), $$ where $X\sim\mathrm{Exp}(\lambda)$. Do you see how to use the memorylessness property of the Poisson process here? (Or more specifically, the memorylessness of the exponential distribution.)

  • $\begingroup$ I'm not very familiar with the property, but I think this means that I am able to reduce that probability to $\mathbb{P}(X \geq 3)$, since what has happened before has no real effect up to this point. Is this correct? $\endgroup$ Nov 17, 2014 at 9:27
  • $\begingroup$ In terms of the exponential distribution, then, does this mean I want to find $\mathbb{P}(X \geq 3)$, where $X \sim Exp(0.8)$, giving me $e^{-(0.8)(3)} = 0.787$? $\endgroup$ Nov 17, 2014 at 9:32
  • $\begingroup$ Yes, everything you wrote is correct. The memorylessness property can be proved easily by writing $\mathbb P(X\ge n+3\mid X\ge n)=\mathbb P(X\ge n+3,X\ge n)/\mathbb(X\ge n)$. $\endgroup$
    – Ian
    Nov 17, 2014 at 9:49

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