I am doing some problems related with the Poisson Process and i have a doubt on one of them. The problem is stated as follows:

A doctor works in an emergency room. The emergencies arrive according a Poisson Process with a rate of $\lambda =0.5$ emergencies per hour. The question is: When the first patient arrived, the doctor took care of him and spent 15 minutes. What is the probability that the doctor will find any patient waiting after finishing with the first one? And exactly with two patients waiting?

As the Poisson Process is memoryless, the time the doctor spent with the first patient does not affect the other ones. But im not sure on how to get the probabilities given the 15 minutes.

Would it be:


$P[N(0.25)=3] = \frac{(0.5*0.25)^3 e^{-0.5 * 0.25}}{3!}$

  • $\begingroup$ Hint: Poisson processes are memoryless. $\endgroup$ – dsaxton Mar 7 '16 at 22:52
  • $\begingroup$ please add the self-study tag to your question and read the wiki for guidelines on posting questions related to homework $\endgroup$ – Marquis de Carabas Mar 7 '16 at 22:56
  • $\begingroup$ If the process is memoryless, then does it means that the solution would be to calculate the probability of having 2 or more arrivals? $\endgroup$ – Daryl K Mar 7 '16 at 23:02
  • $\begingroup$ Here it means that what will happen in the next 15 minutes has the same distribution as any other 15 minute increment, regardless of what has already happened. $\endgroup$ – dsaxton Mar 7 '16 at 23:25
  • $\begingroup$ Just updated the question, as 15min = .25 hours $\endgroup$ – Daryl K Mar 7 '16 at 23:30

The first question asks what the probability is that the doctor will find any patients waiting. One way to do this is to find the probability that he will find no patients waiting. In other words, what is the probability that the next interarrival time is greater than 15 minutes? Since you have a Poisson process, you know the distribution of interarrival times, and hence you can get $P[T_2 > 15]$, which is equivalent to $P[\mbox{no patients waiting}]$. Then, to get the probability of at least one patient waiting is trivial.

For the second part, you seem to be on the right track with the second formula. Remember that $N(t) - N(s) \sim Poisson(\lambda(t-s))$. Here, your $s$ is the time of the first arrival, and $t$ is fifteen minutes after this first arrival, and you're interested in if $N(t) - N(s) = 2$. The distribution of this relies on $\lambda$, which you know, and $t - s$ which you also know.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.