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Suppose we have a helpdesk with tickets arriving at a rate of three per min. Tickets arrival follow a Poisson distribution. How someone can calculate:

a. The probability of the time between the first and the second or the time between the second and the third ticket arrival exceed two mins.
b. The probability of the time interval between two successive ticket arrivals exceeds 2.5 mins, taken into account that there is already passed 1.5 minute without any ticket.

I understand that for the first (a) the $\lambda$ of the Poisson dist is 3 tickets per min, taken into account one min as a unit, while for (b) $\lambda$ is 1.5 and the unit is half minute. But how can I apply the Poisson probability calculation for two successive events? For (b), since Poisson is memoryless, does it matter that 1.5 minute has passed?

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You should know that if the number of arrivals in every time interval has a Poisson distribution then the inter-arrival times are independent identically distributed with an exponential distribution with mean $1/\lambda$ where $\lambda$ is the mean number of arrivals per unit time. This is also the case at any time of the waiting time until the next arrival.

In this case $\lambda=3$ per minute. Now take it from there...

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  • $\begingroup$ why change the $\lambda$? If our unit is one minute then the Poisson parameter $\lambda$ is three, right? $\endgroup$ Dec 19, 2014 at 22:10
  • $\begingroup$ A combination of typo and misreading, yes 3 per minute with a mean inter-arrival time of 1/3 of a minute. Fixed $\endgroup$ Dec 19, 2014 at 22:18
  • $\begingroup$ So, from what I understand I have to work with exponential distribution (and not Poisson) with the same $\lambda = 3$ parameter... $\endgroup$ Dec 20, 2014 at 14:20
  • $\begingroup$ Yes that's right,the question is about waiting times and intervals between events so you will be using the exponential distribution. $\endgroup$ Dec 20, 2014 at 17:23
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The probability that the time until the next arrival is more than $2$ minutes is the probability that the number of arrivals during those $2$ minutes is $0.$ Since the average number of arrivals in $2$ minutes is $6,$ that probability is $\dfrac{6^0 e^{-6}}{0!}= e^{-6}.$

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