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Consider this problem:

The number of telephone calls made to an exchange is Poisson distributed with a mean of 6 calls per hour. Find the probability that a call will be made within 10 minutes given that no call was made in the past 15 minutes.


My question:

I know the 15 minutes interval mentioned does not have any effect since the Poisson distribution has a lack of memory property. And I know the answer is the probability that the interval does not exceed 10 minutes : $1 - e^{-6\frac{10}{60}}$.

But can't we also say that the answer is the probability that exactly 1 call was made within an interval of 10 minutes: $P(1 call) = e^{-1} . 1^1 / 1! = e^{-1} $ ?

And if not, then how can get the probability that exactly 3 calls (for instance) are made within the next 10 minutes, given that no calls were made within the past 15 minutes ?

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Let us call $X$ the number of calls in the next $10$ minutes. You say that by the lack of memory property, then this does not depend on what has happened in the past $15$ minutes, and this is correct.

Additionally, the distribution of $X$ is well known: $$ X\sim\mathrm{Poiss}(1). $$ Specifically, $$ P(X=k)=\frac{1^k}{k!}e^{-1}=\frac{1}{k!}e^{-1}, $$ for every $k\in\mathbb N$. Hence, you are correct in saying that the probability that there is at least a call is $$ P(X\ge1)=1-P(X=0)=1-e^{-1}. $$ Note that this is the probability that there is at least one call, and not exactly one call. This is why you do not get the same result as computing $P(X=1)$. To get the probability that $3$ calls are made, $$ P(X=3)=\frac16e^{-1}. $$ It may also be helpful to note that $$ P(X\ge1)=\sum_{k=1}^\infty P(X=k)=P(X=1)+P(X=2)+P(X=3)+\dots. $$

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  • $\begingroup$ Please replace $\mathrm{Poiss}(e^{-1})$ by $\mathrm{Poiss}(1)$. $\endgroup$ – Did May 17 '14 at 11:39
  • $\begingroup$ @Did, I find it impressive how closely you read every post. Thank you, I have edited the post. $\endgroup$ – Ian May 17 '14 at 11:47

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