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When a random variable follows a Poisson-process, with a known mean / variance, of course the probability of an event occurring at any time can be calculated using the Poisson's pmf, finding $Pr[N(t) = 1]$ (within time $t$, and occurring only once hence the $= 1$).

Though how would one go about knowing that an event has occurred, what was the probability for it to have happened within a given time frame?

For example, if I had a mean of receiving one phone-call per hour, I understand that the probability of receiving a single phone call within a specified span of 10-minutes is $$\frac{e^{-\frac{10}{60}}(\frac{10}{60})^1}{1!}=\frac{e^{-\frac{1}{6}}}6\approx0.141$$ Though if I add the specification that I did, absolutely, receive a phone call within that same hour as the 10-minute span above resides in, what is the probability of said certain phone call having been received in those 10 minutes?

I know that (being Poisson), each time range would have an equal probability, and in this case it would be a probability of 1/6 per 10-minute time span within the hour, in this case. Though how would this be communicated?

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Essentially, it sounds like you are saying that given $N(1\text{ hr}) = 1$, what is the probability that $N(\frac{10}{60}\text{ hr}) = 1$. This translate to $$P(N(10/60) = 1\mid N(1) = 1).$$

It will also be helpful to remember that disjoint blocks of time yield independent Poisson distributions and that $N(t+s)-N(s)\sim \operatorname{Poisson}(t\lambda)$.

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