I'm trying to derive a probability based on a Poisson process. The problem is as follows:
An event fails on average 5% of the time - that over 100 times that it is run it fails 5 times on average.
From this we can say things like the probability it will fail 10% of the time is 1.81% - fairly simple.
Now in a scenario where the the event is only carried out 5 times (not 100) and no failures are seen - what is the probability of that event?
Specifically is it valid to apply the Poisson rule to 1/20th of the mean?
eg: $Pr(exactly\ zero\ failures\ given\ a\ mean\ of\ 0.25) = (e^{-0.25})(0.25^{0})/0! = 77\%$
A Poisson process is a stochastic process over continuous time, while you seem to be describing a process with discrete attempts.
If failures are independent, each with probability $p=0.05$, then the probability of no failures out of five attempts is $$(1-0.05)^5 \approx 0.77378.$$ Your $e^{-0.25} \dfrac{0.25^0}{0!}\approx 0.77880$ is close, but not quite the same: in fact you are using the Poisson approximation to the binomial distribution.