A Poisson distribution is used to calculate the probability that $K$ events occur over time interval $\tau$, assuming a rate $\lambda$.
This is done with the following formula:
$ P = \frac{(\lambda\tau)^k}{k!}*e^{-\lambda\tau}$
This all nice and well. What I would like to know is, how many events have happened at time-point $\tau$, given a probability $P$. I started rearranging the formula and ended up here:
$ \frac{P}{e^{-\lambda\tau}}*k! = (\lambda\tau)^k$
Assuming the following values:
$ P = 0.8$
$ \lambda = 1.6 $
$ \tau = 15 $
The function would be:
$ \frac{0.8}{3.78*10^{-11}}*k! = 24^k $
$2.12*10^{10}*k! = 24^k$
How would I go about finding $k$?
The solution may be an approximation. As long as $ P \geq 0.8 $ the solution would satisfy.