Poisson formula

Radioactive decay is a random process, of course. If we observe decays over an interval of time $dt$ that is very short compared with the lifetime, the number of atoms stays constant to a good approximation. The number of decays over that period of time, on average, should be

\begin{displaymath}
d = \frac{dF}{dt}dt = N(0)\frac{dt}{\tau}\exp(-t/\tau)
= \frac{dt}{\tau} N(0)\, ,
\end{displaymath} (6)

where we are approximating $\exp(-t/\tau) \approx 1$. In practice, because of fluctuations, there may be more or fewer decays. The probability of observing $k$ decays is given by the Poisson distribution:
\begin{displaymath}
P(k) = \frac{d^k}{k!} \exp(-d)
\end{displaymath} (7)

where $k!$ is the factorial $k(k-1)(k-2)\ldots{}1$. If we make $M$ measurements of the decay over a total time period $M dt$, then the number of intervals $dt$ in which exactly $k$ decays occur is given by
\begin{displaymath}
N(k) = M P(k) = M \frac{d^k}{k!} \exp(-d)   .
\end{displaymath} (8)

Because of statistical fluctuations, this statement is true only on average.