In the formula $ e^{-\lambda}\frac{\lambda^k}{k!} $ $\lambda$ is the estimated rate times the time interval, that would be the expected number of occurrences during the given time interval. In this case, the average is $5$ per second and the time period is $1$ second, so $\lambda=5$.
Say we wanted to compute the probability that only $2$ events happen in the next $2$ seconds. Then $\lambda=10$ ($2$ seconds times $5$ per second) and so we get $ e^{-10}\frac{10^2}{2!}=0.00227 $
Explanation:
Let $\lambda$ be the expected number of events in a given time interval. Then the Poisson assumption is that the probability that an event will happen in a given $\frac1n$ slice of that time interval is $\frac\lambda n$. Let's compute the probability that $k$ events happen in that same time interval. The chance that more than one event happens in one $\frac1n$ slice of the time interval is so small that we can ignore it. The binomial estimate says that the probability that we have exactly $k$ events is $ \binom{n}{k}\left(1-\frac \lambda n\right)^{n-k}\left(\frac \lambda n\right)^k $ That is, $\left(1-\frac \lambda n\right)^{n-k}\left(\frac \lambda n\right)^k$ is the probability of $n-k$ non-events and $k$ events. Furthermore, there are $\binom{n}{k}$ different ways to arrange the $k$ events and $n-k$ non-events.
Now, let $n\to\infty$: $ \begin{align} &\lim_{n\to\infty}\binom{n}{k}\left(1-\frac \lambda n\right)^{n-k}\left(\frac \lambda n\right)^k\\ &=\lim_{n\to\infty}\frac nn\frac{n-1}n\frac{n-2}n\dots\frac{n-k+1}{n} \left(1-\frac \lambda n\right)^{n-k}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\frac{\lambda^k}{k!} \end{align} $