Example 6.25. A clerk accepts on average two applications per hour. It is assumed that the flow of applications is a simple one. What is the probability of receiving four applications in four hours?
Solution. For the simplest event flow, we can use the Poisson distribution, which will have the form:

$$
P_{n}(m)=\frac{(\lambda t)^{m}}{m!} e^{-\lambda t},
$$

where, according to the condition, $\lambda=2 ; t=4$ and thus $\lambda t=8$, and $m$ is the number of requests.

The probability of the occurrence of at least four events will be considered as:

$$
\begin{aligned}
& P_{n}(m \geq 4)=1-[P(0)+P(1)+P(2)+P(3)]= \\
& \quad 1-\frac{8^{0}}{0!} e^{-8}+\frac{8}{1!} e^{-8}+\frac{8^{2}}{2!} e^{-8}+\frac{8^{3}}{3!} e^{-8}= \\
& =1-e^{-8} \left(1+8+32+\frac{256}{3}\right)=1-0.0424=0.9576
\end{aligned}
$$