Example 14. A telephone station serves 400 subscribers. For each subscriber, the probability that they will call the station within an hour is 0.01. Find the probabilities of the following events: "5 subscribers will call the station within an hour"; "no more than 4 subscribers will call the station within an hour"; "no fewer than 3 subscribers will call the station within an hour".
Solution. According to the condition $n=400, p=0.01$, then $a=n p=$ $400 \cdot 0.01=4$.

In accordance with formula (3.3.1) we find

$$
\begin{gathered}
P_{400}(5)=\frac{4^{5}}{5!} e^{-4} \approx 0.1563 ; \\
P_{400}(0 \leq k \leq 4)=P_{400}(0)+P_{400}(1)+P_{400}(2)+P_{400}(3)+P_{400}(4) \approx \\
\approx 0.0183+0.0733+0.1465+0.1954+0.1954=0.6289 \\
P_{400}(3 \leq k \leq 400)=1-P_{400}(0 \leq k \leq 2)=1-P_{400}(0)-P_{400}(1)-P_{400}(2)= \\
=1-0.0183-0.0733-0.1465=0.7619 .
\end{gathered}
$$