Example 15. Independent trials are conducted, in each of which event $A$ may occur with a probability of $p=0.01$. Find the probability that in 100 trials, event $A$ will occur respectively $1,2,3,4,5,6$ times, and will not occur at all.
Solution. When calculating the desired probabilities, we will use the recurrence formula

$$
P_{n}(k+1)=\frac{a}{k+1} P_{n}(k)
$$

which is derived as follows:

$$
P_{n}(k+1)=\frac{a^{k+1}}{(k+1)!} e^{-a}=\frac{a}{k+1} \cdot \frac{a^{k}}{k!} e^{-a}=\frac{a}{k+1} P_{n}(k)
$$

When $a=1$, this formula becomes

$$
P_{n}(k+1)=\frac{1}{k+1} P_{n}(k)
$$

If the probability $P_{n}(0)$ is known, then using formula (3.3.9), we can compute $P_{n}(1), P_{n}(2), P_{n}(3)$, and so on.

From the problem statement, it follows that $n=100, p=0.01$, therefore $a=n p=100 \cdot 0.01=1$. Using formula (3.3.1), we find

$$
P_{100}(0)=e^{-1}=0.3679 ; \quad P_{100}(1)=\frac{1^{1}}{1!} e^{-1}=e^{-1}=0.3679
$$

Using formula (3.3.9) sequentially, we obtain:

$$
\begin{aligned}
& P_{100}(2)=\frac{1}{2} P_{100}(1)=0.1839 ; \quad P_{100}(3)=\frac{1}{3} P_{100}(2)=0.0613 \\
& P_{100}(4)=\frac{1}{4} P_{100}(3)=0.0153 ; \quad P_{100}(5)=\frac{1}{5} P_{100}(4)=0.0031 \\
& P_{100}(6)=\frac{1}{6} P_{100}(5)=0.0005
\end{aligned}
$$