Example 11. Independent trials are conducted, in each of which event $A$ can occur with a probability of 0.001. What is the probability that in 2000 trials, event $A$ will occur no less than two and no more than four times.
Solution. From the condition of the problem, it follows that $n=2000, p=0.001$, $a=n p=2000 \cdot 0.001=2, 2 \leq k \leq 4$. Therefore,

$$
P_{n}(2 \leq k \leq 4)=P_{n}(2)+P_{n}(3)+P_{n}(4)=\frac{2^{2}}{2!} e^{-2}+\frac{2^{3}}{3!} e^{-2}+\frac{2^{4}}{4!} e^{-2} \approx 0.541
$$