Example 9. The probability of manufacturing a non-standard part is $p=0.004$. Find the probability that among 1000 parts there will be 5 non-standard ones.
Solution. Here $n=1000, p=0.004, a=n p=1000 \cdot 0.004=4$. Using formula (3.3.1), we find

$$
P_{1000}(5)=\frac{4^{5}}{5!} e^{-4}=\frac{128}{15} \cdot \frac{1}{e^{4}}=8.5333 \cdot 0.0183=0.1562
$$

Example 10. A spinner operates 1000 spindles. The probability of a thread break on one spindle within 1 minute is 0.002. Find the probability that more than three spindles will break within 1 minute.

Solution. According to the conditions, we have: $n=1000, p=0.002$,

$$
\begin{aligned}
& a=n p=1000 \cdot 0.002=2, k>3 \\
& P_{n}(k>3)=1-P_{n}(k \leq 3)=1-P_{n}(0)-P_{n}(1)-P_{n}(2)-P_{n}(3)
\end{aligned}
$$

Using formula (3.3.1), we find:

$$
\begin{aligned}
& P_{n}(k>3)=1-e^{-2} \frac{2^{0}}{0!}-e^{-2} \frac{2}{1!}-e^{-2} \frac{2^{2}}{2!}-e^{-2} \frac{2^{3}}{3!}= \\
& =1-0.1353-0.2707-0.2707-0.1805=0.1428
\end{aligned}
$$