39. A batch of one hundred parts is subject to selective control. The condition for the rejection of the entire batch is the presence of at least one defective part among the four inspected. What is the probability that the batch will not be accepted if it contains $3 \%$ defective parts?
S o l u t i o n. a) Let's denote the events:

$A_{i}$ - the inspected part is defective ( $i=1,2,3,4$ );

$\bar{A}_{i}$ - the inspected part is not defective;

$A$ - the batch of parts will not be accepted (at least one defective part will be found among the four inspected);

$\bar{A}$ - the batch of parts will be accepted (no defective parts will be found among the four inspected).

The event $\bar{A}$ consists in the simultaneous occurrence of all events $\bar{A}_{i}$, therefore $\bar{A}=\bar{A}_{1} \bar{A}_{2} \bar{A}_{3} \bar{A}_{4}$. The events $\bar{A}_{i}$ are dependent; the probability of event $A$ will be found using the theorem of multiplication of probabilities of several dependent events:

$$
\begin{aligned}
& P(\bar{A})=P\left(\bar{A}_{1}\right) \cdot P_{\bar{A}_{1}}\left(\bar{A}_{2}\right) \cdot P_{\bar{A}_{1} \bar{A}_{2}}\left(\bar{A}_{3}\right) \cdot P_{\bar{A}_{1} \bar{A}_{2} \bar{A}_{3}}\left(\bar{A}_{4}\right)= \\
& =\frac{97}{100} \cdot \frac{96}{99} \cdot \frac{95}{98} \cdot \frac{94}{97} \approx 0.8836
\end{aligned}
$$

The required probability of event $A$ is

$$
P(A)=1-P(\bar{A})=1-\frac{97 \cdot 96 \cdot 95 \cdot 94}{100 \cdot 99 \cdot 98 \cdot 97} \approx 0.1164
$$