Example 14. A worker services four identical machines. The probability that any machine will require the worker's attention during an hour is 0.6. Assuming that machine malfunctions are independent, find the probability that during an hour: a) all four machines; b) no machine; c) at least one machine.
Solution. Let $A_{1}, A_{2}, A_{3}, A_{4}$ be the events that the first, second, third, and fourth machines, respectively, will require the worker's attention within an hour. According to the multiplication theorem of probabilities of independent events, the probability that all machines will require the worker's attention within an hour, i.e., that events $A_{1}$, $A_{2}$, $A_{3}$, and $A_{4}$ will occur, is expressed by formula (1.8.13) for $n=4$:

$$
P\left(A_{1} A_{2} A_{3} A_{4}\right)=P\left(A_{1}\right) P\left(A_{2}\right) P\left(A_{3}\right) P\left(A_{4}\right)=0.6 \cdot 0.6 \cdot 0.6 \cdot 0.6=0.1296
$$

The probability that a machine (any one) will not require the worker's attention within an hour can be found using the rule for calculating the probability of the opposite event:

$$
P\left(\bar{A}_{1}\right)=P\left(\bar{A}_{2}\right)=P\left(\bar{A}_{3}\right)=P\left(\bar{A}_{4}\right)=1-0.6=0.4.
$$

Therefore, the probability of event $B$, which consists in none of the machines requiring the worker's attention within an hour, i.e., that events $\overline{A_{1}}$, $\bar{A}_{2}$, $\bar{A}_{3}$, and $\bar{A}_{4}$ will occur, is also expressed by formula (1.8.13) for $n=4$:

$$
\begin{aligned}
& P(B)=P\left(\bar{A}_{1} \bar{A}_{2} \bar{A}_{3} \bar{A}_{4}\right)=P\left(\bar{A}_{1}\right) P\left(\bar{A}_{2}\right) P\left(\bar{A}_{3}\right) P\left(\bar{A}_{4}\right)= \\
& =0.4 \cdot 0.4 \cdot 0.4 \cdot 0.4=0.0256
\end{aligned}
$$

The event that at least one of the four machines will require the worker's attention within an hour and event $B$ are opposite events. Since $P(B)=0.0256$, then

$$
P(\bar{B})=1-P(B)=1-0.0256=0.9744
$$