Example 6.8. When the voltage increases, a break in the electrical circuit may occur due to the failure of one of three elements connected in series; the probabilities of failure of the elements are 0.2; 0.3; 0.4, respectively. Determine the probability that the circuit will not break.
Solution. Let events $A_{1}, A_{2}, A_{3}$ denote the failure of the first, second, and third elements, respectively. Their probabilities are given as: $P\left(A_{1}\right)=0.2$; $P\left(A_{2}\right)=0.3$; $P\left(A_{3}\right)=0.4$. Then the probabilities of the opposite events $\bar{A}_{1}, \bar{A}_{2}, \bar{A}_{3}$ (the first, second, and third elements did not fail) are:

$$
P\left(\overline{A_{1}}\right)=1-P\left(A_{1}\right)=0.8; P\left(A_{2}\right)=0.7; P\left(A_{3}\right)=0.6.
$$

The event $A$, which consists of the circuit not being broken, is the conjunction of independent events $\bar{A}_{1}, \bar{A}_{2}, \bar{A}_{3}: A=\bar{A}_{1} \cdot \bar{A}_{2} \cdot \bar{A}_{3}$. Therefore, by formula (5) we get:

$$
P(A)=P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right) \cdot P\left(\bar{A}_{3}\right)=0.8 \cdot 0.7 \cdot 0.6=0.336
$$