Example 6.7. In problem 6.6, determine the probability of at least one of the two machines operating without failure during time $t$ (event $B$).
## Solution.

The first method. Consider the opposite event $\bar{B}$, which means the failure of both machines during time $t$. Obviously, the event $\bar{B}$ is the conjunction of events $\bar{A}_{1}$ and $\bar{A}_{2}$ - the failures of the first and second machines, i.e., $\bar{B}=\bar{A}_{1} \bar{A}_{2}$. Since events $\bar{A}_{1}$ and $\bar{A}_{2}$ are independent, we have

$$
P(\bar{B})=P\left(\bar{A}_{1}\right) \cdot P\left(\bar{A}_{2}\right)=\left[1-P\left(A_{1}\right)\right]\left[1-P\left(A_{2}\right)\right]=0.1 \cdot 0.2=0.02
$$

From this,

$$
P(B)=1-P(\bar{B})=0.98
$$

The second method. Event $B$ occurs if one of the following three mutually exclusive events happens: either $A_{1} \cdot \bar{A}_{2}$ - the conjunction of events $A_{1}$ and $\bar{A}_{2}$ (the first machine works, the second does not), or $\bar{A}_{1} \cdot A_{2}$ - the conjunction of events $\bar{A}_{1}$ and $A_{2}$ (the first machine does not work, the second does), or $A_{1} \cdot A_{2}$ - the conjunction of events $A_{1}$ and $A_{2}$ (both machines work), i.e.,

$$
B=A_{1} \cdot \bar{A}_{2}+\overline{A_{1}} \cdot A_{2}+A_{1} \cdot A_{2}
$$

By formula (3), we get:

$$
P(B)=P\left(A_{1} \cdot \bar{A}_{2}\right)+P\left(\overline{A_{1}} \cdot A_{2}\right)+P\left(A_{1} \cdot A_{2}\right) .
$$

Since events $A_{1}$ and $A_{2}$, and therefore $A_{1}$ and $\bar{A}_{2}, \bar{A}_{1}$ and $A_{2}$ are independent, we have:

$$
\begin{aligned}
& P(B)=P\left(A_{1}\right) \cdot P\left(\bar{A}_{2}\right)+P\left(\overline{A_{1}}\right) \cdot P\left(A_{2}\right)+P\left(A_{1}\right) \cdot P\left(A_{2}\right)= \\
& =P\left(A_{1}\right)\left[1-P\left(A_{2}\right)\right]+\left[1-P\left(A_{1}\right)\right] P\left(A_{2}\right)+P\left(A_{1}\right) \cdot P\left(A_{2}\right)=0.98 .
\end{aligned}
$$