66. The device consists of two elements operating independently. The probability of the first element failing is 0.2; the probability of the second element failing is 0.3. Find the probability that: a) both elements will fail; b) both elements will work.
Solution. Let event $A$ be the failure of the first element, and event $B$ be the failure of the second element. These events are independent (by condition).

a) The simultaneous occurrence of $A$ and $B$ is the event $A B$. Therefore, $P(A B)=0.2 \cdot 0.3=0.06$.

6) If the first element is working, then the event $\bar{A}$ (the opposite of event $A$ - the failure of this element) occurs; if the second element is working -- event $B$. Let's find the probabilities of events $\bar{A}$ and $\bar{B}$:

$$
P(\bar{A})=1-P(A)=1-0.2=0.8 ; P(\bar{B})=1-P(B)=1-0.3=0.7 .
$$

Then the event that both elements will be working is $\bar{A} \bar{B}$, and thus,

$$
P(\bar{A} \bar{B})=P(\bar{A}) P(\bar{B})=0.8 \cdot 0.7=0.56
$$