Example 6.24. Five televisions have been put on subscription service. It is known that for a group of five televisions, the expected number of failures per year is one. If the televisions have the same probability of working without failure, what is the probability that at least one repair will be needed within a year?
Solution. The random variable is distributed according to the binomial law, since the trials are independent and the probability $P(A)=$ const. Since the expected number of failed televisions is one, the expected number of good ones is:

$$
M\left(X_{A}\right)=n-M\left(X_{\bar{A}}\right)=5-1=4 .
$$

The parameter $P$ is determined from the formula for the expected value, $M(X)=n \cdot p$, i.e., $P=\frac{4}{5}=0.8$. "At least one repair will be required" corresponds to the probability $P_{5}(n5)=1-0.8^{5}=0.67
$$

The problem can also be solved in another way. Since the probability of repairing any television $q=0.2$, and the probability of flawless operation $P=0.8$, the probability of at least one repair out of five televisions is:

$$
P(A)=1-p^{5}=1-0.8^{5}=0.67 .
$$