81. The probability that a lamp will remain defective after 1000 hours of operation is 0.2. What is the probability that out of five lamps, at least three will remain operational after 1000 hours of operation?
Solution. We will consider the burning of each lamp for 1000 hours as a separate experiment. Then we can say that 5 experiments have been conducted. We are interested in the events "3 out of 5 lamps are burning", "4 out of 5 lamps are burning", and "5 out of 5 lamps are burning", i.e., we can find the probability of each of these events using the Bernoulli formula, considering that $p=0.2$ and $q=0.8$:

$$
\begin{gathered}
A_{5,3}=C_{5}^{3} p^{3} q^{2}=\frac{5!}{3!2!} \cdot 0.2^{3} \cdot 0.8^{2}=0.0512 \\
A_{5,4}=C_{5}^{4} p^{4} q^{1}=\frac{5!}{4!} \cdot 0.2^{4} \cdot 0.8=0.0084 \\
A_{5,5}=C_{5}^{5} p^{5} q^{0}=\frac{5!}{5!} \cdot 0.2^{5} \cdot 1=0.00032
\end{gathered}
$$

Then the desired probability will be $0.0512+0.0084+0.00032=$ $=0.0579$.