64. The probability of failure of each device during testing is 0.2. How many such devices need to be tested to assert with a probability of at least 0.9 that at least two devices will fail?
S o l u t i o n. Let the required number of devices be denoted by $n$. The event $A$, which may or may not occur in each trial, consists of a device failing. The probability $p=0.2$; the probability $q=0.8$.

Consider the events:

$B$ - at least two devices fail;

$\bar{B}$ - fewer than two devices fail.

The event $\bar{B}$ consists of either no device failing or one device failing. The probability of event $\bar{B}$ can be found using the Bernoulli formula and the theorem of addition of probabilities of mutually exclusive events:

$$
P(\bar{B})=P_{n}(0)+P_{n}(1)=p^{0} q^{n}+C_{n}^{1} p q^{n-1}=0.8^{n}+n \cdot 0.2 \cdot 0.8^{n-1}=0.8^{n}\left(1+\frac{n}{4}\right)
$$

According to formula (11)

$$
P(B)=1-P(\bar{B})=1-\frac{0.8^{n}(n+4)}{4}
$$

By the condition $P(B) \geq 0.9$. Therefore, to determine $n$, we need to solve the inequality:

$$
1-\frac{0.8^{n}(n+4)}{4} \geq 0.9
$$

We sequentially find

$$
\begin{aligned}
& \frac{0.8^{n}(n+4)}{4} \leq 0.1 \\
& 0.8^{n}(n+4)=0.4 \\
& n \cdot \lg 0.8+\lg (n+4) \leq \lg 0.4 \\
& 0.0969 n \geq \lg (n+4)+0.3979
\end{aligned}
$$

This inequality is satisfied for $n \geq 18$.

Thus, at least 18 devices need to be tested.

## 3.2. The Most Probable Number of Occurrences of an Event in Independent Trials

The most probable number $m_{0}$ (or the most probable frequency) of the occurrence of event $A$ in $n$ independent trials, in each of which it can occur with probability $p$ (and not occur with probability $q=1-p$), is found from the double inequality

$$
n p-q \leq m_{0} \leq n p+p
$$