166. The device consists of three independently operating elements. The probability of failure of each element in one trial is 0.1. Formulate the distribution law of the number of failed elements in one trial.
Solution. The discrete random variable $X$ (the number of failed elements in one experiment) has the following possible values: $x_{1}=0$ (none of the device elements failed), $x_{3}=1$ (one element failed), $x_{3}=2$ (two elements failed), and $x_{4}=3$ (three elements failed).

Since the failures of the elements are independent of each other and the probabilities of failure for each element are equal, the Bernoulli formula is applicable. Considering that, according to the condition, $n=3, p=0.1$ (thus, $q=1-0.1=0.9$), we get:

$$
\begin{aligned}
& P_{3}(0)=q^{3}=0.9^{3}=0.729 ; \quad P_{3}(1)=C_{3}^{1} p q^{2}=3 \cdot 0.1 \cdot 0.9^{2}=0.243 \\
& P_{3}(2)=C_{3}^{2} p^{2} q=3 \cdot 0.1^{2} \cdot 0.9=0.027 ; \quad P_{3}(3)=p^{3}=0.1^{3}=0.001
\end{aligned}
$$

Verification: $0.729+0.243+0.027+0.001=1$.

We write the required binomial distribution law for $X$:

| $X$ | 0 | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: | :---: |
| $p$ | 0.729 | 0.243 | 0.027 | 0.001 |