Example 4. The random variable $X$ has a variance $D(X)=0.001$. What is the probability that the random variable $X$ differs from $M(X)=a$ by more than 0.1?
Solution. By the first Chebyshev's inequality

$$
P(|X-a| \geq 0.1) \leq \frac{D(X)}{(0.1)^{2}}=\frac{0.001}{0.01}=0.1
$$