241. The device consists of 10 independently operating elements: The probability of failure of each element over time $T$ is 0.05. Using Chebyshev's inequality, estimate the probability that the absolute value of the difference between the number of failed elements and the average (expected) number of failures over time $T$ will be: a) less than two; b) not less than two.
Solution. a) Let $X$ be a discrete random variable - the number of failed elements over time $T$. Then

$$
\begin{aligned}
& M(X)=n p=10 \cdot 0.05=0.5 \\
& D(X)=n p q=10 \cdot 0.05 \cdot 0.95=0.475 .
\end{aligned}
$$

Using Chebyshev's inequality:

$$
P(|X-M(X)|<\varepsilon) \geqslant 1-D(X) / \varepsilon^{2}
$$

Substituting $M(X)=0.5 ; D(X)=0.475, \varepsilon=2$, we get

$$
P(|X-0.5|<2) \geqslant 1-0.475 / 4=0.88
$$

b) The events $|X-0.5|<2$ and $|X-0.5| \geqslant 2$ are opposite, so the sum of their probabilities is one. Therefore,

$$
P(|X-0.5| \geqslant 2)<1-0.88=0.12
$$