245. A discrete random variable $X$ is given by the distribution law

$$
\begin{array}{lll}
X & 0.3 & 0.6 \\
p & 0.2 & 0.8
\end{array}
$$

Using Chebyshev's inequality, estimate the probability that $|X-M(X)|<0.2$.
Solution. Let's find the mathematical expectation and variance of the quantity $X$:

$$
\begin{gathered}
M(X)=0.3 \cdot 0.2 + 0.6 \cdot 0.8 = 0.54; \\
D(X)=M\left(X^{2}\right)-[M(X)]^{2}= \\
=\left(0.3^{2} \cdot 0.2 + 0.6^{2} \cdot 0.8\right) - 0.54^{2} = 0.0144.
\end{gathered}
$$

We will use Chebyshev's inequality in the form

$$
P(|\dot{X}-M(X)|<\mathrm{e}) \geqslant 1-D(X) / \mathrm{e}^{2}.
$$

Substituting $M(X)=0.54, D(X)=0.0144, \varepsilon=0.2$, we finally get

$$
P(|X-0.54|<0.2) \geqslant 1-0.0144 / 0.04 = 0.64
$$