218. A discrete random variable $X$ has only two possible values: $x_{1}$ and $x_{2}$, with $x_{2}>x_{1}$. The probability that $X$ will take the value $x_{1}$ is 0.6. Find the distribution law of the variable $X$, if the expected value and variance are known: $M(X)=1.4$; $D(X)=0.24$.
Solution. The sum of the probabilities of all possible values of a discrete random variable is equal to one, so the probability that $X$ will take the value $x_{2}$ is $1-0.6=0.4$.

Let's write the distribution law of $X$:

$$
\begin{array}{lll}
X & x_{1} & x_{2} \\
p & 0.6 & 0.4
\end{array}
$$

To find $x_{1}$ and $x_{2}$, we need to form two equations involving these numbers. For this purpose, we will express the known expected value and variance in terms of $x_{1}$ and $x_{2}$.

Let's find $M(X):$

$$
M(X)=0.6 x_{1}+0.4 x_{2}
$$

According to the condition, $M(X)=1.4$, therefore,

$$
0.6 x_{1}+0.4 x_{2}=1.4
$$

One equation involving $x_{1}$ and $x_{2}$ has been obtained. To get the second equation, we will express the known variance in terms of $x_{1}$ and $x_{2}$.

Let's write the distribution law of $X^{2}$:

$$
\begin{array}{lcc}
X^{2} & x_{1}^{2} & x_{2}^{2} \\
p & 0.6 & 0.4
\end{array}
$$

Let's find $M\left(X^{2}\right)$:

$$
M\left(X^{2}\right)=0.6 x_{1}^{2}+0.4 x_{2}^{2}
$$

Let's find the variance:

$$
D(X)=M\left(X^{2}\right)-[M(X)]^{2}=0.6 x_{1}^{2}+0.4 x_{2}^{2}-1.4^{2}
$$

Substituting $D(X)=0.24$, after elementary transformations, we get

$$
0.6 x_{1}^{2}+0.4 x_{2}^{2}=2.2
$$

Combining (**) and (***), we have the system of equations

$$
\left\{\begin{array}{l}
0.6 x_{i}+0.4 x_{2}=1.4 \\
0.6 x_{1}^{2}+0.4 x_{2}^{2}=2.2
\end{array}\right.
$$

Solving this system, we find two solutions:

$$
x_{i}=1 ; x_{2}=2 \text { and } x_{1}=1.8 ; x_{2}=0.8 .
$$

According to the condition $x_{2}>x_{1}$, only the first solution satisfies the problem:

$$
x_{1}=1 ; \quad x_{2}=2
$$

$(* * * *)$

Substituting (****) into (*), we get the desired distribution law:

$$
\begin{array}{lcc}
X & 1 & 2 \\
p & 0.6 & 0.4
\end{array}
$$