91. A discrete random variable $X$ has three possible values: $x_{1}=1, x_{2}$, and $x_{3}$, with $x_{1}<x_{2}<x_{3}$. The probabilities that $X$ will take the values $x_{1}$ and $x_{2}$ are 0.3 and 0.2, respectively. The expected value of this variable is $M(X)=2.2$, and the variance is $D(X)=0.76$. Find the distribution series of the variable $X$.
Solution. Since the sum of the probabilities of all possible values of a discrete random variable is equal to one, the unknown probability $p_{3}=P\left(X=x_{3}\right)=1-\left(p_{1}+p_{2}\right)=1-(0.3+0.2)=0.5$.

Then the distribution law of the variable $X$ is given as:

$$
X: \begin{array}{cccc}
x_{i} & 1 & x_{2} & x_{3} \\
p_{i} & 0.3 & 0.2 & 0.5
\end{array}
$$

To find the unknown $x_{2}$ and $x_{3}$, we will form two equations linking these numbers. For this purpose, we will express the expected value and variance in terms of $x_{2}$ and $x_{3}$.

Let's find $M(X)$ and $D(X)$:

$$
\begin{gathered}
M(X)=x_{1} p_{1}+x_{2} p_{2}+x_{3} p_{3} \\
D(X)=x_{1}^{2} p_{1}+x_{2}^{2} p_{2}+x_{3}^{2} p_{3}-M^{2}(X)
\end{gathered}
$$

Substituting all known values into the last two equations, and after performing elementary transformations in each equation, we obtain a system of two equations with two unknowns:

$$
\left\{\begin{array}{l}
0.2 x_{2}+0.5 x_{3}=1.9 \\
0.2 x_{2}^{2}+0.5 x_{3}^{2}=5.3
\end{array}\right.
$$

Solving this system, we find two solutions:

$$
\text { 1) } x_{2}=2 ; x_{3}=3 \text { and 2) } x_{2}=24 / 7, x_{3}=17 / 7 \text {. }
$$

By the condition $x_{2}<x_{3}$, only the first solution of the system satisfies the problem: $x_{2}=2 ; x_{3}=3$.

$\because 1$

Substituting the found values of $x_{2}$ and $x_{3}$ into the table (*), we get the required distribution series:

$$
X: \begin{array}{cccc}
x_{i} & 1 & 2 & 3 \\
p_{i} & 0.3 & 0.2 & 0.5
\end{array}
$$