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

$$
\begin{array}{lrrrr}
X & -1 & -2 & 1 & 2 \\
p & 0.3 & 0.1 & 0.2 & 0.4
\end{array}
$$

Find the distribution law of the random variable $Y=X^{2}$.
Solution. Let's find the possible values of $Y$:

$$
\begin{array}{cc}
y_{i}=x_{1}^{2}=(-1)^{2}=1, & y_{2}=x_{2}^{2}=(-2)^{2}=4, \\
y_{3}=x_{3}^{2}=1^{2}=1, & y_{4}=x_{4}^{2}=2^{2}=4 .
\end{array}
$$

Thus, different values of $X$ correspond to the same values of $Y$. This is explained by the fact that the possible values of $X$ belong to an interval on which the function $Y=X^{2}$ is not monotonic.

Let's find the probabilities of the possible values of $Y$. For $Y$ to take the value $Y=1$, it is sufficient for $X$ to take the value $X=-1$ or $X=1$. These two events are mutually exclusive, and their probabilities are 0.3 and 0.2, respectively. Therefore, the probability of the event $Y=1$ by the addition theorem is

$$
P(Y=1)=P(X=-1)+P(X=1)=0.3+0.2=0.5 .
$$

Similarly, we find the probability of the possible value $Y=4$:

$$
P(Y=4)=P(X=-2)+P(X=2)=0.1+0.4=0.5 .
$$

Let's write the required distribution law of the variable $Y$:

$$
\begin{array}{lll}
Y & 1 \\
p & 1 & 4 \\
0.5 & 0.5
\end{array}
$$