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

$$
\begin{array}{cccc}
X & 1 & 3 & 5 \\
p & 0.4 & 0.1 & 0.5
\end{array}
$$

Find the distribution law of the random variable $Y=3X$.
Solution. Let's find the possible values of the quantity $Y=3X$. We have: $y_{1}=3\cdot1=3; y_{2}=3\cdot3=9; y_{n}=3\cdot5=15$. We see that different possible values of $X$ correspond to different values of $Y$. This is explained by the fact that the function $y=\varphi(x)=3x$ is monotonic. Let's find the probabilities of the possible values of $Y$. For $Y=y_{1}=3$, it is sufficient for the quantity $X$ to take the value $x_{i}=1$. The probability of the event $X=1$ is 0.4 by the condition; therefore, the probability of the event $Y=y_{1}=3$ is also 0.4.

Similarly, we will find the probabilities of the other possible values of $Y$:

$$
\begin{aligned}
& P(Y=9)=P(X=3)=0.1; \\
& P(Y=15)=P(X=5)=0.5.
\end{aligned}
$$

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

$$
\begin{array}{rrrr}
Y & 3 & 9 & 15 \\
p & 0.4 & 0.1 & 0.5
\end{array}
$$