Example 4. A discrete random variable $X$ has the following distribution:

| $X$ | 3 | 4 | 5 | 6 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $P$ | $p_{1}$ | 0.15 | $p_{3}$ | 0.25 | 0.35 |

Find the probabilities $p_{1}=P(X=3)$ and $p_{3}=P(X=5)$, given that $p_{3}$ is 4 times $p_{1}$.
Solution. Since

$$
p_{2}+p_{4}+p_{5}=0.15+0.25+0.35=0.75
$$

then based on equality (2.1.2), we conclude that

$$
p_{1}+p_{3}=1-0.75=0.25
$$

Since by the condition $p_{3}=4 p_{1}$, then

$$
p_{1}+p_{3}=p_{1}+4 p_{1}=5 p_{1} .
$$

Thus, $5 p_{1}=0.25$, from which $p_{1}=0.05$; therefore,

$$
p_{3}=4 p_{1}=4 \cdot 0.05=0.20
$$

Therefore, $p_{1}=0.05 ; p_{3}=0.20$.