Example 3. A discrete random variable $X$ has the following distribution:

| $X$ | 0.2 | 0.4 | 0.6 | 0.8 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.2 | 0.4 | $p_{4}$ | 0.1 |

What is the probability $p_{4}=P \quad(X=0.8) ?$

Construct the distribution polygon.
Solution. Since equality (2.2.1) must hold, i.e.,

$$
p_{1}+p_{2}+p_{3}+p_{4}+p_{5}=1
$$

then

$$
\begin{gathered}
p_{4}=1-\left(p_{1}+p_{2}+p_{3}+p_{5}\right)=1-(0.1+0.2+0.4+0.1)=1-0.8=0.2 \\
p_{4}=0.2
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_720b3cbdc89c57e2b629g-086.jpg?height=558&width=688&top_left_y=958&top_left_x=204)

Fig. 2.1

In a rectangular coordinate system, we plot the points \( M_{1}(0.2 ; 0.1) \), \( M_{2}(0.4 ; 0.2) \), \( M_{3}(0.6 ; 0.4) \), \( M_{4}(0.8 ; 0.2) \), \( M_{5}(1 ; 0.1) \); we connect these points with line segments (Fig. 2.1). The broken line \( M_{1} M_{2} M_{3} M_{4} M_{5} \) is the polygon of distribution of the given random variable.