Example 7. Discrete independent random variables are given by the distribution laws:

| $X$ | 5 | 6 |
| :---: | :---: | :---: |
| $P$ | 0.4 | 0.6 |


| $Y$ | 7 | 8 |
| :---: | :---: | :---: |
| $P$ | 0.8 | 0.2 |

Formulate the distribution law of the random variable $Z=X+Y$.
Solution. The possible values of the quantity $Z$ are the sums of each possible value of $X$ with all possible values of $Y$:

$$
z_{1}=5+7=12, z_{2}=5+8=13, z_{3}=6+7=13, z_{4}=6+8=14
$$

Let's find the probabilities of these possible values. For $Z=12$, it is sufficient for the quantity $X$ to take the value $x_{1}=5$ and the quantity $Y$ to take the value $y_{1}=7$. The probabilities of these possible values, as follows from the given distribution laws, are 0.4 and 0.8, respectively. The arguments $X$ and $Y$ are independent, so the events $X=5$ and $Y=7$ are independent; therefore, the probability of their joint occurrence (i.e., the probability of the event $Z=5+7=12$) by the multiplication theorem is $0.4 \cdot 0.8=0.32$.

Similarly, we find:

$$
\begin{aligned}
& P(Z=5+8=13)=0.4 \cdot 0.2=0.08 \\
& P(Z=6+7=13)=0.6 \cdot 0.8=0.48
\end{aligned}
$$

$$
P(Z=6+8=14)=0.6 \cdot 0.2=0.12 \text {. }
$$

The quantity $Z$ takes three different values: $12, 13, 14$. Since the events ($Z=5+8=13$) and ($Z=6+7=13$) are mutually exclusive, then

$$
P(Z=13)=P(Z=5+8=13)+P(Z=6+7=13)=0.08+0.48=0.56
$$

Thus, the quantity $Z$ has the distribution law

| $Z$ | 12 | 13 | 14 |
| :---: | :---: | :---: | :---: |
| $P$ | 0.32 | 0.56 | 0.12 |

Note that $0.32+0.56+0.12=1$, as it should be.