Example 6.2. A die is rolled twice. What is the probability that the sum of the points rolled will be eight?
Solution. Let $A_{ij}$ be the event that $i$ points fall on the first throw, and $j$ points fall on the second throw. Then the 36 events

$$
\begin{array}{cccc}
A_{11}, & A_{12}, & \ldots, & A_{16} \\
A_{21}, & A_{22}, & \ldots, & A_{26} \\
\ldots, & \ldots, & \ldots, & \ldots, \\
A_{61}, & A_{62}, & \ldots, & A_{66}
\end{array}
$$

can be considered as the elementary outcomes of the experiment. Therefore, the number of all elementary outcomes $n=36$. The event $A$ (the sum of the points that fall is eight) is favorable to the outcomes $A_{26}, A_{35}, A_{44}, A_{53}, A_{62}$. Thus, $m=5$. From this, we obtain:

$$
P(A)=\frac{5}{36}
$$