Example 3. Two dice are rolled. Find the probability of event $A$ - "the sum of the points rolled does not exceed four".
Solution. Event $A$ is the sum of three mutually exclusive events $B_{2}, B_{3}, B_{4}$, which means the sum of the points is 2, 3, or 4, respectively. Since

$$
P\left(B_{2}\right)=\frac{1}{36}, \quad P\left(B_{3}\right)=\frac{2}{36}, \quad P\left(B_{4}\right)=\frac{3}{36}
$$

by the theorem of addition of probabilities of mutually exclusive events, we get

$$
P(A)=P\left(B_{2}\right)+P\left(B_{3}\right)+P\left(B_{4}\right)=\frac{1}{36}+\frac{2}{36}+\frac{3}{36}=\frac{6}{36}=\frac{1}{6}
$$

Remark. The same result can be obtained directly. Indeed, event $A$ is favorable to 6 elementary outcomes: $(1,1),(1,2)$, $(2,1),(1,3),(3,1),(2,2)$. The total number of elementary outcomes forming a complete group of events is $n=36$, therefore

$$
P(A)=\frac{6}{36}=\frac{1}{6} .
$$