8. A uniform cube die has the numbers $1,2, \cdots, 6$ marked on its faces. Each time two identical dice are rolled, the sum of the numbers on the top faces is defined as the point number for that roll. Then, the probability that the product of the 3 point numbers obtained from 3 rolls can be divided by 14 is $\qquad$. (Express the answer as a simplified fraction)
Solution: $\frac{1}{3}$.
[Method 1] It is easy to know that in one roll, the probability of rolling a 7 is $\frac{6}{6^{2}}=\frac{1}{6}$, and the probability of rolling an odd (even) number is $\frac{1}{2}$. Therefore, the probability of rolling an odd number but not 7 is $\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$.

In 3 rolls, let "only one roll results in a 7, and the other two rolls include at least one even number" be event $A$, and "two rolls result in a 7, and the other roll results in an even number" be event $B$. Clearly, $A$ and $B$ are mutually exclusive. Therefore, the required event is $C=A+B$.
Since $P(A)=C_{3}^{1} \times \frac{1}{6} \times\left[C_{2}^{1} \times \frac{1}{2} \times \frac{1}{3}+\left(\frac{1}{2}\right)^{2}\right]=\frac{7}{24}$,
$$
P(B)=C_{3}^{2} \times\left(\frac{1}{6}\right)^{2} \times \frac{1}{2}=\frac{1}{24} \text {, }
$$

Thus, $P(C)=P(A+B)=P(A)+P(B)=\frac{7}{24}+\frac{1}{24}=\frac{1}{3}$.
[Method 2] In 3 rolls, let "at least one roll results in an even number" be event $A$, and "at least one roll results in a 7" be event $B$, then the required event is $C=A \cap B$.
Since $\bar{C}=\overline{A \cap B}=\bar{A} \cup \bar{B}$,
$$
\text { Therefore, } \begin{aligned}
P(\bar{C}) & =P(\bar{A})+P(\bar{B})-P(\bar{A} \cap \bar{B}) \\
& =\left(1-\frac{1}{2}\right)^{3}+\left(1-\frac{6}{36}\right)^{3}-\left(\frac{1}{2}-\frac{6}{36}\right)^{3}=\frac{2}{3} .
\end{aligned}
$$

Hence, $P(C)=1-P(\bar{C})=\frac{1}{3}$.