8. For a regular 2015-gon $A_{1} A_{2} \cdots A_{2015}$ inscribed in the unit circle $O$, if any two distinct vertices $A_{i}, A_{j}$ are chosen, then the probability that $\left|\overrightarrow{O A}_{i}+\overrightarrow{O A}_{j}\right| \geq 1$ is $\qquad$ .
Answer $\frac{671}{1007}$. Solution: Since $\left|\overrightarrow{O A_{i}}\right|=\left|\overrightarrow{O A_{j}}\right|=1$, we have
$$
\left|\overrightarrow{O A}_{i}+\overrightarrow{O A}_{j}\right|^{2}=\left|\overrightarrow{O A}_{i}\right|^{2}+\left|\overrightarrow{O A}_{j}\right|^{2}+2 \overrightarrow{O A}_{i} \cdot \overrightarrow{O A_{j}}=2\left(1+\cos \right) \text {. }
$$

Therefore, the necessary and sufficient condition for $\left|\overrightarrow{O A_{i}}+\overrightarrow{O A_{j}}\right| \geq 1$ is $\cos \geq-\frac{1}{2}$, which means the angle between vectors $\overrightarrow{O A}_{i}, \overrightarrow{O A}_{j}$ does not exceed $\frac{2 \pi}{3}$.
For any given vector $\overrightarrow{O A}_{i}$, the number of ways to choose a vector $\overrightarrow{O A}_{j}$ that satisfies $\left|\overrightarrow{O A_{i}}+\overrightarrow{O A_{j}}\right| \geq 1$ is: $\left[\frac{2 \pi}{3} \div \frac{2 \pi}{2015}\right] \times 2=1342$. Hence, the probability that $\left|\overrightarrow{O A_{i}}+\overrightarrow{O A_{j}}\right| \geq 1$ is: $\quad p=\frac{2015 \times 1342}{2015 \times 2014}=\frac{671}{1007}$.