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 \(\left|\overrightarrow{O A_{i}}+\overrightarrow{O A_{j}}\right| \geqslant 1\) is \(\qquad\).
As shown in the figure, $A, B, C$ trisect the unit circle $O$. Let $A_{i}$ coincide with point $A$. According to the problem, $A_{j}$ should be located on $\overparen{A B}$ and $\overparen{A C}$. Since $\left[\frac{2015}{3}\right]=671$, then $\overparen{A B}$ and $\overparen{A C}$ contain $671 \times 2$ vertices of the regular 2015-gon, meaning $A_{j}$ has $671 \times 2$ possible positions. Therefore, the number of ways to choose $A_{i}, A_{j}$ such that $\left|\overrightarrow{O A_{i}}+\overrightarrow{O A_{j}}\right| \geqslant 1$ is $\frac{2015 \times 671 \times 2}{?}=2015 \times 671$.

Let $D=$ “$\left|\overrightarrow{O A_{i}}+\overrightarrow{O A_{j}}\right| \geqslant 1$”. Thus, $P(D)=\frac{2015 \times 671}{C_{2015}^{2}}=\frac{671}{1007}$.