smc

We group this into groups of $3$, because $3|2010$. This means that every residue class mod 3 has an equal probability. If $3|a$, we are done. There is a probability of $\frac{1}{3}$ that that happens. Otherwise, we have $3|bc+b+1$, which means that $b(c+1)\equiv 2(\mathrm{mod}3)$. So either $$b\equiv 1(\mathrm{mod}3),c\equiv 1(\mathrm{mod}3)$$ or $$b\equiv 2(\mathrm{mod}3),c\equiv 0(\mathrm{mod}3)$$ which will lead to the property being true. There is a $\frac{1}{3}\cdot \frac{1}{3}=\frac{1}{9}$ chance for each bundle of cases to be true. Thus, the total for the cases is $\frac{2}{9}$. But we have to multiply by $\frac{2}{3}$ because this only happens with a $\frac{2}{3}$ chance. So the total is actually $\frac{4}{27}$. The grand total is $$\frac{1}{3}+\frac{4}{27}=\boxed{\frac{13}{27}}$$