2010 AMC 12B Problem 16

Factor $abc+ab+a=a(bc+b+1).$ Since $2010$ is a multiple of $3,$ each of $a, b, c$ is uniform modulo $3.$ If $a$ is divisible by $3$ (probability $\tfrac13$), the product is divisible by $3.$ If $a$ is not divisible by $3$ (probability $\tfrac23$), we need $bc+b+1$ to be divisible by $3.$ Checking residues, this holds exactly when $(b,c)\equiv(1,1)$ or $(2,0)\pmod3,$ a probability of $\tfrac13\cdot\tfrac13+\tfrac13\cdot\tfrac13=\tfrac29.$ The total probability is $$\frac13+\frac23\cdot\frac29=\frac13+\frac{4}{27}=\frac{13}{27}.$$ Thus, the correct answer is E.