2010 AMC 10B Problems/Problem 18

First we factor $abc+bc+c$ into $a(b(c+1)+1)$ . For $a(b(c+1)+1)$ to be divisable by three we can either have $a$ be a multiple of 3 or $b(c+1)+1$ be a multiple of three. Adding the probability of these two being divisable by 3 we get that the probability is $frac\{13}{27}$ (Error compiling LaTeX. Unknown error_msg)