jmc

$ab$ (where $a$ and $b$ are digits) is divisible by 3 only when $a+b$ is divisible by 3. Since 3 divides both $a$ and $a+b$, 3 must divide $b$. Therefore, $a$ and $b$ can equal 3 or 6. The probability that $a$ equals 3 or 6 is $\frac{2}{6}=\frac{1}{3}$. The probability that $b$ equals 3 or 6 is $\frac{2}{6}=\frac{1}{3}$. Therefore, The probability that both $a$ and $b$ equal 3 or 6 is ${\left(\frac{1}{3}\right)}^2=\boxed{\frac{1}{9}}$.