jmc

We can choose two numbers in $\left(\genfrac{}{}{0ex}{}{99}{2}\right)=4851$ ways. Two numbers will have a product which is a multiple of 3 if at least one of them is a multiple of 3. We can more easily count the number of ways for the product to not be a multiple of 3: this happens when neither of the numbers is a multiple of 3. There are $\frac{99}{3}=33$ multiples of 3 less than 100, and $99-33=66$ numbers which aren't multiples of 3. The number of ways to choose two of these numbers is $\left(\genfrac{}{}{0ex}{}{66}{2}\right)=2145$, so the number of way to choose two numbers where at least one is a multiple of 3 is $4851-2145=2706$. The final probability is $\frac{2706}{4851}=\boxed{\frac{82}{147}}$.