jmc

There are $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ ways to choose which of the five dice show a 1 or 2. The probability of any one of these occurring is ${\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^3$. So the overall probability is $$10{\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^3=\frac{10\times 2^3}{3^5}=\boxed{\frac{80}{243}}.$$