jmc

There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ ways to choose three of the dice to show prime numbers. Each roll is prime with probability $\frac{1}{2}$ and composite with probability $\frac{1}{3}$, so each arrangement of 3 prime numbers and 3 composite numbers occurs with probability ${\left(\frac{1}{2}\right)}^3{\left(\frac{1}{3}\right)}^3.$ Therefore, the probability that three dice show prime numbers and the rest show composite numbers is $$20\cdot {\left(\frac{1}{2}\right)}^3{\left(\frac{1}{3}\right)}^3=\boxed{\frac{5}{54}}.$$