jmc

There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)$ ways for 3 of the dice to show even numbers and 3 of them to show odd numbers. Each roll is even with probability $\frac{1}{2}$ and odd with probability $\frac{1}{2}$, so each arrangement of 3 odd numbers and 3 even number occurs with probability ${\left(\frac{1}{2}\right)}^6$. Thus, the probability that 3 dice out of 6 show even numbers is $$\left(\genfrac{}{}{0ex}{}{6}{3}\right)\frac{1}{2^6}=\boxed{\frac{5}{16}}.$$