1for third ball to be red it is required that atmost only one red ball will be out...
so probability will be...
2\left( \frac{3}{8}.\frac{2}{7}.\frac{2}{6}\right)+2\left(\frac{3}{8}.\frac{3}{7}.\frac{2}{6}\right)+4\left(\frac{3}{8}.\frac{2}{7}.\frac{1}{6} \right)= \frac{1}{4}