jmc

The beads will all be red at the end of the third draw precisely when two green beads are chosen in the three draws. If the first bead drawn is green, then there will be one green and three red beads in the bag before the second draw. So the probability that green beads are drawn in the first two draws is $$\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}.$$ The probability that a green bead is chosen, then a red bead, and then a green bead, is $$\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{1}{4}=\frac{3}{32}.$$ Finally, the probability that a red bead is chosen then two green beads is $$\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{4}=\frac{1}{16}.$$ The sum of these probabilities is $$\frac{1}{8}+\frac{3}{32}+\frac{1}{16}=\boxed{\frac{9}{32}}.$$