2003 AMC 10B Problem 21

1. The bag always has $4$ beads. Drawing a red and putting back a red changes nothing. Drawing a green and putting back a red lowers the number of greens by one.
2. We start with $2$ greens. To end with $0$ greens after three draws, exactly two of the three draws must be green. (A third green is impossible, since there are none left.)
3. The chance of drawing green depends on how many greens are in the bag right now:

4. The two green draws can happen in three orders. Multiply the chances along each order:
- Green, green, then anything: $\frac{2}{4} \cdot \frac{1}{4} \cdot 1 = \frac{1}{8} = \frac{4}{32}$.
- Green, red, green: after the first green there is $1$ green, so $\frac{2}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{32}$.
- Red, green, green: the first draw leaves $2$ greens, so $\frac{2}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{1}{16} = \frac{2}{32}$.
5. These orders cannot happen together, so add them:
$$\frac{4}{32} + \frac{3}{32} + \frac{2}{32} = \frac{9}{32}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{9}{32}}$.

Everything about the bag is described by one number: how many greens are left. Each draw either lowers that number by one or leaves it alone, and the chances depend only on the current number. List every path from $2$ greens down to $0$, multiply the chances along each path, and add the paths. The main danger is using the starting chances after the bag has changed.

Find the chance of the bad outcomes and take it away from $1$. No green in three draws: the bag never changes, so $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{4}{32}$. Exactly one green, on draw $1$, $2$ or $3$: $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{9}{32} + \frac{6}{32} + \frac{4}{32} = \frac{19}{32}$. So two greens has chance $1 - \frac{4}{32} - \frac{19}{32} = \frac{9}{32}$.