2003 AMC 10B Problem 21

Question

A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?

\dfrac{1}{8}

\dfrac{5}{32}

\dfrac{9}{32}

\dfrac{3}{8}

\dfrac{7}{16}

Answer: C

Concepts:basic probability casework

Difficulty rating: 1600

Small hint:

All beads are red exactly when both green beads are drawn during the three draws

Big hint:

Add the probabilities of the draw sequences GG, GRG, and RGG, tracking how the counts change

Solution:

The bag always holds

4

beads. All are red at the end precisely when both greens are drawn.

Drawing green then green has probability

\dfrac24 \cdot \dfrac14 = \dfrac18.

Green, red, green has probability

\dfrac24 \cdot \dfrac34 \cdot \dfrac14 = \dfrac{3}{32}.

Red, green, green has probability

\dfrac24 \cdot \dfrac24 \cdot \dfrac14 = \dfrac{1}{16}.

The total is

\dfrac18 + \dfrac{3}{32} + \dfrac{1}{16} = \dfrac{9}{32}.

Thus, the correct answer is C.