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?

$\dfrac{1}{8}$

$\dfrac{5}{32}$

$\dfrac{9}{32}$

$\dfrac{3}{8}$

$\dfrac{7}{16}$

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.

2003 AMC 10B Problem 21

Solution:

Problem 21 in Other Years