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2003 AMC 10B Problem 21

Problem 21 of 25HarderProbability & StatisticsProblem-Solving Techniques

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?

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Solution

The bag always holds 44 beads. All are red at the end precisely when both greens are drawn. Drawing green then green has probability 24⋅14=18.\dfrac24 \cdot \dfrac14 = \dfrac18. Green, red, green has probability 24⋅34⋅14=332.\dfrac24 \cdot \dfrac34 \cdot \dfrac14 = \dfrac{3}{32}. Red, green, green has probability 24⋅24⋅14=116.\dfrac24 \cdot \dfrac24 \cdot \dfrac14 = \dfrac{1}{16}. The total is 18+332+116=932.\dfrac18 + \dfrac{3}{32} + \dfrac{1}{16} = \dfrac{9}{32}. Thus, the correct answer is C.
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