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2003 AMC 10B Problem 21

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2003 AMC 10B Problem 21AMC 10 Step by Step
Name:                             Date: September 13, 2026
Conditional Probability & States · difficulty 3 of 5 · about 4 min · position P21-25

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?

  1. A)

    18\frac{1}{8}

  2. B)

    532\frac{5}{32}

  3. C)

    932\frac{9}{32}

  4. D)

    38\frac{3}{8}

  5. E)

    716\frac{7}{16}

Answer

C

Key insight

All beads end up red exactly when both greens are drawn among the three pulls; track the count of greens and note each drawn green is replaced by red.

Solution

The bag always holds four beads. Drawing a red changes nothing; drawing a green replaces it with red, reducing the green count by one. Starting from 22 greens, the bag is all red after three draws exactly when two of the three draws are green.

Track the green count. With gg greens the chance of drawing green is g4\frac{g}{4} and of drawing red is 4−g4\frac{4 - g}{4}. The two green draws can occur at positions {1,2}\{1,2\}, {1,3}\{1,3\}, or {2,3}\{2,3\}:

  • Green, green, anything: 24⋅14⋅1=18\frac{2}{4} \cdot \frac{1}{4} \cdot 1 = \frac{1}{8}.
  • Green, red, green: after one green there is 11 green and 33 reds, so 24⋅34⋅14=332\frac{2}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{3}{32}.
  • Red, green, green: 24⋅24⋅14=116\frac{2}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{1}{16}.

Total:
432+332+232=932.\frac{4}{32} + \frac{3}{32} + \frac{2}{32} = \frac{9}{32}.

The answer is (C) 932\boxed{\textbf{(C)}\ \frac{9}{32}}.

Why this works

The state of the bag is fully described by the number of greens, and each draw moves that state down by one or leaves it alone with probabilities depending only on the current state. Enumerating the paths from state 22 to state 00 in three steps, with the correct probability at each step, is a small Markov-chain computation. The trap is that the probabilities change after each green draw.

Alternative approach

Complementary counting: no greens in three draws has probability (12)3=432\left(\frac{1}{2}\right)^3 = \frac{4}{32}; exactly one green (at draw 11, 22, or 33) has probability 12⋅34⋅34+12⋅12⋅34+12⋅12⋅12=9+6+432=1932\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 + 6 + 4}{32} = \frac{19}{32}. So at least two greens has probability 1−432−1932=9321 - \frac{4}{32} - \frac{19}{32} = \frac{9}{32}.

The trap

Forgetting that a drawn green is replaced by red, so after the first green there is only one green left among four beads (probability 1/4, not 1/2).

Common mistakes

  • Forgetting that a drawn green is replaced by red, so after the first green there is only one green left among four beads (probability 14\frac{1}{4}, not 12\frac{1}{2}).
  • Counting only the case "green then green" (18\frac18) and stopping, or forgetting that a red draw between the two greens still leads to success.

Techniques

Split into exhaustive cases and handle each · Define states/recurrence and iterate