2003 AMC 10B Problems/Problem 21

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Problem

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?

$\textbf{(A) } \frac{1}{8} \qquad\textbf{(B) } \frac{5}{32} \qquad\textbf{(C) } \frac{9}{32} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{7}{16}$

Solution

We can divide the case of all beads in the bag being red after three replacements into three cases.

The first case is in which the two green beads are the first two beads to be chosen. The probability for this is \[\frac{2}{4} \times \frac{1}{4} \times \frac{4}{4} = \frac{1}{8}\]

The second case is in which the green beads are chosen first and third. The probability for this is \[\frac{2}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{32}\]

The third case is in which the green beads are chosen second and third. The probability for this is \[\frac{2}{4} \times \frac{2}{4} \times \frac{1}{4} = \frac{1}{16}\]

Add all these cases together \[\frac{1}{8}+\frac{3}{32}+\frac{1}{16} = \frac{4}{32}+\frac{3}{32}+\frac{2}{32} = \boxed{\textbf{(C) \ } \frac{9}{32}}\]

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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