Difference between revisions of "2003 AMC 10B Problems/Problem 21"
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Revision as of 12:10, 4 July 2013
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?
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
![\[\frac{2}{4} \times \frac{1}{4} \times \frac{4}{4} = \frac{1}{8}\]](http://latex.artofproblemsolving.com/c/7/9/c79094610c98c663ffb3d62502cdcc8d8e4bea3a.png)
The second case is in which the green beads are chosen first and third. The probability for this is
![\[\frac{2}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{32}\]](http://latex.artofproblemsolving.com/7/3/c/73ceeaa46918d0f91fc747ff8f6a3fbc52989dad.png)
The third case is in which the green beads are chosen second and third. The probability for this is
![\[\frac{2}{4} \times \frac{2}{4} \times \frac{1}{4} = \frac{1}{16}\]](http://latex.artofproblemsolving.com/8/9/2/892342fa379797a31251d5885c6957b04e601ee8.png)
Add all these cases together
![\[\frac{1}{8}+\frac{3}{32}+\frac{1}{16} = \frac{4}{32}+\frac{3}{32}+\frac{2}{32} = \boxed{\textbf{(C) \ } \frac{9}{32}}\]](http://latex.artofproblemsolving.com/b/5/8/b582cc507ddc9c4ad70f25708252281aeab43997.png)
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
