Difference between revisions of "2013 AMC 8 Problems/Problem 14"
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==Solution== | ==Solution== | ||
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| + | The probability that both show a green bean is <math>\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}</math>. The probability that both show a red bean is <math>\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}</math>. Therefore the probability is <math>\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=13|num-a=15}} | {{AMC8 box|year=2013|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 08:25, 27 November 2013
Problem
Abe holds 1 green and 1 red jelly bean in his hand. Bea holds 1 green, 1 yellow, and 2 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
Solution
The probability that both show a green bean is 
. The probability that both show a red bean is 
. Therefore the probability is 
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
