Difference between revisions of "2013 AMC 8 Problems/Problem 14"
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==Solution== | ==Solution== | ||
The favorable responses are either they both show a green bean or they both show a red bean. 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> | The favorable responses are either they both show a green bean or they both show a red bean. 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> | ||
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| + | ==Solution 2== | ||
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| + | We can list out all the combinations and we get this: <math>GG, GY, GR_1, GR_2, RG, RY, RR_1, RR_2</math>. There is a total of 8 combinations and 3 that are the same. Hence, we yield the answer <math>\boxed{\textbf{C}}</math>. | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 18:41, 1 January 2023
Problem
Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
Solution
The favorable responses are either they both show a green bean or they both show a red bean. The probability that both show a green bean is 
. The probability that both show a red bean is 
. Therefore the probability is 
Solution 2
We can list out all the combinations and we get this: 
. There is a total of 8 combinations and 3 that are the same. Hence, we yield the answer 
.
Video Solution
https://youtu.be/NMpVIy8QxSY ~savannahsolver
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. 
