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Difference between revisions of "2013 AMC 8 Problems/Problem 14"

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==Problem==
 
==Problem==
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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?
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<math>\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23</math>
  
 
==Solution==
 
==Solution==
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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>.
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==Video Solution==
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https://youtu.be/NMpVIy8QxSY ~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC8 box|year=2013|before=First Problem|num-a=2}}
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{{AMC8 box|year=2013|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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?

$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac38 \qquad \textbf{(D)}\ \frac12 \qquad \textbf{(E)}\ \frac23$

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 $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$. The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$

Solution 2

We can list out all the combinations and we get this: $GG, GY, GR_1, GR_2, RG, RY, RR_1, RR_2$. There is a total of 8 combinations and 3 that are the same. Hence, we yield the answer $\boxed{\textbf{C}}$.

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. AMC logo.png

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