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Difference between revisions of "2002 AMC 8 Problems/Problem 12"

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== Problem ==
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== Problem 12 ==
 
  
 
A board game spinner is divided into three regions labeled <math>A</math>, <math>B</math> and <math>C</math>. The probability of the arrow stopping on region <math>A</math> is <math>\frac{1}{3}</math> and on region <math>B</math> is <math>\frac{1}{2}</math>. The probability of the arrow stopping on region <math>C</math> is:
 
A board game spinner is divided into three regions labeled <math>A</math>, <math>B</math> and <math>C</math>. The probability of the arrow stopping on region <math>A</math> is <math>\frac{1}{3}</math> and on region <math>B</math> is <math>\frac{1}{2}</math>. The probability of the arrow stopping on region <math>C</math> is:
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==Solution==
 
==Solution==
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Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is <math>1-\frac{1}{2}-\frac{1}{3}=\frac{1}{6}</math> C.
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Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is <math>1-\frac{1}{2}-\frac{1}{3}=\boxed{\text{(B)}\ \frac16}</math>.
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==See Also==
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{{AMC8 box|year=2002|num-b=11|num-a=13}}
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{{MAA Notice}}

Revision as of 21:24, 30 October 2016

Problem

A board game spinner is divided into three regions labeled $A$, $B$ and $C$. The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}$. The probability of the arrow stopping on region $C$ is:


$\text{(A)}\ \frac{1}{12}\qquad\text{(B)}\ \frac{1}{6}\qquad\text{(C)}\ \frac{1}{5}\qquad\text{(D)}\ \frac{1}{3}\qquad\text{(E)}\ \frac{2}{5}$

Solution

Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is $1-\frac{1}{2}-\frac{1}{3}=\boxed{\text{(B)}\ \frac16}$.

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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

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