Difference between revisions of "2002 AMC 8 Problems/Problem 12"
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− | + | == Problem == | |
− | == Problem | ||
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== | ||
− | 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}=\ | + | 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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+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/ODY-ycmPvPE | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2002|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:31, 29 October 2024
Problem
A board game spinner is divided into three regions labeled , and . The probability of the arrow stopping on region is and on region is . The probability of the arrow stopping on region is:
Solution
Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is .
Video Solution by WhyMath
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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