Difference between revisions of "2002 AMC 8 Problems/Problem 12"
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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}=\boxed{\text{(B)}\ \frac16}</math>. | 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== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=11|num-a=13}} | {{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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
