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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<math> \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} </math> | <math> \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} </math> | ||
| − | + | ==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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2002|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:24, 30 October 2016
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 
.
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. 
