2002 AMC 8 Problems/Problem 12: Difference between revisions
Giratina150 (talk | contribs) Created page with " == 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 <..." |
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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}=\frac{1}{6}</math> C. | |||
Revision as of 13:49, 11 December 2012
Problem 12
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 
C.
