2002 AMC 8 Problems/Problem 12

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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