Problem 12 · 2002 AMC 8
Medium
Counting & Probability
complementary-counting
A board game spinner is divided into three regions labeled A, B, and C. The probability the arrow stops on region A is 13 and on region B is 12. What is the probability the arrow stops on region C?
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Answer: B — 1/6.
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Hint 1 of 2
The arrow has to land *somewhere*, and A, B, C are the only options — so their three chances can't add up to anything but a whole 1.
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Hint 2 of 2
That makes C the leftover: don't solve for it, just subtract A and B from 1.
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Approach: probabilities of all regions sum to 1
- The three regions cover every outcome with no overlap, so P(A) + P(B) + P(C) = 1. The unknown is just the leftover: P(C) = 1 − 13 − 12.
- Use a common denominator of 6: 66 − 26 − 36 = 16.
- *The principle:* when outcomes split a situation completely (nothing left out, nothing double-counted), their probabilities sum to 1 — so the last unknown piece is always "1 minus the rest."
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