AMC 8 · 2002 · #12

Grade 7 probability
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
probability-basicfraction-arithmeticcomplementary-counting complementary-countingidentify-subproblems ↑ Prerequisites: fraction-arithmetic
📏 Short solution 💡 1 insight
Problem
A spinner is split into three regions A, B, C. The arrow lands on A with probability 13\frac{1}{3} and on B with probability 12\frac{1}{2}. Find the probability that it lands on C.

Pick an answer.

(A)
$\frac{1}{12}$
(B)
$\frac{1}{6}$
(C)
$\frac{1}{5}$
(D)
$\frac{1}{3}$
(E)
$\frac{2}{5}$

AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Region C is exactly the part of the spinner that is not A and not B. Tool #16 (Count the Complement) names this directly: instead of asking "what is P(C)?", ask "what is left over after A and B take their shares of the whole?" Since all probabilities must add to 1, the leftover is P(C) = 1 - [P(A) + P(B)]. Tool #7 (Identify Subproblems) splits the arithmetic into two clean pieces: first add P(A) + P(B) with a common denominator, then subtract that sum from 1.

1STEP 1

Add the two known probabilities over the common denominator 6: A and B together fill 56\frac{5}{6} of the spinner.

P(A) + P(B) = 13\frac{1}{3} + 12\frac{1}{2} = 26\frac{2}{6} + 36\frac{3}{6} = 56\frac{5}{6}
2STEP 2

The three probabilities sum to 1, so C is the leftover: 1 - 56\frac{5}{6} = 16\frac{1}{6}, choice (B).

P(C) = 1 - 56\frac{5}{6} = 66\frac{6}{6} - 56\frac{5}{6} = 16\frac{1}{6} → (B)
Answer
16\frac{1}{6}
Check the sum: 13\frac{1}{3} + 12\frac{1}{2} + 16\frac{1}{6} = 26\frac{2}{6} + 36\frac{3}{6} + 16\frac{1}{6} = 66\frac{6}{6} = 1. The three probabilities add to 1, as required for the only possible outcomes of the spin. Size check: region C should be the smallest region because A (13\frac{1}{3}) and B (12\frac{1}{2}) already take most of the spinner. Indeed 16\frac{1}{6} is smaller than both 13\frac{1}{3} and 12\frac{1}{2}, which matches.
💡Key takeaway

The whole spinner has probability 1, and regions A and B already take 13\frac{1}{3} + 12\frac{1}{2} = 56\frac{5}{6}. Region C is just the leftover: 1 - 56\frac{5}{6} = 16\frac{1}{6}, answer (B).

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