Problem 12 · 2009 AMC 8
Medium
Counting & Probability
enumerate-outcomes
The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
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Answer: D — 7/9.
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Hint 1 of 2
First spinner is all ODD (1, 3, 5); second is all EVEN (2, 4, 6). Odd + even is always odd — so every one of the 9 sums is odd, and you never have to worry about "even, so not prime."
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Hint 2 of 2
Among odd sums the only non-primes are 9, 15, 21, … (odd multiples of 3). The sums here run 3 to 11, so the only danger is a 9. Count BAD outcomes — it's quicker than counting good ones.
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Approach: use parity, then count the few non-primes (complement)
- There are 3 × 3 = 9 equally likely outcomes. Since odd + even = odd, every sum is odd, so the only way to miss "prime" is to land on an odd non-prime in range (3–11) — that's just 9.
- Which pairs sum to 9? Only 3+6 and 5+4 — 2 outcomes.
- So 2 fail and 7 succeed: probability = 7/9 = 7/9.
- Why this transfers: spotting that all sums share a parity slashes the work, and counting the FEW bad cases (complement) beats listing all the good ones.
Another way — list all nine sums:
- Sums: 3, 5, 7, 5, 7, 9, 7, 9, 11. The two 9s are the only composites.
- Prime in 7 of 9 outcomes ⇒ 7/9.
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