Problem 22 · 1994 AJHSME
Stretch
Counting & Probability
parityprobability
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Answer: D — 5/12.
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Hint 1 of 2
Don't list every number-pair and add them — the only thing that decides if a SUM is even is the parity (odd/even) of the two spins. Even + even or odd + odd → even; a mismatch → odd. So shrink each wheel down to just 'P(even)' and 'P(odd).'
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Hint 2 of 2
Get those chances from the AREAS, not the count of numbers. On wheel 1 the '3' fills a big half while 1 and 2 each fill a quarter; wheel 2 is split into three equal thirds.
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Approach: combine the even/odd chances of each wheel
- Wheel 1 by area: the even number 2 owns 1/4, so P(even) = 1/4 and P(odd) = 3/4 (the 3 takes half, the 1 a quarter). Wheel 2 by thirds: evens 6 and 4 give P(even) = 2/3, odd 5 gives P(odd) = 1/3.
- Even sum = both even OR both odd. Multiply within each case, then add: (1/4)(2/3) + (3/4)(1/3) = 2/12 + 3/12 = 5/12.
- Key simplification: for an even/odd-sum question you can throw away the actual numbers and track only parity — that collapses two messy wheels into a 2-outcome problem. Quick check: P(odd sum) = 1 − 5/12 = 7/12, and the two should sum to 1 ✓.
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