Problem 25 · 1987 AJHSME
Stretch
Counting & Probability
paritywithout-replacement
Ten balls numbered 1 to 10 are in a jar. Jack reaches into the jar and randomly removes one of the balls. Then Jill reaches into the jar and randomly removes a different ball. The probability that the sum of the two numbers on the balls removed is even is
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Answer: A — 4⁄9.
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Hint 1 of 2
A sum is even only when the two numbers match in parity (odd+odd or even+even). What's the count of each kind among 1–10?
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Hint 2 of 2
There are 5 odd and 5 even balls. After Jack takes one, only 9 balls remain — so Jill's draw is out of 9, not 10.
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Approach: match Jack's parity (fix the first draw)
- The sum is even exactly when both balls share parity. Whatever Jack pulls, his ball has 4 same-parity partners left (e.g. if he takes an odd, 4 odds remain) out of the 9 balls Jill can choose from.
- So the probability Jill matches is 4⁄9 — and that's the whole answer: 4⁄9.
- Why this transfers: when only the relationship between two draws matters, fix the first draw as 'done' and ask the chance the second one cooperates. No need to enumerate Jack's case at all.
Another way — add the two same-parity cases:
- P(both odd) = (5⁄10)(4⁄9) = 2⁄9 and P(both even) = (5⁄10)(4⁄9) = 2⁄9.
- Total = 2⁄9 + 2⁄9 = 4⁄9, matching the fixed-draw view.
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