Problem 18 · 2019 AMC 8
Medium
Counting & Probability
careful-countingcasework
The faces of each of two fair dice are numbered 1, 2, 3, 5, 7, and 8. When the two dice are tossed, what is the probability that their sum will be an even number?
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Answer: C — 5/9.
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Hint 1 of 2
Don't list all 36 pairs — a sum is even only when the two numbers match in parity (odd+odd or even+even). So all that matters is how many faces are odd vs even.
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Hint 2 of 2
These dice are unusual: faces 1, 3, 5, 7 are odd (4 of them) and only 2, 8 are even (2 of them). Find P(both odd) and P(both even), then add.
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Approach: even sum means matching parity
- A sum is even exactly when both dice are odd or both are even — the actual values don't matter, only odd/even. On this die: odd faces {1, 3, 5, 7} = 4 of 6; even faces {2, 8} = 2 of 6, so P(odd) = 2/3 and P(even) = 1/3.
- Both odd: (2/3)(2/3) = 4/9. Both even: (1/3)(1/3) = 1/9. These can't both happen, so add: 4/9 + 1/9 = 5/9.
- Why this transfers: for parity-of-a-sum questions, throw away the numbers and keep only odd/even — the problem shrinks to a coin-flip count. (Sanity check: a normal 3-odd-3-even die gives exactly 1/2; tilting to 4 odds nudges "both-odd" up, so 5/9 just over 1/2 fits.)
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