Problem 23 · 1992 AJHSME
Stretch
Counting & Probability
casework
If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than 10 is
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Answer: B — 17/36.
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Hint 1 of 3
Two dice give 6 × 6 = 36 equally likely outcomes (treat the dice as different colors so order counts). The whole job is counting how many of those 36 have a product over 10.
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Hint 2 of 3
Organize the count so you don't miss any: fix the FIRST die's value, then ask which second-die values push the product past 10. March through 1, 2, 3, 4, 5, 6 one row at a time.
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Hint 3 of 3
Low first values barely qualify (a 1 never works, a 2 needs a 6), so the winners pile up at the high end — count carefully there.
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Approach: fix the first die, count qualifying partners, then divide by 36
- There are 36 equally likely ordered outcomes. Go die-by-die and count partners giving product > 10: first die 1 → none; 2 → only 6 (gives 12), 1 way; 3 → 4, 5, 6, that's 3; 4 → 3, 4, 5, 6, that's 4; 5 → 3, 4, 5, 6, that's 4; 6 → 2, 3, 4, 5, 6, that's 5.
- Total winners: 1 + 3 + 4 + 4 + 5 = 17. So the probability is 17/36.
- Why count this way: sweeping through one die's values in order is a checklist that guarantees no outcome is double-counted or skipped — the safest method for ‘how many of the 36’ dice questions.
- Watch the boundary: ‘greater than 10’ excludes a product of exactly 10 (like 2×5 or 5×2), so those don't count — a classic trap.
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