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Recognize that the sum of two fair dice ranges from 2 to 12, so the only way to get a sum not less than 12 is to roll a 12 (6+6). Calculate the probability of rolling a 12 (1/36) and subtract from 1 to get 35/36. Alternatively, count all 36 outcomes and subtract the one outcome that sums to 12.
Pro tip: Show that you can solve it both by complement and by direct enumeration, and mention that this is a common trick question where the obvious answer (11/12) is wrong because it ignores the distribution of sums.
State that there are 36 equally likely outcomes when rolling two fair dice (6 sides each).
Clarify that 'sum less than 12' means sum ≤ 11, and the complement is sum ≥ 12, which only occurs for sum = 12.
Count the number of outcomes where sum < 12: all except (6,6), so 35 outcomes.
Divide favorable outcomes by total outcomes: 35/36.
Alternatively, compute P(sum=12) = 1/36, then P(sum<12) = 1 - 1/36 = 35/36.
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