Three fair dice are tossed once. What is the probability that they show different numbers that are in AP?

Explanation

The sequences of 3 numbers from 1 to 6 that form an AP are (1,2,3), (2,3,4), (3,4,5), (4,5,6) and their reverses, plus (1,3,5), (2,4,6) and their reverses. Total favorable sequences = 4 \times 2 + 2 \times 2 = 12. Total possible outcomes for three dice is 6^3 = 216. Probability = \frac{12}{216} = \frac{1}{18}.

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