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Three dice are thrown simultaneously and the sum of the three numbers appearing on the top faces of the dice is found to be 10. The probability that these three numbers are distinct, is
Sum $$= 10$$ with three dice; favourable: probability that all three are distinct.
Unordered partitions of 10 with parts in $$[1,6]$$:
All distinct: $$\{1,3,6\}, \{1,4,5\}, \{2,3,5\}$$ → $$3 \times 6 = 18$$ ordered.
With repeats: $$\{2,2,6\}, \{2,4,4\}, \{3,3,4\}$$ → $$3 \times 3 = 9$$ ordered.
Total sum-10 triples $$= 27$$. P(distinct | sum 10) $$= \dfrac{18}{27} = \dfrac{\mathbf{2}}{\mathbf{3}}$$.
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