Modern Math

Probability

Medium

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

$\frac{2}{3}$

$\frac{3}{5}$

$\frac{1}{2}$

$\frac{3}{4}$

✅ Correct Option: 1

List all unordered triples of dice values $(1$ to $6)$ that sum to $10$ :

Distinct triples: $(1, 3, 6)$ , $(1, 4, 5)$ , $(2, 3, 5)$

Triples with a repeat: $(2, 2, 6)$ , $(2, 4, 4)$ , $(3, 3, 4)$

Count ordered outcomes for each (since the three dice are distinguishable).

A triple with all distinct values gives $3! = 6$ arrangements.

A triple with one repeated value gives $\dfrac{3!}{2!} = 3$ arrangements.

Distinct outcomes $= 3 \times 6 = 18$

Repeated outcomes $= 3 \times 3 = 9$

Total outcomes with sum $10$ $= 18 + 9 = 27$

Required probability $=$

$\dfrac{18}{27} = \dfrac{2}{3}$

Answer $= \dfrac{2}{3}$

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