Three perfect dice are rolled. Under the condition that no two show the same face, what is the probability that one of the faces shown is an ace (one)?

Explanation

The total number of ways to roll three dice such that no two show the same face is P(6,3) = 6 \times 5 \times 4 = 120. The number of ways where exactly one of the faces is a 1 involves choosing which of the 3 dice shows the 1 (3 ways), and then filling the other two dice with distinct numbers from the remaining 5 choices (5 \times 4 = 20 ways). This gives 3 \times 20 = 60 favorable outcomes. The probability is \frac{60}{120} = \frac{1}{2}.

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