Question

Suppose three dice are rolled. Under the condition that no two show the same face, what is the probability that one of the faces is an ace?

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{1}{6}\)

Hint:

Alternatively, since all 3 faces are distinct, we are selecting 3 unique numbers out of 6.
The probability that the number 1 is included in any randomly chosen subset of 3 unique numbers out of 6 is:
\[ P = \frac{\binom{5}{2}}{\binom{6}{3}} = \frac{10}{20} = \frac{1}{2} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem can be solved using conditional probability.
The condition "no two show the same face" means that all three dice show different numbers.
We restrict our sample space only to these outcomes.

Step 2: Detailed Explanation:
Let us find the size of the restricted sample space, \(N(S)\).
Since three dice are rolled and they must all show unique faces:
- The first die can show any of the 6 faces.
- The second die must show a different face, so it has 5 possibilities.
- The third die must show a face different from the first two, leaving 4 possibilities.
Thus:
\[ N(S) = 6 \times 5 \times 4 = 120 \]
Now, let us find the number of favorable outcomes, \(N(A)\), where one of the three unique faces is an "ace" (the number 1).
Since one of the faces must be 1, we must choose the remaining two unique faces from the other 5 numbers \(\{2, 3, 4, 5, 6\}\).
The number of ways to choose 2 numbers from 5 is:
\[ \binom{5}{2} = 10 \]
Once we have our set of 3 unique numbers (which includes the number 1), we can arrange these 3 numbers on the three rolled dice in:
\[ 3! = 6 \text{ ways} \]
Therefore, the total number of favorable outcomes is:
\[ N(A) = \binom{5}{2} \times 3! = 10 \times 6 = 60 \]
The required probability is:
\[ P = \frac{N(A)}{N(S)} = \frac{60}{120} = \frac{1}{2} \]

Step 3: Final Answer:
The correct option is (B).