Probability of Distinct Dice Rolls Including an Ace

The question asks for a conditional probability. We need to find the probability that one of the three dice shows an ace ('1'), given the condition that all three dice show different numbers.

Understanding Conditional Probability

Conditional probability measures the likelihood of an event occurring when another event is known to have already occurred. The formula is:

\( P(A|B) = \frac{P(A \cap B)}{P(B)} \)

Where:

• P(A|B) denotes the probability of event A occurring given that event B has occurred.
• P(A \cap B) is the probability that both events A and B occur together.
• P(B) is the probability of event B occurring.

For problems involving counting outcomes, like this dice rolling scenario, we can use the number of outcomes directly:

\( P(A|B) = \frac{\text{Number of outcomes favourable to both A and B}}{\text{Number of outcomes favourable to B}} \)

In this problem:

• Event B is: No two dice show the same face (all dice are distinct).
• Event A is: One of the faces shown is an ace ('1').

We need to calculate \( P(A|B) \). Let's find the number of outcomes for event B and for the intersection event \( A \cap B \).

Analyzing the Sample Space with Distinct Dice Rolls

First, let's calculate the total number of ways to roll three dice such that no two show the same face (Event B).

• The first die can land on any of the 6 faces.
• The second die must land on a different face than the first die, giving 5 possible outcomes.
• The third die must land on a different face than the first two dice, leaving 4 possible outcomes.

The total number of outcomes where all three dice show distinct faces is:

\( N(B) = 6 \times 5 \times 4 = 120 \)

This count (120) represents the size of our reduced sample space because we are given this condition.

Identifying Favourable Outcomes with an Ace

Now, let's find the number of outcomes that satisfy both conditions: the faces are distinct AND one of the faces is an ace ('1') (Event \( A \cap B \)).

We need outcomes where the three dice show '1' and two other distinct numbers from the set {2, 3, 4, 5, 6}.

We can determine this count using the following steps:

1. Choose the numbers: We must have '1'. We need to choose 2 more distinct numbers from the remaining 5 available faces ({2, 3, 4, 5, 6}). The number of ways to choose these 2 numbers is given by the combination formula:

   \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \)

   So there are 10 possible sets of three distinct numbers that include '1' (e.g., {1, 2, 3}, {1, 2, 4}, ..., {1, 5, 6}).
2. Arrange the numbers: For each set of three distinct numbers (like {1, 2, 3}), there are \(3!\) ways to arrange them across the three dice.

   \( 3! = 3 \times 2 \times 1 = 6 \)

   For example, the set {1, 2, 3} can be arranged as (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1).
3. Total Favourable Outcomes: Multiply the number of ways to choose the numbers by the number of ways to arrange them.

   \( N(A \cap B) = \binom{5}{2} \times 3! = 10 \times 6 = 60 \)

Alternatively, we can think about placing the numbers directly:

• The '1' can appear on the first, second, or third die (3 possibilities).
• Suppose '1' is on the first die. The second die must be different from '1' (5 choices: 2 to 6). The third die must be different from '1' and the second die (4 choices). This gives \( 1 \times 5 \times 4 = 20 \) outcomes.
• Since the '1' could be on any of the three dice, we multiply by 3: \( 3 \times (1 \times 5 \times 4) = 3 \times 20 = 60 \).

Thus, there are 60 outcomes where the three dice show distinct faces and one of them is an ace ('1').

Calculating the Final Probability

We can now calculate the conditional probability \( P(A|B) \) using the formula:

\( P(\text{One face is Ace} | \text{All faces are distinct}) = \frac{N(A \cap B)}{N(B)} \)

Substituting the values we found:

\( P(A|B) = \frac{60}{120} \)

Simplifying the fraction:

\( P(A|B) = \frac{1}{2} \)

The probability that one of the faces shown is an ace, given that no two dice show the same face, is 1/2.