The faces of a cube are labeled from 1 to 6, and when the cube is tossed, each number is...

1. Translate the problem requirements: We need to find the probability that when two dice are rolled, one shows a number that is exactly 2 more than the other. This means we're looking for pairs like \((1,3)\), \((3,1)\), \((2,4)\), \((4,2)\), etc.
2. Identify all favorable outcomes systematically: List all possible pairs where the difference is exactly 2, considering both cases where the first die is higher and where the second die is higher.
3. Count total possible outcomes: Determine the total number of ways two dice can land.
4. Calculate probability using basic ratio: Use the fundamental probability formula of favorable outcomes divided by total outcomes.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in simple terms. We have two dice (cubes with faces labeled 1 through 6), and we're rolling both of them. We want to find the probability that one die shows a number that is exactly 2 more than the other die.

For example, if the first die shows 3 and the second die shows 1, then the first die "exceeds the other by exactly 2" because \(3 - 1 = 2\). Similarly, if the first die shows 1 and the second die shows 3, then the second die exceeds the first by exactly 2.

So we're looking for pairs of numbers where the absolute difference is exactly 2.

Process Skill: TRANSLATE - Converting the phrase "exceeds the number on the other by exactly 2" into mathematical understanding of absolute difference = 2

2. Identify all favorable outcomes systematically

Let's think about this step by step. If two numbers differ by exactly 2, what are all the possible pairs?

Starting with the smaller number:

• If one die shows 1, the other must show 3 (difference = 2)
• If one die shows 2, the other must show 4 (difference = 2)
• If one die shows 3, the other must show 5 (difference = 2)
• If one die shows 4, the other must show 6 (difference = 2)

Now, since we have two dice, we need to consider which die shows which number. Let's call them Die A and Die B:

Favorable outcomes:

• \((1,3)\): Die A shows 1, Die B shows 3
• \((3,1)\): Die A shows 3, Die B shows 1
• \((2,4)\): Die A shows 2, Die B shows 4
• \((4,2)\): Die A shows 4, Die B shows 2
• \((3,5)\): Die A shows 3, Die B shows 5
• \((5,3)\): Die A shows 5, Die B shows 3
• \((4,6)\): Die A shows 4, Die B shows 6
• \((6,4)\): Die A shows 6, Die B shows 4

So we have exactly 8 favorable outcomes.

Process Skill: CONSIDER ALL CASES - Systematically accounting for both directions of the difference

3. Count total possible outcomes

When we roll two dice, each die can show any of 6 numbers (1, 2, 3, 4, 5, or 6). Since the dice are independent, we multiply the possibilities:

Total possible outcomes = \(6 \times 6 = 36\)

We can think of this as a grid where each cell represents one possible outcome when rolling both dice.

4. Calculate probability using basic ratio

Now we can use the fundamental probability formula:

Probability = Number of favorable outcomes ÷ Total number of possible outcomes

Probability = \(8 \div 36\)

To simplify this fraction, we find the greatest common divisor of 8 and 36:
\(8 = 2^3\)
\(36 = 2^2 \times 3^2\)
\(\mathrm{GCD} = 2^2 = 4\)

So: \(8 \div 36 = (8 \div 4) \div (36 \div 4) = 2 \div 9 = \frac{2}{9}\)

4. Final Answer

The probability that one cube exceeds the other by exactly 2 is \(\frac{2}{9}\).

Verifying against the answer choices, this matches choice D: \(\frac{2}{9}\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "either cube exceeds the other"

Students often interpret this as only one direction of difference (e.g., only first die > second die by 2) rather than understanding it means the absolute difference should be 2. They miss that "either" means we need to consider both Die A exceeding Die B AND Die B exceeding Die A.

2. Confusing "exactly 2" with "at least 2"

Some students might incorrectly include cases where the difference is 3, 4, or 5, thinking the problem asks for differences of 2 or more. The word "exactly" is crucial but easily overlooked.

Errors while executing the approach

1. Incomplete enumeration of favorable outcomes

Students often list only the pairs where the first die is smaller: \((1,3)\), \((2,4)\), \((3,5)\), \((4,6)\) and forget to include the reverse cases: \((3,1)\), \((4,2)\), \((5,3)\), \((6,4)\). This leads to counting only 4 favorable outcomes instead of 8.

2. Arithmetic errors in fraction simplification

When simplifying \(\frac{8}{36}\), students may incorrectly reduce it. Common mistakes include getting \(\frac{4}{18}\), \(\frac{1}{9}\), or failing to simplify at all, rather than the correct \(\frac{2}{9}\).

Errors while selecting the answer

1. Selecting \(\frac{1}{9}\) instead of \(\frac{2}{9}\)

Students who only counted 4 favorable outcomes (missing the bidirectional nature) would get \(\frac{4}{36} = \frac{1}{9}\) and select choice E instead of the correct choice D (\(\frac{2}{9}\)).