Two fair dice are thrown together. Find the probability that one die shows a multiple of 2 and the other die shows a multiple of 3 (order doesn’t matter).

Introduction / Context:We are asked for the probability that, in a single toss of two fair dice, the pair of outcomes has one value that is a multiple of 2 and the other value that is a multiple of 3. Order does not matter, so pairs like (6,6) should be counted once overall, not twice.

Given Data / Assumptions:

• Each die has faces {1,2,3,4,5,6} with equal probability.
• Multiples: of 2 are {2,4,6}; of 3 are {3,6}.
• Sample space size = 36 ordered pairs.

Concept / Approach:Use inclusion–exclusion over the two ordered scenarios: (Die1 multiple of 2 & Die2 multiple of 3) OR (Die1 multiple of 3 & Die2 multiple of 2). Subtract the overlap where both dice show a common multiple of 2 and 3, namely 6.

Step-by-Step Solution:

Verification / Alternative check:Counting favorable ordered pairs directly: {2,3},{2,6},{4,3},{4,6},{6,3},{6,6} and the symmetric ones where roles swap; careful de-duplication yields 11 favorable outcomes out of 36.

Why Other Options Are Wrong:1/3 double-counts (6,6) without inclusion–exclusion. 1/4 and 7/26 come from ad-hoc approximations, not exact counting.

Common Pitfalls:Double-counting (6,6) because it satisfies both role assignments; forgetting that order is part of the 36 equally likely ordered outcomes.

Final Answer:11/36