Difficulty: Easy
Correct Answer: 12/36
Explanation:
Introduction / Context:We classify die faces by residues modulo 3 and use independence. A 1–6 die has two faces of each residue class 0,1,2 (since 3 and 6 are 0 mod 3; 1,4 are 1 mod 3; 2,5 are 2 mod 3).
Given Data / Assumptions:Die outcomes are independent and uniform on {1,…,6}.
Concept / Approach:A sum is divisible by 3 when residues add to 0 mod 3: (0,0), (1,2), or (2,1). Each residue probability per die is 2/6.
Step-by-Step Solution:
P(0,0) = (2/6)*(2/6) = 4/36P(1,2) = (2/6)*(2/6) = 4/36P(2,1) = (2/6)*(2/6) = 4/36Total = 12/36 = 1/3Verification / Alternative check:Explicit counting yields 12 favorable ordered pairs out of 36 total, confirming 1/3.
Why Other Options Are Wrong:11/36 and 13/36 are off by ±1 pair; 17/36 is too large.
Common Pitfalls:Forgetting that there are two faces in each residue class; double-counting permutations.
Final Answer:12/36
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