Two fair dice are thrown. What is the probability that the sum is 3, 5, or 11 (any of the three totals)?

Introduction / Context:We aggregate favorable outcomes for three different target sums and divide by 36.

Given Data / Assumptions:

• Two fair dice → 36 equally likely ordered pairs.
• Target sums: 3, 5, 11.

Concept / Approach:Count outcomes for each sum and add them.

Step-by-Step Solution:Sum 3: (1,2), (2,1) → 2 outcomes.Sum 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes.Sum 11: (5,6), (6,5) → 2 outcomes.Total favorable = 2 + 4 + 2 = 8.Probability = 8 / 36 = 2 / 9.

Verification / Alternative check:Check that counts align with standard sum-distribution table for two dice.

Why Other Options Are Wrong:5/36 applies to sum 8; 1/9 is 4/36 (too small); 19/36 is much too large; 7/36 undercounts by one outcome.

Common Pitfalls:Missing symmetric pairs (e.g., both (2,3) and (3,2)).

Final Answer:2/9