Three dice rolled: at least one shows 2: Three fair six-faced dice are rolled. How many outcomes in the 6^3 sample space have at least one die showing 2?

Introduction / Context:
Counting “at least one” events on dice is best done via the complement rule: subtract the count with no 2s from the total sample space of 6^3 equiprobable outcomes.

Given Data / Assumptions:

• Three independent fair dice, faces {1,…,6}.
• We count ordered triples (d1, d2, d3).

Concept / Approach:

• Total outcomes = 6^3.
• No-2 outcomes: each die has 5 allowed faces (exclude 2) → 5^3.
• At least one 2 = 6^3 − 5^3.

Step-by-Step Solution:

Verification / Alternative check:
Inclusion–exclusion by exact number of 2s (1, 2, 3) yields the same 91; complement is simpler.

Why Other Options Are Wrong:

• 36 and 81 are too small; 116 exceeds total remaining after subtraction.

Common Pitfalls:

• Mixing “count of outcomes” with “probability.”

Final Answer:
91