Three fair coins are tossed simultaneously. What is the probability of getting at most two heads?

Introduction / Context:
“At most two heads” means 0, 1, or 2 heads. It is often fastest to use the complement: 1 minus the probability of 3 heads.

Given Data / Assumptions:

• Three independent fair coin tosses.
• Each toss: P(H) = 1/2.

Concept / Approach:
P(at most 2H) = 1 − P(3H). Since P(3H) = (1/2)^3, subtract from 1.

Step-by-Step Solution:
P(3 heads) = (1/2)^3 = 1/8.Therefore P(at most two heads) = 1 − 1/8 = 7/8.

Verification / Alternative check (enumeration):
Outcomes with ≤2 heads: 7 out of 8 ({TTT, HTT, THT, TTH, HHT, HTH, THH}). Only HHH is excluded.

Why Other Options Are Wrong:
1/4 and 3/8 undercount; 1/2 is the probability of exactly 1 or exactly 2 heads combined for two coins, not three.

Common Pitfalls:
Counting mistakes when listing outcomes; using “at least” instead of “at most”.

Final Answer:
7/8