Five fair coins are tossed simultaneously. What is the probability of obtaining at least one tail?

Introduction / Context:
For multiple independent coin tosses, it is often fastest to compute the complement of the desired event. “At least one tail” is the complement of “no tails,” i.e., all heads.

Given Data / Assumptions:

• Five independent fair coin tosses.
• P(H) = P(T) = 1/2.

Concept / Approach:
P(≥1 T) = 1 − P(all heads) = 1 − (1/2)^5.

Step-by-Step Solution:
P(all heads) = (1/2)^5 = 1/32.Therefore, P(≥1 tail) = 1 − 1/32 = 31/32.

Verification / Alternative check:
Enumerating outcomes confirms that only one of the 32 equally likely outcomes has zero tails.

Why Other Options Are Wrong:
1/32 is the complement event; 5/32 is the probability of exactly one tail, not at least one; 1/5 is unrelated.

Common Pitfalls:
Misreading “at least one” as “exactly one.”

Final Answer:
31/32