Two unbiased coins are tossed simultaneously. What is the probability of getting at least one head?

Introduction / Context:
“At least one” events are conveniently handled using complements. For two fair coins, we want the probability of at least one head in the pair of tosses.

Given Data / Assumptions:

• Two unbiased coins; outcomes are HH, HT, TH, TT.
• Each outcome has probability 1/4.
• Event: At least one head occurs.

Concept / Approach:
P(at least one head) = 1 − P(no heads). “No heads” means both coins show tails, i.e., TT.

Step-by-Step Solution:
P(TT) = 1/4.Therefore P(at least one head) = 1 − 1/4 = 3/4.

Verification / Alternative check:
Direct counting: favorable outcomes are HH, HT, TH — 3 outcomes out of 4 gives 3/4.

Why Other Options Are Wrong:
1/2 and 2/3 underestimate; 1/3 is the probability of exactly one tail, not the target event; 1/4 is P(no heads), i.e., the complement.

Common Pitfalls:
Confusing “at least one head” with “exactly one head”. The former includes HH as well.

Final Answer:
3/4