Two unbiased coins are tossed simultaneously. What is the probability that at most one head appears (that is, zero or one head)?

Introduction / Context:
This problem focuses on basic probability with a small and simple sample space. Tossing two unbiased coins is one of the most familiar examples in probability. The question asks for the probability of getting at most one head, which means the number of heads can be zero or one, but not two. This tests understanding of sample spaces and counting favourable outcomes correctly.

Given Data / Assumptions:

• Two unbiased (fair) coins are tossed simultaneously.
• Each coin has two outcomes: Head (H) or Tail (T).
• All four possible ordered outcomes are equally likely.
• We require the probability of getting at most one head (0 or 1 head).

Concept / Approach:
For two coin tosses, it is easiest to list all possible outcomes. Then we count how many outcomes satisfy the condition "at most one head". At most one head means the allowed numbers of heads are 0 and 1. We then divide the count of favourable outcomes by the total number of possible outcomes in the sample space.

Step-by-Step Solution:
List the sample space for two coin tosses: {HH, HT, TH, TT}. Total number of possible outcomes = 4. Outcomes with zero heads: TT (this has 0 heads). Outcomes with exactly one head: HT and TH (each has 1 head). So the favourable outcomes are: TT, HT, TH. Number of favourable outcomes = 3. Probability = favourable outcomes / total outcomes = 3 / 4.

Verification / Alternative check:
An alternative way is to recognize that the only outcome that does not satisfy the requirement is HH, which has two heads. The probability of HH is 1 out of 4. Therefore, the probability of at most one head is 1 minus the probability of two heads: 1 - 1/4 = 3/4. This matches the previous calculation exactly.

Why Other Options Are Wrong:
1/2: This would imply that half the outcomes are favourable, but we have 3 favourable outcomes out of 4, not 2 out of 4.
3/2: This is greater than 1 and therefore cannot be a probability value at all.
1/6: This is much too small and does not reflect the actual count of favourable outcomes in a four outcome space.
None of these: This is incorrect because 3/4 is present and is correct.

Common Pitfalls:
Many learners misinterpret the phrase "at most one head" and think it means exactly one head, missing the zero head case. Another mistake is to forget that the order of outcomes matters in the sample space, which always has four outcomes for two unbiased coins. Also, probabilities greater than 1 should be immediately rejected, since they are invalid in any probability problem.

Final Answer:
Thus, the probability that at most one head appears when two unbiased coins are tossed is 3/4.