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

Introduction / Context:
This problem is related to coin toss probability and uses the phrase at most two heads, which means zero, one or two heads allowed, but not three. The easiest way is to consider the complement event of getting three heads and then subtract this probability from 1.

Given Data / Assumptions:

• Three fair coins are tossed at the same time.
• Each coin has equal chance of head or tail.
• Total outcomes = 2^3 = 8, all equally likely.
• Event of interest: at most two heads (0, 1 or 2 heads).

Concept / Approach:
At most two heads is the complement of exactly three heads. Therefore P(at most two heads) = 1 minus P(three heads). Since each of the eight outcomes is equally likely, counting the single outcome with three heads is straightforward. This approach avoids listing all 8 outcomes in detail.

Step-by-Step Solution:
Total outcomes = 2^3 = 8. Outcome with three heads is HHH, so there is exactly 1 favourable outcome for three heads. P(three heads) = 1 / 8. P(at most two heads) = 1 - P(three heads) = 1 - 1 / 8 = 7 / 8.

Verification / Alternative check:
We can directly count outcomes with 0, 1 or 2 heads. Zero heads: TTT (1 outcome). One head: HTT, THT, TTH (3 outcomes). Two heads: HHT, HTH, THH (3 outcomes). Total with at most two heads = 1 + 3 + 3 = 7 outcomes. Probability = 7 / 8, which agrees with the complement method.

Why Other Options Are Wrong:
3/4 would mean 6 favourable outcomes, but there are 7. 1/2 implies only 4 favourable outcomes out of 8, which is too small. 1/4 is the probability of getting exactly one head in some cases, not at most two heads.

Common Pitfalls:
A common error is misinterpreting at most two heads as exactly two heads, which would give 3/8. Others may forget to include the zero head case. Always translate at most as less than or equal to the specified number and consider using the complement if it simplifies the counting.

Final Answer:
Therefore, the probability of getting at most two heads is 7/8.