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

### Concept & Logic "At most" conditions are often best solved using the complement rule. "At most two heads" includes 0, 1, or 2 heads. The only case it does NOT include is 3 heads. $$P(E) = 1 - P(\text{Complement of } E)$$ ### Step-by-Step Solution 1. **Find Sample Space:** Total number of outcomes for 3 coins is $n(S) = 2^3 = 8$. 2. **Define Complement Event:** The event $E$ is "at most 2 heads". The complement $E'$ is "more than 2 heads", which means exactly 3 heads. $E' = \{HHH\}$ $n(E') = 1$. 3. **Calculate Complement Probability:** $P(E') = \frac{1}{8}$ 4. **Calculate Final Probability:** $P(E) = 1 - P(E') = 1 - \frac{1}{8} = \frac{7}{8}$ ### Exam Strategy & Shortcut Always use $1 - P(\text{all heads})$ when asked for "at most $(n-1)$ heads" in $n$ coin tosses. It saves listing out all 7 valid outcomes and eliminates counting errors. ### Common Pitfall A common mistake is thinking "at most two" means strictly less than two (0 or 1), omitting the exact 2 heads scenario. ### Final Answer Therefore, the correct answer is **$\frac{7}{8}$**.