Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

### Concept & Formula To find the probability, we first list out the sample space for tossing three coins and then count the favorable outcomes that satisfy the condition. $$P(E) = \frac{n(E)}{n(S)}$$ ### Step-by-Step Solution 1. **Find Sample Space:** When three coins are tossed, the possible outcomes are: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ Total outcomes, $n(S) = 2^3 = 8$. 2. **Identify Favorable Outcomes:** We want "at least 2 heads", which means exactly 2 heads or exactly 3 heads. $E = \{HHT, HTH, THH, HHH\}$ Number of favorable outcomes, $n(E) = 4$. 3. **Calculate Probability:** $P(E) = \frac{n(E)}{n(S)} = \frac{4}{8} = \frac{1}{2}$ ### Exam Strategy & Shortcut By symmetry, the probability of getting at least 2 heads (2 or 3) is equal to the probability of getting at least 2 tails (2 or 3). Since these two groups cover all 8 possible outcomes equally and don't overlap, each group must represent half of the total probability. Thus, it is $\frac{1}{2}$. ### Common Pitfall Misinterpreting "at least 2" as "exactly 2" and forgetting to include the case with 3 heads ($HHH$). ### Final Answer Therefore, the correct answer is **$\frac{1}{2}$**.