In a simultaneous throw of two coins, the probability of getting at least one head is

### Concept & Formula The probability of an event $E$ is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes in the sample space $S$. $$P(E) = \frac{n(E)}{n(S)}$$ ### Step-by-Step Solution 1. **Find Sample Space:** When two coins are tossed simultaneously, the possible outcomes are (Head, Head), (Head, Tail), (Tail, Head), and (Tail, Tail). $S = \{HH, HT, TH, TT\}$ Total number of outcomes, $n(S) = 4$. 2. **Identify Favorable Outcomes:** The event is getting "at least one head", which means we want outcomes with 1 head or 2 heads. $E = \{HT, TH, HH\}$ Number of favorable outcomes, $n(E) = 3$. 3. **Calculate Probability:** $P(E) = \frac{n(E)}{n(S)} = \frac{3}{4}$ ### Exam Strategy & Shortcut Instead of counting "at least one", you can use the complement rule: $P(\text{at least one head}) = 1 - P(\text{no heads})$. The only outcome with no heads is $TT$ (1 outcome). So, $P(\text{no heads}) = \frac{1}{4}$. $P(\text{at least one head}) = 1 - \frac{1}{4} = \frac{3}{4}$. ### Common Pitfall Students often count $\{HT\}$ and $\{TH\}$ as a single outcome, mistakenly thinking there are only 3 total outcomes (two heads, two tails, one of each), which leads to an incorrect probability. ### Final Answer Therefore, the correct answer is **$\frac{3}{4}$**.