This solution calculates the probability of obtaining at least one head when tossing three coins simultaneously.
Each coin toss has two possible outcomes: Heads (H) or Tails (T). When tossing three coins, the total number of possible outcomes is calculated as:
Total Outcomes = $2^n$, where $n$ is the number of coins.
In this case, $n=3$, so the total number of outcomes is $2^3 = 8$. These outcomes are: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
It's often easier to calculate the probability of the event *not* happening and subtract it from 1. The complement of getting "at least one head" is getting "no heads" at all.
The only outcome with no heads is getting all tails (TTT).
The probability of getting tails on a single fair coin toss is $\frac{1}{2}$. Since the tosses are independent, the probability of getting three tails (TTT) is:
$P(\text{No Heads}) = P(\text{TTT}) = P(T) \times P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
The probability of the event we are interested in (at least one head) is 1 minus the probability of its complement (no heads):
$P(\text{At least one head}) = 1 - P(\text{No Heads})$
$P(\text{At least one head}) = 1 - \frac{1}{8} = \frac{8}{8} - \frac{1}{8} = \frac{7}{8}$
The probability of getting at least one head when tossing three coins is $\frac{7}{8}$.
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