Probability of At Least One Head in Three Coin Tosses
This solution calculates the probability of obtaining at least one head when tossing three coins simultaneously.
Total Possible Outcomes
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.
Calculating Probability Using the Complement
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.
Complementary Event: No Heads
The only outcome with no heads is getting all tails (TTT).
Probability of the Complementary Event
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:
$P(\text{No Heads}) = P(\text{TTT}) = P(T) \times P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$
Probability of At Least One Head
The probability of the event we are interested in (at least one head) is 1 minus the probability of its complement (no heads):
$P(\text{At least one head}) = 1 - P(\text{No Heads})$
$P(\text{At least one head}) = 1 - \frac{1}{8} = \frac{8}{8} - \frac{1}{8} = \frac{7}{8}$
Conclusion
The probability of getting at least one head when tossing three coins is $\frac{7}{8}$.
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Two dice are thrown simultaneously. What is the probability of getting the same number on both the dice?
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