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Question

In tossing three coins at a time, the probability of getting at least one heads is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$\frac{7}{8}$

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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