A fair coin is tossed 6 times. Find probability of at least one tail.
Explanation
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Statement
The probability of getting at least one success in repeated independent trials can be calculated using the complement rule:
\[ P(\text{at least one success}) = 1 - (1-p)^n \]
Here, \(p\) is the probability of success in one trial, and \(n\) is the number of trials.
Why it’s true (short reason)
It’s often easier to calculate the probability of the opposite event.
The opposite of “at least one success” is “no successes.”
The probability of “no successes” is \((1-p)^n\), since every trial must fail.
Therefore, the probability of at least one success is \(1 - (1-p)^n\).
Recipe (how to use it)
Identify the probability of success in one trial, \(p\).
Work out the probability of failure: \(1-p\).
Raise this to the power \(n\), the number of trials.
Subtract from 1 to get the probability of at least one success.
Spotting it
Use this formula when:
You are asked for the probability of “at least one” success, event, or outcome.
The trials are independent (outcomes do not affect each other).
You know the probability of a single success and the total number of trials.
Common pairings
Binomial probability, where “at least one” is a common case.
Geometric probability, for waiting times until first success.
Complementary probability in general (e.g. “at least one head” vs “no heads”).
Mini examples
Coin tosses: Toss a fair coin 3 times. Probability of at least one head?
\(p=0.5, n=3.\)
\(P = 1 - (1-0.5)^3 = 1 - 0.125 = 0.875.\)
Dice rolls: Roll a die 4 times. Probability of at least one six?
\(p=1/6, n=4.\)
\(P = 1 - (5/6)^4 \approx 0.5177.\)
Pitfalls
Forgetting to subtract from 1.
Using \(p^n\) instead of \((1-p)^n\) for the complement.
Applying the formula when trials are not independent.
Mixing up “at least one” with “exactly one.”
Exam strategy
Underline the phrase “at least one” in the question.
Check the independence of events before applying the formula.
Always calculate failure probability first (\(1-p\)).
Use a calculator for powers to avoid mistakes with large \(n\).
Summary
The complement rule is a powerful shortcut: instead of adding probabilities for one success, two successes, and so on, you calculate the easier opposite event. With \(P(\text{at least one}) = 1 - (1-p)^n\), you can quickly handle problems involving multiple independent trials.
Worked examples
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A fair coin is tossed 3 times. Find the probability of at least one head.
\( p=0.5, n=3 \)
\( P = 1 - (1-0.5)^3 \)
\( P = 1 - 0.125 \)
\( P = 0.875 \)
Answer:
0.875
A die is rolled 4 times. Find the probability of at least one six.
\( p=1/6, n=4 \)
\( P = 1 - (5/6)^4 \)
P ≈ 1 - 0.4823
P ≈ 0.5177
Answer:
0.5177
A fair coin is tossed 5 times. Find the probability of at least one tail.
\( p=0.5, n=5 \)
\( P = 1 - (1-0.5)^5 \)
\( P = 1 - 0.03125 \)
\( P = 0.96875 \)
Answer:
0.969
A card is drawn from a deck and replaced. This is done 10 times. Find the probability of at least one ace.
\( p=4/52=1/13, n=10 \)
\( P = 1 - (12/13)^10 \)
P ≈ 0.555
Answer:
0.555
A die is rolled 8 times. Find the probability of at least one even number.
\( p=3/6=0.5, n=8 \)
\( P = 1 - (0.5)^8 \)
\( P = 1 - 0.0039 \)
P ≈ 0.996
Answer:
0.996
A machine has probability 0.1 of failing on any given day. Over 7 days, find the probability it fails at least once.
\( p=0.1, n=7 \)
\( P = 1 - (0.9)^7 \)
P ≈ 1 - 0.478
P ≈ 0.522
Answer:
0.522
\( A coin is biased with P(head)=0.7. Tossed 3 times, find probability of at least one head. \)
\( p=0.7, n=3 \)
\( P = 1 - (0.3)^3 \)
\( P = 1 - 0.027 \)
\( P = 0.973 \)
Answer:
0.973
A die is rolled 2 times. Find probability of at least one five.
\( p=1/6, n=2 \)
\( P = 1 - (5/6)^2 \)
\( P = 1 - 25/36 \)
\( P = 11/36 ≈ 0.306 \)
Answer:
0.306
A spinner has 1/4 chance of landing on red. It is spun 6 times. Find probability of at least one red.
\( p=0.25, n=6 \)
\( P = 1 - (0.75)^6 \)
P ≈ 1 - 0.178
P ≈ 0.822
Answer:
0.822
A basketball player scores a 3-point shot with probability 0.4. In 5 attempts, find probability of at least one score.