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P(at least one success)

P(none at all)
As Fraction
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How to Calculate "At Least One" Probability

When you have n independent attempts, each with the same probability p of success, the probability that at least one of them succeeds is found with the complement rule: subtract the probability that all of them fail from 1. P(at least one)=1(1p)nP(\text{at least one}) = 1 - (1-p)^n

p: the probability of success on a single attempt, as a decimal.

n: the number of independent attempts.

1-p: the probability of failure on a single attempt.

Why the Complement Trick Works

Calculating "at least one success" directly means adding up P(exactly 1 success) + P(exactly 2 successes) + P(exactly 3 successes) and so on, all the way up to P(exactly n successes) — every one of those needs its own binomial term, and the list gets long fast as n grows. The complement sidesteps all of that. There is exactly one way to get zero successes: every single attempt has to fail, and since the attempts are independent, those failure probabilities simply multiply together as (1p)n(1-p)^n — one clean term instead of n separate ones. Because "at least one" and "zero successes" are the only two possible outcomes and together account for 100% of the outcomes, subtracting that one term from 1 gives the exact same answer as the long addition, with a fraction of the work.

Independent Attempts

The formula assumes every attempt is independent — the outcome of one attempt has no effect on the probability of any other attempt succeeding or failing, and every attempt shares the same probability p. This fits scenarios like repeated coin flips, repeated dice rolls, or repeated random draws with replacement. It does not fit scenarios where the probability itself changes from attempt to attempt, such as drawing cards from a deck without putting them back (see the Card Draw calculator for that case).

Worked Examples

  1. 10% chance, 5 attempts. p=0.1p=0.1, n=5n=5: P(none)=0.95=59.049%P(\text{none}) = 0.9^5 = 59.049\%, so P(at least one)=10.59049=40.951%P(\text{at least one}) = 1 - 0.59049 = 40.951\%.
  2. 50% chance, 3 attempts. p=0.5p=0.5, n=3n=3: P(none)=0.53=12.5%P(\text{none}) = 0.5^3 = 12.5\%, so P(at least one)=10.125=87.5%P(\text{at least one}) = 1 - 0.125 = 87.5\%.
  3. A rare 1% event, 100 tries. p=0.01p=0.01, n=100n=100: P(none)=0.9910036.6%P(\text{none}) = 0.99^{100} \approx 36.6\%, so P(at least one)63.4%P(\text{at least one}) \approx 63.4\% — far from guaranteed, even after 100 tries.
  4. 20% chance, 10 attempts. p=0.2p=0.2, n=10n=10: P(none)=0.81010.74%P(\text{none}) = 0.8^{10} \approx 10.74\%, so P(at least one)89.26%P(\text{at least one}) \approx 89.26\%.
  5. 5% chance, 20 attempts. p=0.05p=0.05, n=20n=20: P(none)=0.952035.85%P(\text{none}) = 0.95^{20} \approx 35.85\%, so P(at least one)64.15%P(\text{at least one}) \approx 64.15\%.

At Least One Terms You Should Know

Independent Trial — an attempt whose outcome doesn't affect, and isn't affected by, any other attempt in the sequence.

Complement Rule — the probability of an event happening equals 1 minus the probability that it doesn't happen; the fastest route to an "at least one" answer.

Success Probability (p) — the chance of success on a single attempt, expressed as a decimal between 0 and 1.

Cumulative Probability — the combined probability of an event happening across a whole series of attempts rather than just one; "at least one" is a form of cumulative probability.

Frequently Asked Questions

What does "at least one" mean in probability?

"At least one" means one success or more out of a number of independent attempts — it includes exactly one success, exactly two, exactly three, and so on, all the way up to every attempt succeeding. The only outcome it excludes is zero successes, which is why it's fastest to calculate as 1 minus the probability of zero successes.

Why do we calculate the complement?

Because directly adding up P(exactly 1) + P(exactly 2) + ... + P(exactly n) gets combinatorially messy fast, especially as n grows. The complement — P(zero successes) — is always a single clean term, (1-p)^n, so subtracting that one term from 1 gives the exact same answer with dramatically less work.

Does increasing the number of attempts always increase the probability?

Yes, as long as the per-attempt probability is greater than zero. Each additional attempt shrinks P(none) further, since (1-p)^n keeps getting smaller as n grows, which means P(at least one) always climbs — though it only approaches 100% and never quite reaches it for a finite number of attempts unless p itself equals 1.

What if each attempt has a different probability of success?

This calculator assumes every attempt shares the same probability p. If your attempts have different probabilities, the complement rule still works, but you multiply each attempt's own (1 minus its probability) term individually — P(none) = (1-p1) x (1-p2) x (1-p3) x ... — then subtract that product from 1.

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