Probability of At Least One Calculator

Find the chance that an event happens at least once across repeated independent tries — or flip the question around and find how many tries you need before the odds reach a target you choose.

Before You Calculate

  • Enter the per-trial probability as a decimal: a 5% chance per try is p = 0.05.
  • The formula assumes the tries are independent and share the same p — one outcome must not change the next.
  • “At least one” includes every count from 1 up to n, which is why the complement trick below is so much faster than adding those cases up.

The Complement Rule

Computing “at least one” directly means adding the probabilities of exactly 1, exactly 2, …, up to exactly n successes. The shortcut is to notice that the only outcome not included is zero successes:

P(at least one) = 1 − P(none)

P(none) = (1 − p)ⁿ

P(at least one) = 1 − (1 − p)ⁿ

Trials for a target probability T:

n = ⌈ ln(1 − T) / ln(1 − p) ⌉

Each independent trial fails with probability 1 − p, so n trials all fail with probability (1 − p)ⁿ, and everything else — the entire “at least one” event — is the complement. The trials-needed formula is the same equation solved for n and rounded up to the next whole trial.

Intuition: Small Chances Add Up, But Not Linearly

A tempting wrong answer is to multiply p by n: ten tries at 5% “should” give 50%. The true answer is 1 − 0.95¹⁰ ≈ 40.1%, because successes can overlap — some of those imagined 50 points double-count runs where the event would have happened twice. The gap widens as np grows: at 30 tries the naive rule says 150%, while the real probability is 78.5%. For very small p and small np the naive product is a decent approximation, which is exactly when overlaps are rare.

The other direction of the intuition also matters: rare events become likely given enough tries. A 1-in-100 risk per attempt crosses a 50% overall chance at just 69 attempts, which is why “it has never happened yet” is weak evidence of safety for frequently repeated processes.

Worked Example: Defects, Lotteries, and Bugs

Suppose each unit off a production line has a 5% chance of a defect and an inspector samples 10 units. What is the chance the sample contains at least one defect?

  1. Probability a single unit is clean: 1 − 0.05 = 0.95.
  2. Probability all ten are clean: 0.95¹⁰ ≈ 0.598737.
  3. Complement: 1 − 0.598737 ≈ 0.401263 — about a 40% chance of catching at least one defect.

Now invert it: how many units must be inspected for a 90% chance of finding at least one defect? n = ln(0.1) / ln(0.95) ≈ 44.9, so 45 units, which achieves 90.06%. Both computations match the calculator's two modes, and the same arithmetic covers lottery tickets, retry logic in software, and the chance a flaky test fails at least once in a night of runs.

Frequently Asked Questions

What is the formula for the probability of at least one success?

P(at least one) = 1 - (1 - p)^n, where p is the per-trial probability and n is the number of independent trials. It comes from the complement rule: the only way to avoid 'at least one' is to fail every single time, and n independent failures have probability (1 - p)^n.

Why can't I just multiply the probability by the number of trials?

Multiplying p by n double-counts the scenarios where the event happens more than once, so it overshoots - and it can even exceed 1, which is impossible for a probability. The product np is actually the expected number of successes, not the chance of at least one. The two are close only when p is small and np is well below 1.

How many trials do I need for a 50% or 90% chance of at least one success?

Solve 1 - (1 - p)^n >= T for n, which gives n = ln(1 - T) / ln(1 - p), rounded up. For a 1% event, that is 69 trials for a 50% chance and 230 trials for 90%. The calculator's second mode performs this computation and also reports the exact probability achieved at that whole number of trials.

Does the formula work if the trials are not independent?

No. Independence is what lets the failure probabilities multiply into (1 - p)^n. If one failure makes another more likely (defects clustering in a bad batch) the true 'at least one' probability differs from the formula. Sampling without replacement from a small population is a common case - the hypergeometric distribution calculator handles it exactly.

How do I find the probability of at least two successes?

Extend the complement to exclude both zero and one success: P(at least 2) = 1 - (1 - p)^n - n*p*(1 - p)^(n-1). For higher thresholds it is easier to use the binomial distribution calculator on this site, which returns P(X >= k) for any k directly.

Does a failure make a success more likely on the next try?

Not for independent trials - the coin, die, or lottery has no memory, and p stays the same on every attempt. What grows with more attempts is the overall probability that at least one succeeds somewhere in the batch. Expecting a win to be 'due' after a string of losses is the gambler's fallacy.

Related reading: the binomial distribution calculator generalizes this page to “at least k” for any k, and the geometric distribution calculator answers the closely related question of when the first success arrives.