Calculate the probability of at least one success in multiple independent trials. Covers coin flips, dice rolls, defect rates, and any repeated event.
The "probability of at least one" is one of the most practical probability calculations in everyday life and engineering. Whether you're rolling dice, testing for defective parts, estimating the chance of rain over a week, or assessing security vulnerabilities, the complement rule makes this calculation elegant and straightforward.
This calculator computes P(at least one success) = 1 - P(no successes) = 1 - (1-p)ⁿ, where p is the probability of success on a single trial and n is the number of independent trials. It also shows the full binomial distribution for exactly k successes, expected value, and standard deviation.
From quality control (what's the chance of finding at least one defect in a batch?) to gaming (what's the chance of rolling at least one six in four dice?) to reliability engineering (what's the chance of at least one system failure in a year?), this tool handles them all with visual breakdowns and comparison tables.
The complement rule is the most efficient way to solve "at least one" problems. This calculator automates it and shows the full distribution for deeper analysis. It is useful when you need a fast answer for sampling, reliability, or game-probability questions without expanding the whole binomial expression by hand. That saves time when you need a quick risk estimate from only two inputs.
P(at least 1) = 1 - (1-p)ⁿ. P(exactly k) = C(n,k) × pᵏ × (1-p)ⁿ⁻ᵏ. Expected value: E = n × p. Standard deviation: σ = √(n × p × (1-p)).
Result: P(at least one 6) = 51.8%
Rolling 4 dice, each with 1/6 chance of a 6: P(≥1 six) = 1 - (5/6)⁴ = 51.8%. You're slightly more likely than not to get at least one six.
The complement rule states P(A) = 1 - P(not A). For "at least one" problems, P(at least one success) = 1 - P(zero successes). Since P(zero successes in n independent trials) = (1-p)ⁿ, the formula becomes P(≥1) = 1 - (1-p)ⁿ. This single expression replaces what would otherwise require summing n terms.
Quality control: Sampling inspection plans use this formula to determine how many items to inspect to achieve a desired detection probability. Security: The probability of at least one successful attack attempt over n attempts guides defense strategy. Medicine: The chance of at least one false positive in multiple diagnostic tests determines screening protocols.
The full binomial distribution gives P(exactly k successes) = C(n,k) × pᵏ × (1-p)ⁿ⁻ᵏ. The "at least one" probability is the sum of P(1) through P(n), which equals 1 - P(0). The distribution's shape depends on n and p: symmetric when p ≈ 0.5, right-skewed when p is small, and left-skewed when p is large.
Calculating P(at least 1) directly requires summing P(1) + P(2) + ... + P(n). The complement P(0 successes) is a single calculation: (1-p)ⁿ, making it much simpler.
Each trial's outcome doesn't affect the others. Coin flips are independent; drawing cards without replacement is not (use hypergeometric distribution instead).
Solve n = ln(1-0.95) / ln(1-p). For p=0.1, you need about 29 trials. For p=0.01, about 299 trials.
This calculator assumes equal probability per trial. For varying probabilities, P(at least 1) = 1 - ∏(1-pᵢ), which requires a different calculation.
The birthday problem uses the complement rule: P(at least one shared birthday) = 1 - P(all different). With 23 people, it exceeds 50%.
If a defect rate is 2% and you inspect 50 items, P(at least 1 defect) = 1 - 0.98⁵⁰ = 63.6%. This guides sampling plan design.