Binomial probability: exactly one is not at least one
Compare an exact success count with the complement of no successes.
Method
A binomial model counts successes in a fixed number of independent trials with the same success probability p. Exactly one success excludes two, three and more. At least one includes every positive count. For the latter, calculating one minus the probability of no successes is often shorter than adding every possible positive count.
Worked example
Consider four independent inspection checks, each with an assumed 25% chance of detecting a defect. Exactly one detection has probability 42.1875%, while at least one has probability 68.359375%. Find C(4,1) in Permutation and combination, then use Scientific calculator for the powers and multiplication in the equations below. These are assumed example probabilities, not measured detection performance.
n = 4; p = 0.25; k = 1
P(X = 1) = C(4,1) × 0.25 × 0.75^3 = 0.421875
P(X ≥ 1) = 1 − 0.75^4 = 0.68359375
42.1875% ≠ 68.359375%
Checks and limits
Do not add the four 25% chances: the resulting 100% is not the chance of at least one detection. Independence and unchanged probability are essential. Checks affected by the same defect or shared conditions may be correlated, so this model may be unsuitable. Probability of either event uses a supplied overlap; it is not a general binomial cumulative calculator.
Related calculators
Source: OpenStax