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Binomial Probability

Also known as exactly k successes

P(X=k)=(nk)pk(1−p) n−kP(X = k) = \binom{n}{k} p^{k} (1 - p)^{\,n-k}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

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Three ideas multiply here: p^k for the successes, (1 − p)^(n−k) for the failures, and the binomial coefficient for the number of orders in which they can occur. Exactly two heads in five fair tosses: C(5,2) × 0.5² × 0.5³ = 10 × 0.03125 = 0.3125. The distribution is Jacob Bernoulli's, worked out before his death in 1705 and published in Ars Conjectandi in 1713 by his nephew Nicolaus, together with the first law of large numbers.

Enter p as a decimal from 0 to 1, or switch that input to %. Three conditions must hold or the answer is meaningless: fixed n, independent trials, and a constant p — sampling without replacement from a small population violates the last two, and that is the hypergeometric distribution's job instead. There is no closed form for n, k or p given the probability, so this calculator solves only for P. And read the question carefully: "exactly k" is one term, while "at least k" means summing several. Three sixes in ten rolls of a fair die: C(10,3) × (1/6)³ × (5/6)⁷ ≈ 0.155.

Binomial Probability
P(X=k)=(nk)pk(1−p) n−kP(X = k) = \binom{n}{k} p^{k} (1 - p)^{\,n-k}
Where
  • P(X=k)P(X = k)= Probability of exactly k successes
  • nn= Number of trials
  • kk= Number of successes
  • pp= Probability of success per trial
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