LifeTally
Exactly k successes: binomial probability
What is the probability of exactly k successes in n independent trials?
Enter a trial count, an exact success count and the same success probability for every binary trial.
Trials, successes and probability
Enter whole n and k from 0 to 1000, with k ≤ n. Enter the per-trial probability from 0 to 100%; fractions such as 0.5 are allowed here, not 1/2 notation. All three values must be valid, including at zero.
How the estimate works
Assume a fixed number of independent trials, two outcomes per trial, and the same supplied success probability each time. Write p as the entered percentage divided by 100. C(n,k) counts the positions occupied by the k successes. Exactly k means one count, not at least or at most k.
p = % ÷ 100
P(X = k) = C(n,k) × pᵏ × (1 − p)ⁿ⁻ᵏ
C(n,k) = n! ÷ [k! × (n − k)!]The 1000-trial cap is our processing limit, not a mathematical bound or a recommended sample size. This model does not estimate p, test independence, calculate cumulative probabilities, handle changing probabilities or dependent draws, or guarantee an actual outcome.
NIST: binomial probability mass function (English) ↗Check the filled example
Our chosen example: n = 7, k = 2 and 40% per trial. C(7,2) = 21, so 21 × 0.4² × 0.6⁵ = 0.2612736, or 26.12736%. The display rounds these values.
Accepted input values
- Number of trials n
- 7
- Exact number of successes k
- 2
- Success probability per trial (%)
- 40
Plan the next step
Use the standard calculator to check the arithmetic of a small example.
Open the standard calculator →