Exactly five successes
For n=10, p=0.5, and x=5, C(10,5)×0.5¹⁰=0.24609375, or 24.609375%.
Calculate exact, cumulative, tail, and range probabilities for a binomial model, with its expected value, variance, standard deviation, and distribution.
A binomial distribution models the number of successes in a fixed number of independent trials when every trial has two outcomes and the same success probability.
The model requires fixed n, constant p, independent trials, and a consistent definition of success. If those conditions do not fit, binomial results may not represent the situation.
P(X=x) = C(n,x)pˣ(1−p)ⁿ⁻ˣμ = npσ² = np(1−p)σ = √(np(1−p))P(X≤x) = Σ P(X=k), k=0…xFor n=10, p=0.5, and x=5, C(10,5)×0.5¹⁰=0.24609375, or 24.609375%.
For n=5 and p=0.2, 1−0.8⁵=0.67232, or 67.232%.
For n=100 and p=0.2, mean=20, variance=16, and standard deviation=4.
Choose the number of trials before observing outcomes.
The success probability must remain the same on each trial.
One trial's outcome must not change another trial's probability.
Exactly uses one distribution point. At most sums from zero through x. At least sums from x through n. Fewer than excludes x, while more than excludes x in the other direction.
The distribution is generated with a recurrence centered at its mode, then normalized. This avoids converting enormous combination values to floating point and multiplying them by tiny powers. Exact combinations are still shown for educational output.
Calculations support up to 10,000 trials. Full tables and charts are shown through 200 trials; larger calculations show a focused 41-row window and omit the chart to avoid excessive DOM and rendering work.
It is the probability of a specified number or range of successes in a fixed set of independent, constant-probability trials.
It needs fixed trials, two outcomes per trial, constant success probability, and independent trials.
n is trials, p is success probability per trial, and x is the number of successes.
Use C(n,x)p^x(1−p)^(n−x).
Subtract the probability of zero successes from one.
At most x includes every success count from zero through x.
The expected success count is np; it need not be a possible whole-number outcome.
Binomial variance is np(1−p).
Independence is an assumption required by the binomial model.
Not in a binomial model; p must remain constant.
No. This calculator numerically evaluates the binomial distribution itself.
Yes, when flips are modeled as independent with constant head probability.