BEO

Probability

Binomial Exact Outcome Calculator

Calculate the probability of exactly k successes in n independent equal-probability trials. Inspect the combination count, exact probability mass, expected successes, spread, observed trial share, and repeated-group expectation with a complete substitution.

Independent trialsExact successes of interestSuccess probability per trialEntered success-count comparisonNumber of identical trial groups

Exact binomial probability (decimal)-

Combinations of trials choose successes-

Expected successes per group-

Binomial variance-

Binomial standard deviation-

Exact success count as trial share-

Exact successes minus expected successes-

Expected groups with exact count-

Expected successes minus entered comparison-

Decision view

Exact binomial probability-mass distribution

Exact binomial probability-mass distributionEvery possible success count is plotted, with the selected count, expected center, and entered comparison highlighted.

Exact scenario comparisonSuccess probability per trial changes while all other entered assumptions remain constant.

Success probability per trial	Exact binomial probability (decimal)	Combinations of trials choose successes	Expected successes per group	Binomial variance	Binomial standard deviation	Exact success count as trial share	Exact successes minus expected successes	Expected groups with exact count	Expected successes minus entered comparison

How to use Binomial Exact Outcome Calculator

Enter a fixed trial count and an exact success count from zero through n.
Enter one constant per-trial success probability.
Use the exact probability for one count and the mean/SD for distribution context.
Enter repeated groups only to scale the exact-count expectation.

Calculator guide

Understanding Binomial Exact Outcome Calculator

A binomial exact-outcome probability is one point of a discrete distribution. This page calculates that single probability while also showing the distribution mean, variance, standard deviation, and the expected number of repeated groups with exactly that count.

Exact means one count The output is P(X=k), not a tail.

Arrangements matter C(n,k) counts all valid sequences.

Mean can be fractional np is a long-run average.

Groups scale mass G x P(X=k) is an expected repeated-group count.

Calculation method

How the calculation works

Evaluate the exact binomial mass at one success count using combinations and independent identical-trial probability, then show moments and repeated-group expectation. In the Binomial Exact Outcome Calculator, the live scenario varies success probability per trial and tracks exact binomial probability (decimal) while the remaining results preserve the reconciliation path. Count the ways to place k successes among n trials, multiply by p raised to k, and multiply by failure probability raised to n-k. Use np and np(1-p) for the distribution mean and variance.

Detailed calculation process

Count arrangements and weight one exact success total

The defaults ask for exactly 5 successes in 12 independent trials when each trial succeeds with probability 0.4.

General formula: P(X=k) = C(n,k) p^k (1-p)^(n-k); E[X] = np; Var(X) = np(1-p); SD(X) = sqrt[np(1-p)] C(n,k) counts success placements. The probability powers weight each placement, and the moment formulas describe the entire binomial distribution rather than only the selected count.

What each symbol means

n Fixed number of trials per group, measured as trials.

k Exact success count of interest, measured as successes.

p Constant success probability per trial, unitless.

C(n,k) Number of distinct ways to arrange k successes among n trials.

X Random success count across the n trials.

G Entered number of identical groups used for the expected-group count.

Worked substitution with the default inputs

1. Count success arrangements: C(12,5) = 12! / (5! x 7!) = 792 There are 792 trial sequences containing exactly five successes.

2. Weight one arrangement: 0.4^5 x 0.6^7 = 0.0002866545 Every qualifying sequence has five success probabilities and seven failure probabilities.

3. Calculate exact probability mass: P(X=5) = 792 x 0.4^5 x 0.6^7 = 0.227030 = 22.703% This is the probability of exactly five, not five or more.

4. Calculate center and spread: E[X] = 12x0.4 = 4.8; Var(X) = 12x0.4x0.6 = 2.88; SD = sqrt(2.88) = 1.69706 The selected count of five is 0.2 above the distribution mean.

5. Scale to repeated groups: 10 x 0.227030 = 2.2703 expected groups; observed share = 5/12 x 100 = 41.667% Expected group count may be fractional because it is a long-run average.

Exactly five successes has probability about 22.703%; the distribution mean is 4.8, SD is 1.697, and ten identical groups would average about 2.270 groups with exactly five successes.

Distribution position

See one count inside the full binomial shape

The highlighted bar is meaningful only in relation to neighboring counts and the expected center.

Selected count The exact k being evaluated.

Probability mass Height of the selected distribution bar.

Expected count The center np.

Spread The binomial standard deviation.

Worked situations

Practical examples

Twelve trials with p = 0.4 have an expected count of 4.8 successes.
Exactly five successes occurs with about 22.703% probability.
Ten repeated groups average 2.270 groups with exactly five successes.

Better inputs

Useful tips

Distinguish exact, cumulative, and tail probabilities.
Verify trials have approximately constant p and meaningful independence.
Report n, k, and p beside the probability.

Before relying on the result

Limitations and common mistakes

The binomial model requires a fixed trial count, two outcomes, constant p, and independent trials.
The calculator does not provide cumulative probability or overdispersion adjustments.
Dependence, heterogeneous probabilities, stopping rules, and clustered trials violate the model.

Reference

Key terms

Probability mass: Probability assigned to one discrete count.
Combination: Number of unordered placements of k successes among n trials.
Binomial mean: Expected success count np.

Important note

Calculated directly from the entered values using the displayed formula and rounding settings.

Frequently asked questions

Is this the probability of at least five successes?

No. It is the probability of exactly five; a tail probability would sum several counts.

Why is the expected count 4.8?

Expectation averages many repeated groups and therefore does not have to be an integer.

What if k is greater than n?

That is not a valid binomial count because successes cannot exceed trials.

Can p change by trial?

Not in the ordinary binomial model used here.