Binomial Probability

Binomial probability describes the number of successes in a fixed number of independent trials when each trial has the same success probability. “Success” means the event being counted; it does not have to be a favourable result.

The model combines two ideas: a specified success-and-failure sequence has a product probability, and combinations count how many sequences contain the requested number of successes.

On this page

Check the conditions before calculating

Let X count successes in n trials. The binomial model requires a fixed non-negative integer n, two categories per trial, mutually independent trials, and a common success probability p. For a nontrivial model below, n is positive and $0<p<1$ .

Write

X\sim\operatorname{Bin}(n,p).

A fair die can produce a binary trial by classifying six as success and every other face as failure, giving $p=1/6$ . The original experiment need not have only two raw outcomes; it must have two categories for the event being counted.

A changing success probability or dependence generally produces a different count distribution. The NIST binomial reference gives the probability formula and its mean and variance.

Derive the probability of an exact count

For five independent trials with success probability $0.3$ , one specified sequence containing two successes and three failures has probability

(0.3)^2(0.7)^3=0.03087.

There are $\binom52=10$ ways to choose the two success positions. These sequences are mutually exclusive, so add their equal probabilities:

P(X=2)=\binom52(0.3)^2(0.7)^3=0.3087.

In general,

P(X=k)=\binom nk p^k(1-p)^{n-k},\qquad k=0,1,\ldots,n.

The combination counts the arrangements of successes; the powers give the probability of one arrangement. Omitting the combination factor calculates only one specified sequence, except in cases where just one sequence is possible.

Read the complete distribution

For $X\sim\operatorname{Bin}(5,0.3)$ , the exact terminating-decimal probabilities are:

Read the complete distribution: reference table
Successes k	$P(X=k)$	$P(X\le k)$
0	0.16807	0.16807
1	0.36015	0.52822
2	0.30870	0.83692
3	0.13230	0.96922
4	0.02835	0.99757
5	0.00243	1.00000

Ten sequences contain exactly two successes

The distinct count values have different probabilities. Bars begin at zero; the highlighted k=2 bar has mass 0.30870.

Diagram explanation and exact data

Each sequence with two successes has probability 0.03087. There are ten such sequences, giving total 0.3087. The six count masses sum to one and their weighted mean is 1.5.

Ten sequences contain exactly two successes: exact reference data
k	Exact probability	Decimal
0	16807/100000	0.16807
1	7203/20000	0.36015
2	3087/10000	0.30870
3	1323/10000	0.13230
4	567/20000	0.02835
5	243/100000	0.00243

The masses sum to one. The most likely individual count is one, although the mean will be 1.5. Success-count values are not equally likely simply because they are listed in consecutive order.

Translate exact, at most and at least carefully

“At most two” includes 0, 1 and 2:

P(X\le2)=0.16807+0.36015+0.30870=0.83692.

“At least three” is its complement:

P(X\ge3)=1-P(X\le2)=0.16308.

“At least one” instead excludes only zero:

P(X\ge1)=1-(0.7)^5=0.83193.

For an integer count, $P(X\ge k)=1-P(X\le k-1)$ . Subtracting $P(X\le k)$ would exclude k itself and answer “more than k”. When using a calculator or software CDF, write the event in symbols before entering the cutoff.

Interpret the mean and spread

For the binomial model,

E[X]=np,\qquad \operatorname{Var}(X)=np(1-p),\qquad \sigma=\sqrt{np(1-p)}.

With $n=5$ and $p=0.3$ , the expected count is 1.5, the variance is 1.05 and the standard deviation is approximately 1.0247 successes. A non-integer expectation is compatible with an integer-valued variable.

The expectation follows by adding the expectations of the five 0–1 success indicators. The usual variance formula uses independence as well as the common probability. Neither formula is a guarantee that an individual five-trial run will be near the mean.

Compare with sampling without replacement

Two draws from three A and two B tokens have the same marginal A probability $3/5$ at each position, but the draws are dependent without replacement. Their A-count probabilities are $1/10$ , $3/5$ and $3/10$ .

A binomial model with $n=2$ and $p=3/5$ would instead give

P(X=0)=\frac4{25},\quad P(X=1)=\frac{12}{25},\quad P(X=2)=\frac9{25}.

Those are the probabilities under replacement with fresh independent selection, not the original no-replacement experiment. The correct no-replacement model is hypergeometric. A large population with a small sampling fraction may justify a binomial approximation for some purposes, but that is an approximation with conditions, not exact independence.

Model checks and edge cases

If trials continue until the first success, their number is not fixed in advance; a geometric waiting model may apply instead. If individual trials have different success probabilities, the simple binomial formula does not apply even if the trials are independent.

At $p=0$ , the count is zero with probability one. At $p=1$ , it is n with probability one. Handle these endpoint models directly rather than relying on an unexplained $0^0$ calculation. If $n=0$ , there are no trials and the success count is zero.

An estimated p is another source of uncertainty. Entering a sample proportion into a binomial formula treats that estimate as if it were the model parameter for the calculation; it does not automatically account for uncertainty in p. Confidence Intervals addresses that additional inferential problem.

Practice questions

For $X\sim\operatorname{Bin}(5,0.3)$ , find $P(X=2)$ .
For that model, find the probability of at least one success.
In four independent trials with success probability $0.2$ , find the probability of at most one success.

Check your answers

$\binom52(0.3)^2(0.7)^3=0.3087$ .
$1-P(X=0)=1-(0.7)^5=0.83193$ .
Add zero and one: $(0.8)^4+\binom41(0.2)(0.8)^3=0.4096+0.4096=0.8192$ .