Binomial Distribution

The Probabilistic Experiment Behind binomial distribution

Notation Used

Parameters

Probability Mass Function (PMF) and Support (Range)

Cumulative Distribution Function (CDF)

Expected Value (Mean)

Variance and Standard Deviation

Applications and Examples

Binomial Distribution: Repeated Trials with Fixed Count

The binomial distribution arises from repeating the same Bernoulli experiment a fixed number of times under identical conditions. Each trial is independent, each has exactly two outcomes (success or failure), and the probability of success remains constant across trials. The experiment does not ask when success happens, but rather how many successes occur in the predetermined number of trials.

The Probabilistic Experiment Behind binomial distribution

The binomial distribution models the number of successes obtained when the same experiment is repeated a fixed number of times under identical conditions. Each repetition — called a Bernoulli trial — has only two outcomes (success or failure), and the probability of success remains constant across trials. Crucially, the trials are independent.

What distinguishes the binomial distribution is that the number of trials is fixed in advance, and the random variable counts how many successes occur, not when they occur. The order of outcomes does not matter, only the total count does. This makes the binomial distribution a natural model for controlled, repeatable processes.

The binomial distribution captures variability caused by chance alone when conditions are stable. It breaks down if probabilities change from trial to trial, if trials influence each other, or if the number of trials is not predetermined.

Learn More About Distinguishing Discrete Distributions →

Example:

Flipping a coin

10

times and counting how many times it lands heads. Each flip is independent, the probability of heads stays the same, and the total number of flips is fixed before starting.

See More Examples of Binomial Distribution →

Notation Used

X \sim \text{Bin}(n, p)

X \sim \text{B}(n, p)

— distribution of the random variable.

\text{Binomial}(n, p)

— used to denote the distribution itself (not the random variable).

B(n,p)

— occasionally used in theoretical or formal contexts (less common).

P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}

— probability mass function (PMF), where:

n

— number of trials

p

— probability of success on each trial

k

— number of successes (

k = 0, 1, 2, \ldots, n

)

\binom{n}{k} = \frac{n!}{k!(n-k)!}

— binomial coefficient

Alternative notations:

q = 1 - p

— probability of failure, so PMF can be written as

P(X = k) = \binom{n}{k} p^k q^{n-k}

X \sim \text{Binomial}(n, p)

— more verbose but explicit

Key properties:

E[X] = np

— expected value (mean)

\text{Var}(X) = np(1-p)

— variance

See All Probability Symbols and Notations →

Parameters

𝑛

: fixed number of independent trials;

𝑝

: probability of success in each trial;

This distribution models the number of successes when repeating the same binary experiment

𝑛

times under identical conditions. The two parameters fully describe the setup:

𝑛

gives the structure — how many attempts, and

𝑝

defines the behavior of each — what chance success has.

It’s useful to compare with the negative binomial, where instead of fixing how many trials you run, you fix how many successes you want and ask: how many trials will it take? Both deal with repeated binary outcomes, but what’s held constant — trials vs. successes — flips.

Probability Mass Function (PMF) and Support (Range)

The probability mass function (PMF) of a binomial distribution is given by:

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, 2, \ldots, n

where

\binom{n}{k} = \frac{n!}{k!(n-k)!}

is the binomial coefficient.

Fixed Number of Trials: The binomial distribution models the number of successes in

n

independent trials, where each trial has only two possible outcomes: success or failure.

Support (Range of the Random Variable):
* The random variable

X

can take on values from

0

n

(inclusive).
* These represent the possible number of successes:

0, 1, 2, \ldots, n

.
* The support is thus a finite set of

n+1

non-negative integers.

Logic Behind the Formula:
*

\binom{n}{k}

: The number of ways to choose

k

successes from

n

trials
*

p^k

: The probability of getting exactly

k

successes
*

(1-p)^{n-k}

: The probability of getting exactly

n-k

failures
* The total probability sums to 1:

\sum_{k=0}^{n} P(X = k) = \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} = 1

* This follows from the binomial theorem:

(p + (1-p))^n = 1^n = 1

Binomial Distribution

Successes in n independent trials with probability p each

Number of Trials (n): 10

Success Probability (p): 0.50

Explanation

The binomial distribution models the number of successes in $n$ independent trials, each with probability $p$ of success. The probability mass function is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ , where $\binom{n}{k}$ is the binomial coefficient. The expected value is $E[X] = np$ and the variance is $\text{Var}(X) = np(1-p)$ . This distribution is commonly used for modeling coin flips (number of heads in $n$ tosses), quality control testing (defective items in a batch), and clinical trial success rates.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of a binomial distribution is given by:

F_X(k) = P(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}

Where:

n

= number of trials

p

= probability of success on each trial

k

= number of successes (where

0 \leq k \leq n

)

\binom{n}{i}

= binomial coefficient "

n

choose

i

Intuition Behind the Formula

Definition: The CDF gives the probability of observing

k

or fewer successes in

n

trials.

Summation of Probabilities:
Since we want the probability of getting *at most*

k

successes, we sum the individual probabilities from 0 successes up to

k

successes:

P(X \leq k) = P(X=0) + P(X=1) + P(X=2) + \cdots + P(X=k)

Discrete Nature: Unlike continuous distributions, the binomial CDF is a step function that jumps at each integer value from 0 to

n

. Between integers, the CDF remains constant.

Complementary Probabilities:
For calculations involving "more than

k

successes," use:

P(X > k) = 1 - P(X \leq k) = 1 - F_X(k)

For "at least

k

successes":

P(X \geq k) = 1 - P(X \leq k-1) = 1 - F_X(k-1)

Binomial Distribution CDF

CDF for successes in n independent trials

Number of Trials (n): 10

Success Probability (p): 0.50

CDF Explanation

The binomial CDF is

F(k) = P(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}

. This represents the probability of getting

k

or fewer successes in

n

trials. The CDF starts at

F(0) = (1-p)^n

(the probability of zero successes) and increases with each additional possible outcome, reaching 1.0 at

k = n

. The shape of the CDF reflects the underlying binomial distribution: symmetric when

p = 0.5

, right-skewed when

p < 0.5

, and left-skewed when

p > 0.5

Expected Value (Mean)

As explained in the general case for calculating expected value, the expected value of a discrete random variable is computed as a weighted sum where each possible value is multiplied by its probability:

E[X] = \sum_{x} x \cdot P(X = x)

For the binomial distribution, we apply this general formula to the specific probability mass function of this distribution.

Formula

E[X] = np

Where:

n

= number of independent trials

p

= probability of success on each trial

Derivation and Intuition

The binomial random variable

X

can be expressed as the sum of

n

independent Bernoulli random variables:

X = X_1 + X_2 + \cdots + X_n

, where each

X_i

represents the outcome of trial

i

(1 for success, 0 for failure).

Each Bernoulli variable has expected value

E[X_i] = p

.

By the linearity of expectation:

E[X] = E[X_1 + X_2 + \cdots + X_n] = E[X_1] + E[X_2] + \cdots + E[X_n] = p + p + \cdots + p = np

The result

E[X] = np

captures a simple and intuitive fact: if you perform

n

trials, each with probability

p

of success, you expect on average to get

np

successes. This is the most natural interpretation of "average number of successes."

Example

Consider flipping a fair coin 10 times, where

n = 10

and

p = 0.5

E[X] = 10 \times 0.5 = 5

The expected number of heads in 10 flips is 5, which aligns perfectly with our intuition about a fair coin.

Variance and Standard Deviation

The variance of a discrete random variable measures how spread out the values are around the expected value. It is computed as:

\mathrm{Var}(X) = \mathbb{E}[(X - \mu)^2] = \sum_{x} (x - \mu)^2 P(X = x)

Or using the shortcut formula:

\mathrm{Var}(X) = \mathbb{E}[X^2] - \mu^2

For the binomial distribution, we apply this formula to derive the variance.

Formula

\mathrm{Var}(X) = np(1-p)

Where:

n

= number of independent trials

p

= probability of success on each trial

(1-p) = q

= probability of failure on each trial

Derivation and Intuition

Since the binomial random variable can be expressed as the sum of

n

independent Bernoulli variables:

X = X_1 + X_2 + \cdots + X_n

, and for independent variables variances add:

\mathrm{Var}(X) = \mathrm{Var}(X_1) + \mathrm{Var}(X_2) + \cdots + \mathrm{Var}(X_n)

Each Bernoulli variable has variance

\mathrm{Var}(X_i) = p(1-p)

.

Therefore:

\mathrm{Var}(X) = np(1-p)

The result

\mathrm{Var}(X) = np(1-p)

shows that variance depends on both the number of trials and the probability structure. The term

p(1-p)

is maximized when

p = 0.5

(a fair coin), meaning maximum variability occurs when success and failure are equally likely. As

p

approaches 0 or 1, variance decreases because outcomes become more predictable.

Standard Deviation

\sigma = \sqrt{np(1-p)}

Example

Consider flipping a fair coin 100 times, where

n = 100

and

p = 0.5

\mathrm{Var}(X) = 100 \times 0.5 \times 0.5 = 25

\sigma = \sqrt{25} = 5

The variance of 25 and standard deviation of 5 indicate that while we expect 50 heads on average, typical outcomes will deviate by about 5 heads in either direction.

Applications and Examples

Practical Example

Suppose you flip a fair coin

n = 5

times, where the probability of heads (success) is

p = 0.5

. The probability of getting exactly

k = 3

heads is:

P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^{5-3} = 10 \cdot 0.125 \cdot 0.25 = 0.3125

This means there's a 31.25% chance of getting exactly 3 heads in 5 coin flips.

The possible outcomes range from

k = 0

(no heads) to

k = 5

(all heads), with probabilities determined by the formula above.