In this post, we’ll discuss Binomial Random Variables.

**Prerequisite : ** Random Variables

A specific type of **discrete** random variable that counts how often a particular event occurs in a fixed number of tries or trials.

For a variable to be a binomial random variable, ALL of the following conditions must be met:

- There are a fixed number of trials (a fixed sample size).
- On each trial, the event of interest either occurs or does not.
- The probability of occurrence (or not) is the same on each trial.
- Trials are independent of one another.

Mathematical Notations

n = number of trials p = probability of success in each trial k = number of success in n trials

Now we try to find out probability of k success in n trials.

Here probability of success in each trial is p independent of other trials.

So we first choose k trials in which there will be success and in rest n-k trials there will be failure. Number of ways to do so is

Since all n events are independent, hence probability of k success in n trials is equivalent to multiplication of probability for each trial.

Here its k success and n-k failures, So probability for each way to achieve k success and n-k failure is

Hence final probability is

(number of ways to achieve k success and n-k failures) * (probability for each way to achieve k success and n-k failure)

Then Binomial Random Variable Probability is given by:

Let X be binomial random variable with number of trials n and probability of success in each trial be p.

Expected number of success is given by

E[X] = np

Variance of number of success is given by

Var[X] = np(1-p)

**Example 1** : Consider a random experiment in which a biased coin (probability of head = 1/3) is thrown for 10 times. Find probabilit that number of heads appearing will be 5.

Solution :

Let X be binomial random variable with n = 10 and p = 1/3 P(X=5) = ?

Here is a C++ code for the same

## C++

// C++ program to compute Binomial Probability #include <iostream> #include <cmath> using namespace std; // function to calculate nCr i.e., number of // ways to choose r out of n objects int nCr(int n, int r) { // Since nCr is same as nC(n-r) // To decrease number of iterations if (r > n / 2) r = n - r; int answer = 1; for (int i = 1; i <= r; i++) { answer *= (n - r + i); answer /= i; } return answer; } // function to calculate binomial r.v. probability float binomialProbability(int n, int k, float p) { return nCr(n, k) * pow(p, k) * pow(1 - p, n - k); } // Driver code int main() { int n = 10; int k = 5; float p = 1.0 / 3; float probability = binomialProbability(n, k, p); cout << "Probability of " << k; cout << " heads when a coin is tossed " << n; cout << " times where probability of each head is " << p << endl; cout << " is = " << probability << endl; }

## Java

// Java program to compute Binomial Probability import java.util.*; class GFG { // function to calculate nCr i.e., number of // ways to choose r out of n objects static int nCr(int n, int r) { // Since nCr is same as nC(n-r) // To decrease number of iterations if (r > n / 2) r = n - r; int answer = 1; for (int i = 1; i <= r; i++) { answer *= (n - r + i); answer /= i; } return answer; } // function to calculate binomial r.v. probability static float binomialProbability(int n, int k, float p) { return nCr(n, k) * (float)Math.pow(p, k) * (float)Math.pow(1 - p, n - k); } // Driver code public static void main(String[] args) { int n = 10; int k = 5; float p = (float)1.0 / 3; float probability = binomialProbability(n, k, p); System.out.print("Probability of " +k); System.out.print(" heads when a coin is tossed " +n); System.out.println(" times where probability of each head is " +p); System.out.println( " is = " + probability ); } } /* This code is contributed by Mr. Somesh Awasthi */

Output:

Probability of 5 heads when a coin is tossed 10 times where probability of each head is 0.333333 is = 0.136565

**Reference** :

stat200

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