The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials, each with the same probability of success. It is defined by two parameters: the number of trials (n) and the probability of success in each trial (p). The formula to calculate the probability of exactly k successes is **P(X=k) = ( ^{n}_{k})p^{k}(1−p)^{n-k}**. Key properties include mean (np) and variance (npq, where q = 1 – p). This distribution is fundamental for analyzing binary outcomes and is frequently tested in competitive exams like

**IIT-JEE**,

**GATE**,

**GRE**, and

**CAT**, where solving related problems helps understand real-life applications like quality control and decision-making under uncertainty.

Table of Contents

## What is Binomial Distribution?

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent and identical trials, where each trial has only two possible outcomes: success or failure. The probability of success remains constant across trials, and each trial is independent of others. The distribution is used to calculate the likelihood of achieving a specific number of successes in a given set of trials. Important terms related to binomial distribution are:

**Trial**: A single event with two possible outcomes (success/failure).**Success (p)**: The probability of the desired outcome in a single trial.**Failure (q)**: The probability of the undesired outcome, where q=1−p.**Number of Trials (n)**: The total number of independent trials or experiments conducted.**Number of Successes (k)**: The number of successful outcomes in the trials.**Binomial Coefficient****(**^{n}_{k}**)**: A combination that calculates the number of ways to choose k successes from n trials, represented as**n!/k!(n−k)!**.**Probability Mass Function (PMF)**: The formula used to calculate the probability of exactly k successes:**P(X=k) = (**^{n}_{k}**)p**^{k}**(1−p)**.^{n-k}**Mean (np)**: The expected number of successes in n trials.**Variance (npq)**: A measure of the spread or variability in the number of successes.

**Must Read: Commercial Maths**

## Properties of Binomial Distribution

The **binomial distribution** has several key properties that describe its behavior and characteristics. These properties are crucial in understanding the distribution’s shape, central tendency, and variability

**Two Possible Outcomes**: Each trial has only two possible outcomes: success (with probability p) and failure (with probability q=1−p).**Fixed Number of Trials (n)**: The number of trials, n, is fixed and predetermined. Each trial is independent of others, meaning the outcome of one trial does not affect the others.**Constant Probability (p)**: The probability of success, p, remains the same for all trials.**Discrete Distribution**: The binomial distribution is discrete, meaning it takes only integer values (the number of successes) ranging from 0 to n.**Mean (Expected Value)**: The mean or expected value of the binomial distribution is given by:**μ = np**

This represents the average number of successes expected in n trials.**Variance**: The variance, which measures the spread of the distribution, is:**σ**^{2 }= npq

where q = 1−p is the probability of failure. This shows how much the number of successes is expected to deviate from the mean.**Standard Deviation**: The standard deviation is the square root of the variance:**σ = √npq****Symmetry and Skewness**: The binomial distribution becomes more symmetric as n increases and when p is close to 0.5. If p is much greater than 0.5, the distribution is skewed to the left (negatively skewed), while if p is much less than 0.5, it is skewed to the right (positively skewed).**Maximum Probability**: The most probable value (the mode) of the binomial distribution is around k = np (the mean), but it may vary slightly depending on whether p is closer to 0 or 1.**Relationship with Normal Distribution**: For large n, the binomial distribution approaches a normal distribution, especially when p is close to 0.5. This is known as the**normal approximation**to the binomial distribution, often applied when np ≥ 5 and nq ≥ 5.

## Formula of Binomial Distribution

The formula of the binomial distribution gives the probability of obtaining exactly **k** successes in **n** independent trials, where each trial has two possible outcomes (success or failure) and the probability of success is **p**.

**Binomial Distribution Formula:**

**P(X=k) = (**^{n}_{k}**)p**^{k}**(1−p)**^{n-k}

Where:

- P(X=k): The probability of getting exactly
**k**successes in**n**trials. - n: The total number of trials.
- k: The number of successes.
- p: The probability of success in a single trial.
- 1−p or q: The probability of failure in a single trial.
- (
^{n}): The binomial coefficient, representing the number of ways to choose_{k}**k**successes from**n**trials, is calculated as:

**(**^{n}_{k}**) = n!/k!(n−k)! **

where n! is the factorial of n.

**Explanation:**

- The term p
^{k}represents the probability of getting k successes. - The term (1−p)
^{n-k}represents the probability of getting n – k failures. - The binomial coefficient (
^{n}_{k}) accounts for the different combinations in which the k successes can occur in n trials.

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## Solved Examples of Binomial Distribution

Here are 5 solved examples of binomial distribution to illustrate its application:

**Example 1: Tossing a Coin**

You toss a fair coin 5 times. What is the probability of getting exactly 3 heads?

**Solution**:- Number of trials n = 5
- Probability of success (head) p=0.5
- Probability of failure (tail) q = 1−p = 0.5
- Number of successes k = 3

Using the binomial distribution formula:

P(X=3) = (^{5}_{3})(0.5)^{3}(0.5)^{5-3}

P(X=3) = {5!/3!(5−3)!}(0.5)^{5 }= 10×0.03125 = 0.3125

So, the probability of getting exactly 3 heads is **0.3125**.

**Example 2: Quality Control**

A factory produces light bulbs, and 95% of them are good. If you randomly select 10 bulbs, what is the probability that exactly 9 of them are good?

**Solution**:- Number of trials n = 10
- Probability of success (good bulb) p = 0.95
- Probability of failure (bad bulb) q = 0.05
- Number of successes k = 9

Using the formula:

P(X=9) = (^{10}_{9})(0.95)^{9}(0.05)^{1}

P(X=9) = 10×(0.95)9×(0.05)

P(X=9) ≈ 10×0.6302×0.05 = 0.3151

So, the probability of exactly 9 good bulbs is **0.3151**.

**Example 3: Exam Questions**

In a multiple-choice exam, each question has 4 options, with only one correct answer. If you randomly guess on 6 questions, what is the probability of getting exactly 2 correct answers?

**Solution**:- Number of trials n = 6
- Probability of success (correct answer) p = ¼ = 0.25
- Probability of failure (wrong answer) q = 0.75
- Number of successes k = 2

Using the formula:

P(X=2) = (^{6}_{2})(0.25)^{2}(0.75)^{4}

P(X=2) = {6!/2!(6−2)!}×(0.25)^{2}×(0.75)^{4 }= 15×0.0625×0.3164P

P(X=2) ≈ 15×0.01977 = 0.2966

The probability of getting exactly 2 correct answers is **0.2966**.

**Example 4: Defective Products**

A shipment contains 12 items, and there is a 20% chance that any item is defective. If you randomly check 8 items, what is the probability that exactly 2 are defective?

**Solution**:- Number of trials n = 8
- Probability of success (defective item) p = 0.2
- Probability of failure (non-defective) q = 0.8
- Number of successes k = 2

Using the formula:

P(X=2) = (^{8}_{2})(0.2)^{2}(0.8)^{6}

P(X=2) = 28×(0.2)^{2}×(0.8)^{6 }= 28×0.04×0.2621

P(X=2) ≈ 28×0.010484 = 0.2936

The probability of finding exactly 2 defective items is **0.2936**.

**Example 5: Drug Trial**

In a drug trial, a new treatment has a 70% success rate. If 10 patients are treated, what is the probability that exactly 7 patients will recover?

**Solution**:- Number of trials n = 10
- Probability of success (recovery) p = 0.7
- Probability of failure (no recovery) q = 0.3
- Number of successes k = 7

Using the formula:

P(X=7) = (^{10}_{7})(0.7)^{7}(0.3)^{3}

P(X=7) = {10!7!(10−7)!}×(0.7)^{7}×(0.3)^{3 }= 120×0.0823543×0.027

P(X=7) ≈ 120×0.002224 = 0.2669

The probability of exactly 7 patients recovering is **0.2669**.

## FAQs

**What are the 4 factors for binomial distribution?**The four factors for binomial distribution are:

— **Fixed number of trials (n):** The experiment must have a predetermined number of trials.**–**– I**ndependent trials:** Each trial must be independent of the others.

— **Two possible outcomes:** Each trial must have only two possible outcomes (success or failure).

— **Constant probability of success (p):** The probability of success must remain the same for each trial.

**Why is it called binomial distribution?**

Binomial distribution is called so because it deals with experiments having only two possible outcomes – success or failure, and the probability of success remains constant across trials.

**What is the full formula of binomial distribution?**

The full formula of the binomial distribution is: P(X=k) = (nk)pk(1−p)n-k

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