The Probability Distribution Formula describes how probabilities are assigned to each possible value of a random variable, forming the mathematical backbone of statistics and probability theory. Covered in NCERT Class 12 Mathematics (Chapter 13 — Probability), this formula is equally critical for JEE Main, JEE Advanced, and NEET aspirants. This article covers every key expression, a complete formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications — everything you need in one place.
Key Probability Distribution Formulas at a Glance
Quick reference for the most important probability distribution formulas.
- Sum of all probabilities: \( \sum_{i} P(X = x_i) = 1 \)
- Mean (Expected Value): \( \mu = E(X) = \sum_{i} x_i \cdot P(x_i) \)
- Variance: \( \sigma^2 = \sum_{i} x_i^2 \cdot P(x_i) – \mu^2 \)
- Standard Deviation: \( \sigma = \sqrt{\text{Var}(X)} \)
- Binomial Distribution: \( P(X = r) = \binom{n}{r} p^r q^{n-r} \)
- Poisson Distribution: \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \)
- Normal Distribution PDF: \( f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \)
What is Probability Distribution Formula?
The Probability Distribution Formula is a mathematical rule that links every possible outcome of a random variable to its corresponding probability. A random variable \( X \) can be discrete (taking countable values) or continuous (taking values over an interval). For a discrete random variable, the probability distribution is presented as a table or function listing each value \( x_i \) and its probability \( P(X = x_i) \).
This concept is introduced in NCERT Class 12 Mathematics, Chapter 13 (Probability). It builds directly on the ideas of sample spaces and events from earlier chapters. The fundamental condition that every valid probability distribution must satisfy is that all probabilities are non-negative and their sum equals exactly 1.
There are several types of probability distributions. The most important ones for CBSE and competitive exams are the discrete probability distribution, the Binomial distribution, the Poisson distribution, and the Normal (Gaussian) distribution. Each has its own formula, conditions, and applications. Understanding all of them is essential for scoring full marks in Class 12 board exams and for cracking JEE/NEET questions on statistics and probability.
Probability Distribution Formula — Expression and Variables
Discrete Probability Distribution
For a discrete random variable \( X \) taking values \( x_1, x_2, \ldots, x_n \), the probability distribution must satisfy:
\[ P(X = x_i) \geq 0 \quad \text{and} \quad \sum_{i=1}^{n} P(X = x_i) = 1 \]
Mean (Expected Value)
\[ E(X) = \mu = \sum_{i=1}^{n} x_i \cdot P(x_i) \]
Variance
\[ \text{Var}(X) = \sigma^2 = E(X^2) – [E(X)]^2 = \sum_{i=1}^{n} x_i^2 \cdot P(x_i) – \mu^2 \]
Standard Deviation
\[ \sigma = \sqrt{\text{Var}(X)} \]
| Symbol | Quantity | SI Unit / Remark |
|---|---|---|
| \( X \) | Random Variable | Dimensionless |
| \( x_i \) | Possible values of X | Dimensionless |
| \( P(x_i) \) | Probability of \( X = x_i \) | Dimensionless (0 to 1) |
| \( E(X) \) or \( \mu \) | Mean / Expected Value | Same unit as X |
| \( \sigma^2 \) | Variance | Square of unit of X |
| \( \sigma \) | Standard Deviation | Same unit as X |
| \( n \) | Number of trials (Binomial) | Dimensionless (positive integer) |
| \( p \) | Probability of success | Dimensionless (0 to 1) |
| \( q \) | Probability of failure = \( 1 – p \) | Dimensionless (0 to 1) |
| \( \lambda \) | Average rate (Poisson) | Dimensionless (positive real) |
Derivation of Mean and Variance
The mean \( E(X) \) is derived by weighting each outcome by its probability and summing. This is the probability-weighted average. For variance, we first compute \( E(X^2) = \sum x_i^2 P(x_i) \). We then apply the identity \( \text{Var}(X) = E(X^2) – [E(X)]^2 \). This identity comes from expanding \( E[(X – \mu)^2] \) and simplifying. Standard deviation is simply the positive square root of variance. These three measures fully characterise the centre and spread of any discrete distribution.
Complete Probability Distribution Formula Sheet
| Formula Name | Expression | Key Variables | Conditions | NCERT Chapter |
|---|---|---|---|---|
| Validity Condition | \( \sum P(x_i) = 1,\; P(x_i) \geq 0 \) | \( P(x_i) \) = probability of each value | Must hold for any distribution | Class 12, Ch 13 |
| Mean (Expected Value) | \( E(X) = \sum x_i P(x_i) \) | \( x_i \) = values, \( P(x_i) \) = probabilities | Discrete random variable | Class 12, Ch 13 |
| Variance | \( \sigma^2 = \sum x_i^2 P(x_i) – \mu^2 \) | \( \mu \) = mean | Discrete random variable | Class 12, Ch 13 |
| Standard Deviation | \( \sigma = \sqrt{\text{Var}(X)} \) | \( \text{Var}(X) \) = variance | Always non-negative | Class 12, Ch 13 |
| Binomial Distribution PMF | \( P(X=r) = \binom{n}{r} p^r q^{n-r} \) | \( n \) = trials, \( p \) = success prob, \( q=1-p \) | Fixed \( n \), independent trials, \( p \) constant | Class 12, Ch 13 |
| Binomial Mean | \( \mu = np \) | \( n \) = trials, \( p \) = success probability | Binomial distribution | Class 12, Ch 13 |
| Binomial Variance | \( \sigma^2 = npq \) | \( q = 1-p \) | Binomial distribution | Class 12, Ch 13 |
| Poisson Distribution PMF | \( P(X=k) = \dfrac{e^{-\lambda}\lambda^k}{k!} \) | \( \lambda \) = mean rate, \( k \) = number of events | Large \( n \), small \( p \), \( \lambda = np \) | Class 12 / JEE |
| Poisson Mean and Variance | \( E(X) = \text{Var}(X) = \lambda \) | \( \lambda \) = rate parameter | Poisson distribution only | JEE / Advanced |
| Normal Distribution PDF | \( f(x) = \dfrac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \) | \( \mu \) = mean, \( \sigma \) = std deviation | Continuous, symmetric about \( \mu \) | JEE Advanced / NEET |
| Cumulative Distribution Function (CDF) | \( F(x) = P(X \leq x) = \sum_{x_i \leq x} P(x_i) \) | \( x \) = value up to which probability is summed | Discrete random variable | Class 12, Ch 13 |
Probability Distribution Formula — Solved Examples
Example 1 (Class 10-12 Level — Verify and Find Mean)
Problem: A random variable \( X \) has the following probability distribution. Find the value of \( k \) and then calculate the mean \( E(X) \).
| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X) | 0.1 | k | 0.3 | 0.2 |
Given: \( P(0) = 0.1,\; P(1) = k,\; P(2) = 0.3,\; P(3) = 0.2 \)
Step 1: Use the validity condition: \( \sum P(x_i) = 1 \)
\[ 0.1 + k + 0.3 + 0.2 = 1 \]
Step 2: Solve for \( k \):
\[ k = 1 – 0.6 = 0.4 \]
Step 3: Calculate the mean using \( E(X) = \sum x_i P(x_i) \):
\[ E(X) = 0(0.1) + 1(0.4) + 2(0.3) + 3(0.2) \]
\[ E(X) = 0 + 0.4 + 0.6 + 0.6 = 1.6 \]
Answer
\( k = 0.4 \) and Mean \( E(X) = 1.6 \)
Example 2 (Class 12 Level — Variance and Standard Deviation)
Problem: A random variable \( X \) has the distribution below. Find the variance and standard deviation of \( X \).
| X | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X) | 0.2 | 0.3 | 0.3 | 0.2 |
Given: \( P(1)=0.2,\; P(2)=0.3,\; P(3)=0.3,\; P(4)=0.2 \)
Step 1: Verify the distribution: \( 0.2+0.3+0.3+0.2 = 1 \). Valid.
Step 2: Find \( E(X) \):
\[ E(X) = 1(0.2) + 2(0.3) + 3(0.3) + 4(0.2) = 0.2 + 0.6 + 0.9 + 0.8 = 2.5 \]
Step 3: Find \( E(X^2) \):
\[ E(X^2) = 1^2(0.2) + 2^2(0.3) + 3^2(0.3) + 4^2(0.2) \]
\[ = 0.2 + 1.2 + 2.7 + 3.2 = 7.3 \]
Step 4: Apply the variance formula:
\[ \text{Var}(X) = E(X^2) – [E(X)]^2 = 7.3 – (2.5)^2 = 7.3 – 6.25 = 1.05 \]
Step 5: Find standard deviation:
\[ \sigma = \sqrt{1.05} \approx 1.025 \]
Answer
Variance \( = 1.05 \) and Standard Deviation \( \approx 1.025 \)
Example 3 (JEE/NEET Level — Binomial Distribution)
Problem: A fair coin is tossed 6 times. Find the probability of getting exactly 4 heads. Also find the mean and variance of the number of heads.
Given: \( n = 6,\; p = \frac{1}{2}\; (\text{head}),\; q = \frac{1}{2},\; r = 4 \)
Step 1: Identify the distribution. Each toss is independent with constant \( p \). This is a Binomial distribution: \( X \sim B(6, \frac{1}{2}) \).
Step 2: Apply the Binomial PMF:
\[ P(X = 4) = \binom{6}{4} \left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^2 \]
Step 3: Calculate \( \binom{6}{4} = \frac{6!}{4! \cdot 2!} = 15 \)
Step 4: Substitute:
\[ P(X=4) = 15 \times \frac{1}{16} \times \frac{1}{4} = \frac{15}{64} \]
Step 5: Mean and Variance using Binomial formulas:
\[ \mu = np = 6 \times \frac{1}{2} = 3 \]
\[ \sigma^2 = npq = 6 \times \frac{1}{2} \times \frac{1}{2} = \frac{3}{2} = 1.5 \]
Answer
\( P(X=4) = \dfrac{15}{64} \approx 0.234 \), Mean \( = 3 \), Variance \( = 1.5 \)
CBSE Exam Tips 2025-26
- Always verify the distribution first. Before computing mean or variance, confirm that \( \sum P(x_i) = 1 \) and all probabilities are non-negative. CBSE awards one step mark for this check.
- Use the shortcut variance formula. Always use \( \text{Var}(X) = E(X^2) – [E(X)]^2 \). It is faster than computing \( \sum (x_i – \mu)^2 P(x_i) \) directly.
- Learn Binomial conditions by heart. CBSE 2025-26 papers frequently ask you to justify why a situation follows a Binomial distribution. State all four conditions clearly in your answer.
- Write the distribution table neatly. For 3-mark questions, a well-organised probability distribution table earns full presentation marks. We recommend drawing it before solving.
- Distinguish between \( E(X^2) \) and \( [E(X)]^2 \). These are two different quantities. Mixing them up is the single most common error in board exams. Compute each separately.
- Practice standard Binomial values. For CBSE 2025-26, questions often use \( n = 5, 6, 8 \) and \( p = \frac{1}{2}, \frac{1}{3}, \frac{1}{4} \). Memorise \( \binom{n}{r} \) for these values to save time.
Common Mistakes to Avoid
- Mistake 1 — Forgetting to check \( \sum P(x_i) = 1 \): Students often skip the validity check and jump straight to calculations. Always verify the sum equals 1 before proceeding. If the sum is not 1, the distribution is invalid.
- Mistake 2 — Confusing \( E(X^2) \) with \( [E(X)]^2 \): These are entirely different. \( E(X^2) = \sum x_i^2 P(x_i) \) while \( [E(X)]^2 = \left(\sum x_i P(x_i)\right)^2 \). Compute them separately every time.
- Mistake 3 — Using wrong \( q \) in Binomial: Remember that \( q = 1 – p \), not \( q = p \). Also confirm \( p + q = 1 \) before substituting into \( P(X=r) = \binom{n}{r} p^r q^{n-r} \).
- Mistake 4 — Applying Binomial when conditions fail: Binomial distribution requires fixed \( n \), independent trials, only two outcomes, and constant \( p \). If any condition fails, the distribution is not Binomial. Do not force-fit the formula.
- Mistake 5 — Negative variance: Variance can never be negative. If you get a negative value for \( \text{Var}(X) \), you have made an arithmetic error. Recheck \( E(X^2) \) and \( [E(X)]^2 \) immediately.
JEE/NEET Application of Probability Distribution Formula
In our experience, JEE aspirants encounter the Probability Distribution Formula in at least 2-3 questions per paper. The Binomial distribution is the most heavily tested type in both JEE Main and JEE Advanced. NEET tests basic discrete distributions and expected value calculations in its Biology-linked statistics questions.
Pattern 1 — Finding Missing Probability and Mean
JEE Main frequently gives a table with one unknown probability value \( k \). You must first find \( k \) using \( \sum P(x_i) = 1 \), then compute \( E(X) \) or \( \text{Var}(X) \). These are typically 4-mark single-correct questions. Speed and accuracy in arithmetic are decisive.
Pattern 2 — Binomial Distribution Problems
JEE Advanced tests Binomial distribution in multi-step problems. A typical question gives you \( P(X = r) \) and asks you to find \( n \) and \( p \), or to compare probabilities \( P(X = r) \) and \( P(X = r+1) \). The key identity used is:
\[ \frac{P(X = r+1)}{P(X = r)} = \frac{(n-r)p}{(r+1)q} \]
This ratio helps identify the mode of the Binomial distribution. Our experts suggest practising at least 10 such ratio-based problems before the exam.
Pattern 3 — Poisson Approximation
When \( n \) is very large and \( p \) is very small (with \( np = \lambda \) finite), JEE Advanced problems use the Poisson approximation to the Binomial. The formula \( P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!} \) appears in integer-type and paragraph-based questions. For NEET, this level of depth is not required, but understanding the concept of expected value from a distribution is tested in data interpretation questions.
FAQs on Probability Distribution Formula
Explore more related topics on ncertbooks.net. Visit our Algebra Formulas hub for a complete collection of formulas. You may also find these articles helpful: Bayes Theorem Formula, Antiderivative Formula, and Implicit Differentiation Formula. For the official NCERT syllabus, refer to ncert.nic.in.