Probability and Statistics Formulas For JEE 2026
Probability & Statistics is an important topic in JEE Mathematics and usually contributes 2–3 questions every year. It covers basic concepts like mean, median, mode, variance, and standard deviation, which help students understand and analyze data more effectively. The chapter also introduces the fundamentals of probability, making it easier to calculate the likelihood of different outcomes.
In addition, students learn topics such as conditional probability, Bayes’ theorem, and binomial distribution, which are frequently asked in exams. Since many questions are based on direct application of formulas, regular practice is very important. With proper revision, this topic can become a scoring area. For quick revision, students can also use a well-organized JEE Mains Maths Formula PDF to go through important formulas and concepts easily.
Statistics: Measures of Central Tendency
Mean (Average)
Mean Ungrouped Data
For observations $$x_1, x_2, \ldots, x_n$$:
$$\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{\sum_{i=1}^{n} x_i}{n}$$Mean Grouped Data (Frequency Distribution)
If values $$x_1, x_2, \ldots, x_k$$ have frequencies $$f_1, f_2, \ldots, f_k$$:
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$For continuous grouped data, $$x_i$$ is the class midpoint.
Median
Median Ungrouped Data
Arrange data in ascending order. If $$n$$ is the number of observations:
- If $$n$$ is odd: Median $$= \left(\dfrac{n+1}{2}\right)^{\text{th}}$$ observation
- If $$n$$ is even: Median $$= \dfrac{1}{2}\left[\left(\dfrac{n}{2}\right)^{\text{th}} + \left(\dfrac{n}{2}+1\right)^{\text{th}}\right]$$
Mode
The value that occurs most frequently in the data set. A data set can have no mode, one mode (unimodal), or multiple modes.
Tip: Mean is affected by extreme values (outliers), median is not. If a data set is symmetric, mean = median = mode.
Statistics: Measures of Dispersion
Range
$$\text{Range} = x_{\max} - x_{\min}$$
Mean Deviation
About the mean (ungrouped data):
$$\text{MD}(\bar{x}) = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$$About the mean (grouped data):
$$\text{MD}(\bar{x}) = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}$$Variance and Standard Deviation
Variance and Standard Deviation — Ungrouped Data
$$\sigma^2 = \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n}$$Shortcut formula (easier for computation):
$$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$ $$\sigma = \sqrt{\sigma^2}$$Variance — Grouped Data
$$\sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \left(\frac{\sum f_i x_i}{\sum f_i}\right)^2$$Worked Example
Find the variance and standard deviation of: 4, 7, 13, 16.
$$\bar{x} = \dfrac{40}{4} = 10$$. $$\sum x_i^2 = 16 + 49 + 169 + 256 = 490$$
$$\sigma^2 = \dfrac{490}{4} - 100 = 122.5 - 100 = \textbf{22.5}$$
$$\sigma = \sqrt{22.5} \approx \textbf{4.74}$$
Effect of Linear Transformation
If each observation $$x_i$$ is transformed to $$y_i = ax_i + b$$:
- $$\bar{y} = a\bar{x} + b$$ (mean shifts and scales)
- $$\sigma_y^2 = a^2 \sigma_x^2$$ (variance scales by $$a^2$$; adding $$b$$ has no effect)
- $$\sigma_y = |a|\,\sigma_x$$ (SD scales by $$|a|$$)
Combined Mean and Variance
For two groups with sizes $$n_1$$, $$n_2$$, means $$\bar{x}_1$$, $$\bar{x}_2$$, and variances $$\sigma_1^2$$, $$\sigma_2^2$$:
Combined mean:
$$\bar{x} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$Combined variance:
$$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$$where $$d_1 = \bar{x}_1 - \bar{x}$$ and $$d_2 = \bar{x}_2 - \bar{x}$$.
Tip: The shortcut formula $$\sigma^2 = \dfrac{\sum x_i^2}{n} - \bar{x}^2$$ is much faster than computing each $$(x_i - \bar{x})^2$$. In JEE, always use the shortcut.
Probability: Basic Concepts
Classical Definition of Probability
If all outcomes are equally likely and the sample space has $$n(S)$$ outcomes:
$$P(A) = \frac{n(A)}{n(S)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$$Always: $$0 \leq P(A) \leq 1$$ and $$P(A') = 1 - P(A)$$.
Types of Events
- Sure event: $$S$$ itself. $$P(S) = 1$$.
- Impossible event: $$\emptyset$$. $$P(\emptyset) = 0$$.
- Complementary event: $$A' = S - A$$.
- Mutually exclusive events: $$A \cap B = \emptyset$$ (cannot happen simultaneously).
- Exhaustive events: $$A_1 \cup A_2 \cup \cdots \cup A_n = S$$.
Addition Theorem
For any two events $$A$$ and $$B$$:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$If $$A$$ and $$B$$ are mutually exclusive:
$$P(A \cup B) = P(A) + P(B)$$For three events:
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$
Worked Example
A card is drawn from a standard deck. Find the probability that it is a king or a heart.
$$P(\text{King} \cup \text{Heart}) = \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \boldsymbol{\dfrac{4}{13}}$$
Conditional Probability
The probability of event $$A$$ given that event $$B$$ has occurred:
$$P(A|B) = \dfrac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0$$
Multiplication Theorem
For three events:
$$P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|A \cap B)$$Worked Example
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.
$$P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \boldsymbol{\dfrac{5}{14}}$$
Independent Events
Events $$A$$ and $$B$$ are independent if:
$$P(A \cap B) = P(A) \cdot P(B)$$- If $$A$$ and $$B$$ are independent, so are: $$A$$ and $$B'$$, $$A'$$ and $$B$$, $$A'$$ and $$B'$$.
- Independence $$\neq$$ mutually exclusive. If $$P(A) > 0$$ and $$P(B) > 0$$, mutually exclusive events are never independent.
- For $$n$$ independent events: $$P(A_1 \cap A_2 \cap \cdots \cap A_n) = P(A_1) \cdot P(A_2) \cdots P(A_n)$$
Tip: Do not confuse independent and mutually exclusive. Mutually exclusive means $$P(A \cap B) = 0$$; independent means $$P(A \cap B) = P(A) \cdot P(B)$$. Two events with non-zero probabilities cannot be both.
Bayes' Theorem
If $$A_1, A_2, \ldots, A_n$$ are mutually exclusive and exhaustive events, then for any event $$B$$ with $$P(B) > 0$$:
$$P(A_i | B) = \frac{P(A_i) \cdot P(B|A_i)}{\sum_{j=1}^{n} P(A_j) \cdot P(B|A_j)}$$The denominator is the Total Probability:
$$P(B) = \sum_{j=1}^{n} P(A_j) \cdot P(B|A_j)$$Worked Example
Factory has 3 machines: $$M_1$$ produces 50%, $$M_2$$ produces 30%, $$M_3$$ produces 20% of items. Defective rates: $$M_1$$: 3%, $$M_2$$: 4%, $$M_3$$: 5%. An item is found defective. Find the probability it came from $$M_1$$.
$$P(D) = 0.5(0.03) + 0.3(0.04) + 0.2(0.05) = 0.015 + 0.012 + 0.010 = 0.037$$
$$P(M_1|D) = \dfrac{0.5 \times 0.03}{0.037} = \dfrac{0.015}{0.037} = \boldsymbol{\dfrac{15}{37}}$$
Tip: For Bayes' theorem problems, always identify the "hypotheses" ($$A_i$$) and the "evidence" ($$B$$). Compute the total probability $$P(B)$$ first, then apply the formula.
Random Variables
Mean (Expected Value) and Variance of a Random Variable
Mean (Expected value):
$$E(X) = \mu = \sum_{i=1}^{n} x_i \cdot P(X = x_i)$$Variance:
$$\text{Var}(X) = E(X^2) - [E(X)]^2$$where $$E(X^2) = \sum x_i^2 \cdot P(X = x_i)$$
Standard Deviation: $$\sigma = \sqrt{\text{Var}(X)}$$
Binomial Distribution
Binomial Probability
The probability of getting exactly $$r$$ successes in $$n$$ independent trials:
$$P(X = r) = \binom{n}{r} p^r q^{n-r}, \quad r = 0, 1, 2, \ldots, n$$where $$p$$ is the probability of success and $$q = 1 - p$$.
Mean, Variance, and Mode of Binomial Distribution
- Mean: $$E(X) = np$$
- Variance: $$\text{Var}(X) = npq$$
- Standard Deviation: $$\sigma = \sqrt{npq}$$
- Mode: If $$(n+1)p$$ is an integer, two modes: $$(n+1)p$$ and $$(n+1)p - 1$$. Otherwise, mode $$= \lfloor(n+1)p\rfloor$$.
Worked Example
A fair coin is tossed 10 times. Find the probability of getting exactly 6 heads.
$$n = 10$$, $$p = \dfrac{1}{2}$$, $$r = 6$$.
$$P(X=6) = \binom{10}{6}\left(\dfrac{1}{2}\right)^{10} = \dfrac{210}{1024} = \boldsymbol{\dfrac{105}{512}}$$
Worked Example
The probability that a student passes a test is $$\dfrac{3}{4}$$. If 8 students take the test, find the mean and variance.
Mean $$= np = 8 \times \dfrac{3}{4} = \textbf{6}$$. Variance $$= npq = 8 \times \dfrac{3}{4} \times \dfrac{1}{4} = \boldsymbol{\dfrac{3}{2}}$$
Tip: For "at least" problems, use $$P(X \geq k) = 1 - P(X < k)$$. Computing the complement is almost always faster.
Quick Reference
Statistics Summary
| Measure | Formula |
|---|---|
| Mean (ungrouped) | $$\bar{x} = \dfrac{\sum x_i}{n}$$ |
| Mean (grouped) | $$\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$$ |
| Variance (shortcut) | $$\sigma^2 = \dfrac{\sum x_i^2}{n} - \bar{x}^2$$ |
| Standard deviation | $$\sigma = \sqrt{\sigma^2}$$ |
Probability Summary
| Rule | Formula |
|---|---|
| Addition | $$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$ |
| Conditional | $$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$$ |
| Multiplication | $$P(A \cap B) = P(A) \cdot P(B|A)$$ |
| Independence | $$P(A \cap B) = P(A) \cdot P(B)$$ |
| Bayes' theorem | $$P(A_i|B) = \dfrac{P(A_i)P(B|A_i)}{\sum P(A_j)P(B|A_j)}$$ |
| Binomial $$P(X=r)$$ | $$\binom{n}{r}p^r q^{n-r}$$ |
| Binomial mean | $$np$$ |
| Binomial variance | $$npq$$ |
Probability and Statistics Formulas For JEE 2026: Conclusion
Probability and statistics is an important and scoring topic in mathematics that helps students understand data and predict outcomes effectively. By mastering concepts like mean, variance, probability rules, and distributions, students can solve a wide range of problems with better accuracy. Since many questions are formula-based, consistent practice and revision play a key role in improving performance.
To score well, students should focus on applying concepts correctly and avoiding common mistakes such as confusion between different probability rules. Regular revision of formulas and practicing different types of questions will improve both speed and confidence. With the right strategy, this chapter can become one of the easiest scoring areas in the exam.
