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Probability and Statistics Formulas For JEE 2027
Probability and Statistics formulas for JEE 2027 cover mean, median, mode, variance, standard deviation, probability rules, conditional probability, Bayes’ theorem and binomial distribution. Statistics helps students summarise data and measure its spread, while probability helps calculate the likelihood of events. Understanding these concepts and the conditions for using each formula is essential for solving questions accurately.
This guide includes key formulas, important properties and worked examples for revision. Use a JEE Formula Sheet PDF to review essential results, then practise applying them through previous years’ questions and a JEE Mains test series. Reviewing your mistakes can help you distinguish between probability rules, strengthen your understanding and improve accuracy under timed conditions.
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 2027: Conclusion
Probability and Statistics preparation for JEE 2027 requires a clear understanding of statistical measures, probability rules, conditional probability and distributions. Use a JEE Mains Maths Formula PDF to revise essential results and the conditions for applying them. Knowing when to use each formula can help you avoid mistakes and solve problems more accurately.
Solve JEE Mains previous year papers to practise different question types and strengthen your understanding. Pay particular attention to distinguishing independent events from mutually exclusive events and choosing the correct probability rule. Regular revision, timed practice and careful review of mistakes can help you build confidence and improve your speed.
