How much of JEE Main comes from Statistics and Probability?

JEEGenie estimates about 5% of JEE Main marks come from Statistics and Probability, based on recent papers. The share varies from year to year.

Which formulas matter most in Statistics and Probability?

The Statistics and Probability sheet on this page has 23 formula entries, starting with Mean (Arithmetic Mean), Median, Mode, Range, Mean Deviation. Each one comes with when to use it and the trap to avoid.

What topics does Statistics and Probability cover for JEE?

Statistics and Probability covers Measures of Central Tendency, Measures of Dispersion, Probability Basics, Conditional Probability, Bayes' Theorem, Binomial Distribution.

How should I revise Statistics and Probability before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Statistics and Probability questions on JEEGenie.

How many previous-year questions are there on Statistics and Probability?

JEEGenie has 586 previous-year questions on Statistics and Probability, and this page shows a few with worked explanations.

Is practice on Statistics and Probability free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Statistics and Probability for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 5% of JEE Main marks come from Statistics and Probability. This page covers 23 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Forgetting to use class marks (midpoints) instead of class boundaries in grouped data.

Key points to know

Mean (Arithmetic Mean). The sum of all observations divided by the number of observations. For grouped data, use class marks and frequencies.
Median. The middle value when data is arranged in order; for grouped data, use cumulative frequency to locate median class.
Mode. The value that appears most frequently; for grouped data, use modal class and formula.
Range. Difference between maximum and minimum values; simplest measure of dispersion.
Mean Deviation. Average of absolute deviations from mean or median; measures dispersion robustly.
Probability - Basic Terms. Probability measures likelihood of an event, from 0 (impossible) to 1 (certain). Sample space is set of all outcomes.
Mutually Exclusive and Exhaustive Events. Mutually exclusive: no common outcomes. Exhaustive: union covers entire sample space.
Random Variable and Probability Distribution. A random variable assigns a number to each outcome; probability distribution lists probabilities for each value.
Binomial Distribution. Models number of successes in n independent Bernoulli trials with constant success probability p.
Conditional Probability and Total Probability Theorem. If events partition sample space, total probability of B is weighted sum of conditional probabilities.

Formula sheet

Name	Formula	When to use
Mean (Arithmetic Mean)	$\overset{ˉ}{x} = \frac{\sum x_{i}}{n}$ (ungrouped), $\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$ (grouped)	To find the central tendency of a dataset; baseline for variance and standard deviation calculations.
Median	$Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$ (grouped)	When data has outliers or is skewed; preferred over mean for ordinal data.
Mode	$Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$ (grouped)	To find the most common value; useful for categorical or discrete data.
Range	$Range = \max - \min$	Quick estimate of spread; used in quality control or small datasets.
Mean Deviation	$MD = \frac{\sum ∣ x_{i} - \overset{ˉ}{x} ∣}{n}$ (ungrouped), $MD = \frac{\sum f_{i} ∣ x_{i} - \overset{ˉ}{x} ∣}{\sum f_{i}}$ (grouped)	When outliers are present but you need a simple dispersion measure.
Variance and Standard Deviation	$σ^{2} = \frac{\sum (x_{i} - \overset{ˉ}{x})^{2}}{n}$ (population), $s^{2} = \frac{\sum (x_{i} - \overset{ˉ}{x})^{2}}{n - 1}$ (sample), $σ = \sqrt{σ^{2}}$	To quantify spread; basis for many statistical tests and probability distributions.
Combined Mean and Variance	${\overset{ˉ}{x}}_{12} = \frac{n_{1} {\overset{ˉ}{x}}_{1} + n_{2} {\overset{ˉ}{x}}_{2}}{n_{1} + n_{2}}$ , $σ_{12}^{2} = \frac{n_{1} (σ_{1}^{2} + d_{1}^{2}) + n_{2} (σ_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}}$ where $d_{i} = {\overset{ˉ}{x}}_{i} - {\overset{ˉ}{x}}_{12}$	When given summary statistics of subgroups and need overall statistics.
Shortcut Method for Variance	$σ^{2} = \frac{\sum x_{i}^{2}}{n} - {(\frac{\sum x_{i}}{n})}^{2}$	When raw data is available; saves time in calculations.
Probability - Basic Terms	$P (E) = \frac{n (E)}{n (S)}$ (equally likely outcomes)	Foundation for all probability problems; define sample space first.
Addition Theorem of Probability	$P (A \cup B) = P (A) + P (B) - P (A \cap B)$	When finding probability of at least one of two events occurring.
Mutually Exclusive and Exhaustive Events	$P (A \cap B) = 0$ (mutually exclusive), $P (A \cup B) = 1$ (exhaustive)	To simplify probability calculations; check if events partition sample space.
Conditional Probability	$P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ , $P (B) \neq 0$	When additional information is given; basis for Bayes' theorem.
Multiplication Theorem and Independent Events	$P (A \cap B) = P (A) \cdot P (B)$ (independent), $P (A \cap B) = P (A) \cdot P (B ∣ A)$ (general)	When events are independent (e.g., repeated trials); otherwise use conditional.
Bayes' Theorem	$P (A_{i} ∣ B) = \frac{P (A_{i}) \cdot P (B ∣ A_{i})}{\sum_{j} P (A_{j}) \cdot P (B ∣ A_{j})}$	When you have prior probabilities and new evidence; reverse conditional probability.
Random Variable and Probability Distribution	$P (X = x_{i}) = p_{i}$ , with $\sum p_{i} = 1$ , $p_{i} \geq 0$	To model random experiments numerically; basis for expectation and variance.
Expectation (Mean) of a Random Variable	$E (X) = \sum x_{i} p_{i}$	To find average outcome in the long run; used in games, insurance, etc.
Variance of a Random Variable	$V a r (X) = E (X^{2}) - [E (X)]^{2} = \sum (x_{i} - μ)^{2} p_{i}$	To quantify risk or variability; needed for binomial and other distributions.
Binomial Distribution	$P (X = k) = (\binom{n}{k}) p^{k} (1 - p)^{n - k}$ , $k = 0, 1, \dots, n$	When experiment has exactly two outcomes, repeated n times independently.
Mean and Variance of Binomial Distribution	$μ = n p$ , $σ^{2} = n p (1 - p)$	Quickly compute expected value and spread for binomial random variable.
Using Complement and Symmetry	$P (at least one) = 1 - P (no success)$	When direct calculation of many probabilities is tedious; common in binomial problems.
Probability Tree Diagrams	No formula; use tree to compute $P (A \cap B) = P (A) \cdot P (B ∣ A)$	When experiment has sequential steps (e.g., drawing without replacement).
Conditional Probability and Total Probability Theorem	$P (B) = \sum P (A_{i}) \cdot P (B ∣ A_{i})$	When B can occur via multiple mutually exclusive routes.
Solving for Missing Frequencies in Statistics	Set $\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$ or median formula equal to given value.	When frequency distribution has missing values and a measure of central tendency is given.

Open the Statistics and Probability formula page

What JEE asks

Mean (Arithmetic Mean). JEE often asks to compute mean from frequency distribution or combine means of two groups.
Median. JEE may ask to find median from ogive or from a frequency distribution with missing frequencies.
Mode. JEE tests empirical relation: Mode = 3 Median - 2 Mean, often to check consistency.
Range. Rarely asked directly; may appear in comparison of variability.
Mean Deviation. JEE may ask to compute MD from mean or median, especially for discrete data.
Variance and Standard Deviation. JEE often combines variance with mean, or asks to find combined variance of two groups.
Combined Mean and Variance. JEE frequently asks to compute combined variance from two groups' means and variances.
Shortcut Method for Variance. JEE expects this efficient method for variance in both ungrouped and grouped data.
Probability - Basic Terms. JEE tests basic counting and probability with dice, cards, coins.
Addition Theorem of Probability. JEE often asks for probability of union with given probabilities, sometimes with conditional.
Mutually Exclusive and Exhaustive Events. JEE tests these concepts in word problems, e.g., drawing cards or balls.
Conditional Probability. JEE frequently asks conditional probability in multi-stage experiments or with given tables.
Multiplication Theorem and Independent Events. JEE tests independence in problems like drawing with replacement or coin tosses.
Bayes' Theorem. JEE often asks Bayes' theorem in medical tests, defective items, or multi-stage experiments.
Random Variable and Probability Distribution. JEE asks to find missing probability or verify if given distribution is valid.
Expectation (Mean) of a Random Variable. JEE often asks expectation for simple distributions like dice sums or number of heads.
Variance of a Random Variable. JEE tests variance calculation for discrete distributions, often with expectation.
Binomial Distribution. JEE frequently asks for mean $n p$ and variance $n p (1 - p)$ , or probability of at least k successes.
Mean and Variance of Binomial Distribution. JEE often asks to find n or p given mean and variance.
Using Complement and Symmetry. JEE uses this in dice, coin, and card problems to save time.
Probability Tree Diagrams. JEE may ask for probability of a specific path or total probability using tree.
Conditional Probability and Total Probability Theorem. JEE uses this as a step in Bayes' theorem or in multi-stage problems.
Solving for Missing Frequencies in Statistics. JEE often asks to find missing frequency using mean or median.

Common traps

Forgetting to use class marks (midpoints) instead of class boundaries in grouped data. Mean (Arithmetic Mean)
Forgetting to sort data before finding median in ungrouped case. Median
Assuming mode is unique; data can be multimodal. Mode
Ignoring outliers that drastically change range. Range
Forgetting absolute values; using squared deviations instead. Mean Deviation
Dividing by n-1 when JEE says 'variance': unless 'sample variance' is explicitly stated, use the population formula (divide by n). Variance and Standard Deviation
Simply averaging variances without adjusting for mean differences. Combined Mean and Variance
Forgetting to square the mean; miscomputing sum of squares. Shortcut Method for Variance
Assuming outcomes are equally likely when they are not. Probability - Basic Terms
Forgetting to subtract intersection when events are not mutually exclusive. Addition Theorem of Probability
Assuming mutually exclusive events are also exhaustive (not necessarily). Mutually Exclusive and Exhaustive Events
Forgetting to check $P (B) > 0$ ; dividing by zero. Conditional Probability
Assuming independence without checking; using product for dependent events. Multiplication Theorem and Independent Events
Forgetting to normalize by total probability of B. Bayes' Theorem

Topics in Statistics and Probability

Measures of Central Tendency
Measures of Dispersion
Probability Basics
Conditional Probability
Bayes' Theorem
Binomial Distribution

Previous-year questions on Statistics and Probability

JEEGenie has 586 previous-year questions on Statistics and Probability. A few worked examples follow.

Year	Questions
2026	34
2025	32
2024	44
2023	53
2022	48
2021	60
2020	34
2019	46
2018	13
2017	15
2016	11
2015	14
2014	11
2013	10
2012	13
2011	8
2010	6
2009	8
2008	5
2007	8
2006	6
2005	7
2004	8
2003	8
2002	7
2001	2
2000	2
1999	3
1998	7
1997	3
1996	2
1995	3
1994	4
1993	3
1992	3
1991	4
1990	1
1989	4
1988	4
1987	4
1986	4
1985	4
1984	4
1983	3
1982	3
1981	3
1980	2
1979	3
1978	2

Question 1 medium

Let a, b, c

\in

{1, 2, 3, 4}. If the probability, that

a x^{2} + 2 \sqrt{2} b x + c > 0

for all

x \in R

, is

\frac{m}{n}

\gcd (m, n) = 1

, then

m + n

is equal to ________.

Answer: 81

Explanation. For the quadratic to be positive for all real

x

, we need -

a > 0

(which is always true since

a \in {1, 2, 3, 4}

), - discriminant

< 0

(2 \sqrt{2} b)^{2} - 4 a c < 0 ⟹ 8 b^{2} - 4 a c < 0 ⟹ 2 b^{2} < a c .

Total number of choices for

(a, b, c)

4^{3} = 64

. Count favorable triples

(a, b, c)

with

2 b^{2} < a c

. -

b = 1

2 < a c

a c \geq 3

. Pairs

(a, c)

from

{1, 2, 3, 4}

with product

\geq 3

: total

16

pairs minus those with product

\leq 2

(1, 1), (1, 2), (2, 1) \to 3

bad, so

13

good. -

b = 2

8 < a c

. Product

\geq 9

. Pairs:

(3, 3), (3, 4), (4, 3), (4, 4) \to 4

good. -

b = 3

18 < a c

. Max product

4 \cdot 4 = 16

, so

0

good. -

b = 4

32 < a c

, also

0

good. Total favorable =

13 + 4 = 17

. Probability =

\frac{17}{64}

, so

m = 17

n = 64

m + n = 81

81

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 hard

Three distinct numbers are selected randomly from the set

{1, 2, 3, \dots, 40}

. If the probability, that the selected numbers are in an increasing G.P., is

\frac{m}{n}, \gcd (m, n) = 1

, then

m + n

is equal to __________ .

Answer: 2477

Explanation. We need triples

a < b < c

from

{1, \dots, 40}

that form a geometric progression. Let the common ratio be

r = \frac{p}{q}

in lowest terms,

p > q > 0

\gcd (p, q) = 1

. Then the numbers are

a, a \frac{p}{q}, a \frac{p^{2}}{q^{2}}

. For these to be integers,

q^{2} ∣ a

. Write

a = k q^{2}

, so the triple is

k q^{2}, k p q, k p^{2} .

We need

k p^{2} \leq 40

. **Step 1:** Count possible

(p, q)

with

p > q \geq 1

\gcd (p, q) = 1

, and

p^{2} \leq 40

. Since

p^{2} \leq 40

p \leq 6

. List coprime pairs: -

p = 2

q = 1 \to p^{2} = 4

p = 3

q = 1, 2 \to p^{2} = 9

p = 4

q = 1, 3 \to p^{2} = 16

p = 5

q = 1, 2, 3, 4 \to p^{2} = 25

p = 6

q = 1, 5 \to p^{2} = 36

Total pairs:

1 + 2 + 2 + 4 + 2 = 11

. **Step 2:** For each

(p, q)

, count

k

such that

k p^{2} \leq 40

. -

p = 2

⌊ 40 / 4 ⌋ = 10

p = 3

⌊ 40 / 9 ⌋ = 4

(two pairs, so

2 \times 4 = 8

) -

p = 4

⌊ 40 / 16 ⌋ = 2

(two pairs, so

4

) -

p = 5

⌊ 40 / 25 ⌋ = 1

(four pairs, so

4

) -

p = 6

⌊ 40 / 36 ⌋ = 1

(two pairs, so

2

) Total triples:

10 + 8 + 4 + 4 + 2 = 28

. **Step 3:** However, each increasing G.P. is counted exactly once because

p > q

and

a

is the smallest term. So number of favorable selections =

28

. **Step 4:** Total ways to choose 3 distinct numbers from 40:

(\binom{40}{3}) = \frac{40 \cdot 39 \cdot 38}{6} = 9880.

**Step 5:** Probability =

\frac{28}{9880} = \frac{7}{2470}

. Simplify:

\gcd (7, 2470) = 1

, so

m = 7, n = 2470

. Thus

m + n = 7 + 2470 = 2477.

2477

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 1

Question 3 easy

A company has two plants

A

and

B

to manufacture motorcycles.

60 %

motorcycles are manufactured at plant

A

and the remaining are manufactured at plant

B .80 %

of the motorcycles manufactured at plant

A

are rated of the standard quality, while

90 %

of the motorcycles manufactured at plant

B

are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If

p

is the probability that it was manufactured at plant

B

, then

126 p

Answer: A

Explanation. Given:

P (A) = 0.6, P (B) = 0.4 P (Standard ∣ A) = 0.80, P (Standard ∣ B) = 0.90

Step 1: Find the total probability of a standard-quality motorcycle using the law of total probability.

P (Standard) = P (A) \cdot P (Standard ∣ A) + P (B) \cdot P (Standard ∣ B) P (Standard) = (0.6) (0.8) + (0.4) (0.9) = 0.48 + 0.36 = 0.84

Step 2: Apply Bayes' theorem to find

p = P (B ∣ Standard)

p = P (B ∣ Standard) = \frac{P (B) \cdot P (Standard ∣ B)}{P (Standard)} = \frac{(0.4) (0.9)}{0.84} = \frac{0.36}{0.84} = \frac{36}{84} = \frac{3}{7}

Step 3: Compute

126 p

126 p = 126 \times \frac{3}{7} = 18 \times 3 = 54

From NTA/IIT papers: JEE Main 2024 · 6 Apr Shift 1

Question 4 medium

A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is

\frac{k}{3^{11}}

, then

k

is equal to :

Answer: C

Explanation. **Principle:** Repeated independent Bernoulli trials — use the binomial distribution. For one throw of two dice, total = 5 occurs as

(1, 4), (2, 3), (3, 2), (4, 1)

: 4 outcomes out of 36.

p = \frac{4}{36} = \frac{1}{9}, q = 1 - \frac{1}{9} = \frac{8}{9}

With

n = 5

throws,

X

= number of successes:

P (X = r) = (\binom{5}{r}) {(\frac{1}{9})}^{r} {(\frac{8}{9})}^{5 - r}

We need at least 4 successes:

P (X \geq 4) = P (4) + P (5)

P (4) = (\binom{5}{4}) {(\frac{1}{9})}^{4} (\frac{8}{9}) = \frac{5 \cdot 8}{9^{5}} = \frac{40}{9^{5}}

P (5) = {(\frac{1}{9})}^{5} = \frac{1}{9^{5}}

P (X \geq 4) = \frac{40 + 1}{9^{5}} = \frac{41}{9^{5}}

Now

9^{5} = (3^{2})^{5} = 3^{10}

, so:

P (X \geq 4) = \frac{41}{3^{10}} = \frac{123}{3^{11}}

Comparing with

\frac{k}{3^{11}}

gives

k = 123

. The other options fail: 82, 164, 75 do not equal

3 \times 41 = 123

, which is forced by multiplying numerator and denominator by 3. Answer: (C)

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

Suppose that Box-I contains 8 red, 3 blue and 5 green balls, Box-II contains 24 red, 9 blue and 15 green balls, Box-III contains 1 blue, 12 green and 3 yellow balls, Box-IV contains 10 green, 16 orange and 6 white balls. A ball is chosen randomly from Box-I; call this ball

b

. If

b

is red then a ball is chosen randomly from Box-II, if

b

is blue then a ball is chosen randomly from Box-III, and if

b

is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

$\frac{15}{256}$
$\frac{3}{16}$
$\frac{5}{52}$
$\frac{1}{8}$

Answer: C

Explanation. We need

P (W ∣ G) = \frac{P (W \cap G)}{P (G)}

where

W

= one of the chosen balls is white,

G

= at least one chosen ball is green. **Step 1: Probabilities from Box-I** Total balls in Box-I =

8 + 3 + 5 = 16

P (R) = \frac{8}{16} = \frac{1}{2}, P (B) = \frac{3}{16}, P (G_{1}) = \frac{5}{16} .

**Step 2: Cases for

G

** - If first ball is red: second from Box-II (24R, 9B, 15G; total 48). Green appears only if second is green: probability

\frac{15}{48} = \frac{5}{16}

. Contribution:

\frac{1}{2} \cdot \frac{5}{16} = \frac{5}{32}

. - If first ball is blue: second from Box-III (1B, 12G, 3Y; total 16). Green appears if second is green: probability

\frac{12}{16} = \frac{3}{4}

. Contribution:

\frac{3}{16} \cdot \frac{3}{4} = \frac{9}{64}

. - If first ball is green: second from Box-IV (10G, 16O, 6W; total 32). At least one green is already true. Probability = 1. Contribution:

\frac{5}{16} \cdot 1 = \frac{5}{16} = \frac{20}{64}

. **Step 3: Total

P (G)

P (G) = \frac{5}{32} + \frac{9}{64} + \frac{20}{64} = \frac{10}{64} + \frac{9}{64} + \frac{20}{64} = \frac{39}{64} .

**Step 4: Cases for

W \cap G

** White can only come from Box-IV, so first ball must be green. Then second ball is white with probability

\frac{6}{32} = \frac{3}{16}

. Thus

P (W \cap G) = \frac{5}{16} \cdot \frac{3}{16} = \frac{15}{256} .

**Step 5: Conditional probability**

P (W ∣ G) = \frac{15 / 256}{39 / 64} = \frac{15}{256} \cdot \frac{64}{39} = \frac{15 \cdot 64}{256 \cdot 39} = \frac{15}{4 \cdot 39} = \frac{15}{156} = \frac{5}{52} .

C) \frac{5}{52}

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Statistics and Probability?: JEEGenie estimates about 5% of JEE Main marks come from Statistics and Probability, based on recent papers. The share varies from year to year.
Which formulas matter most in Statistics and Probability?: The Statistics and Probability sheet on this page has 23 formula entries, starting with Mean (Arithmetic Mean), Median, Mode, Range, Mean Deviation. Each one comes with when to use it and the trap to avoid.
What topics does Statistics and Probability cover for JEE?: Statistics and Probability covers Measures of Central Tendency, Measures of Dispersion, Probability Basics, Conditional Probability, Bayes' Theorem, Binomial Distribution.
How should I revise Statistics and Probability before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Statistics and Probability questions on JEEGenie.
How many previous-year questions are there on Statistics and Probability?: JEEGenie has 586 previous-year questions on Statistics and Probability, and this page shows a few with worked explanations.
Is practice on Statistics and Probability free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.