Probability JEE Main Questions: PYQs, Weightage, Formulas & Practice

Probability JEE Main Questions: PYQs, Weightage, Formulas & Practice

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JEE Main Application Date:31 Oct' 25 - 27 Nov' 25

Shivani PooniaUpdated on 01 Nov 2025, 02:09 PM IST

Probability JEE Main Questions: If you want to take admission into top IITs then maths is one such subject which you simply cannot ignore. In JEE Mains Mathematics, probability is one of the most important topics that tests the understanding of students in logic, principles and analytical thinking. Every year, there is a good amount of weightage from this chapter. Questions are mostly moderate level in difficulty and are application based. These questions are designed to assess how well students can apply probability concepts in real life and maths situations. It is considered one of the high scoring topics in the JEE Mains exam so you must be strong in formula and concepts. JEE Main 2026 registration has already started and students can register from 31 October 2025 to 27 November 2025, session 1 is scheduled from 21 to 30 January 2026.

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This Story also Contains

  1. Main Topics of Probability for JEE Mains Questions 2026
  2. Previous Year Questions JEE Mains: Probability
  3. Why Probability is Important in JEE Mains
  4. JEE Main Previous Year Maths Questions With Solutions
Probability JEE Main Questions: PYQs, Weightage, Formulas & Practice
Probability JEE Main Questions: PYQs, Weightage, Formulas & Practice

In this article, important questions from Probability are provided that will help you understand the types of questions asked in the JEE Main exam. Questions from concepts like Discrete Random Variable and Variance, Mean deviation about Mean, Standard deviation and Variance are covered. They help to improve conceptual clarity and speed during the exam. Go through the questions carefully, revise the formulas related to these questions, and analyse the logic behind every solution to ensure solid preparation for JEE Mains Mathematics.

Main Topics of Probability for JEE Mains Questions 2026

Probability in JEE Mains covers a variety of concepts that test your reasoning and mathematical understanding. These questions are designed to check how to apply formulas and solve questions. The following are some main topics of Probability JEE Mains questions with solutions asked in previous years:

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Topic

Key Focus Area

Terms In Probability

Basic definitions: experiment, outcome, sample space, event

Set Theoretical Notations Of Probability

Union, intersection, complement, Venn diagram representation

Algebra of Events

Laws of probability, mutually exclusive & exhaustive events

Multiplication Theorem on Probability

Formulas, proof, applications

Independent Event in Probability

Definition, examples, difference from mutually exclusive events

Conditional Probability: Definition, Formula, Properties And Examples

Definition, formula, properties, solved examples

Total Probability Theorem and Bayes' Theorem

Applications in problem-solving, real-life probability questions

Random Variables and its Probability Distributions

Discrete random variables, mean & variance

Bernoulli Trials and Binomial Distribution

Properties, probability mass function, examples

Also refer

Previous Year Questions JEE Mains: Probability

Solving previous year questions from probability is one of the best ways to prepare for the JEE Main exam. These questions help you to understand the exam pattern and the types of questions asked. Given below some previous year questions of JEE Mains from Probability:

Question: A box contains 10 pens, of which 3 are defective. A sample of 2 pens is drawn at random, and let X denote the number of defective pens. Then the variance of X is

(1) $\frac{11}{15}$

(2) $\frac{28}{75}$

(3) $\frac{2}{15}$

(4) $\frac{3}{5}$

Solution: Discrete Random Variable and Variance Calculation

The concept involves using a discrete random variable $X$ with a given probability distribution $P(X = x_i)$. The mean (expected value) is calculated using:

$
\mu = \sum x_i P(x_i)
$

The variance is given by:

$
\text{Var}(X) = \sum P(x_i)(x_i - \mu)^2
$

xx = 0x = 1x = 2
$\mathrm{P}(\mathrm{x})$$\frac{{ }^7 \mathrm{C}_2}{{{ }^{10} \mathrm{C}_2}}$$\frac{{ }^{7} \mathrm{C}_1 \ ^{3}C_1 }{^{10} \mathrm{C}_2}$$\frac{{ }^3 \mathrm{C}_2}{{ }^{10} \mathrm{C}_2}$
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$\begin{aligned} & \mu=\Sigma x_i P\left(x_i\right)=0+\frac{7}{15}+\frac{2}{15}=\frac{3}{5} \\ & \text { Variance }(x)=\Sigma P_i\left(x_i-\mu\right)^2=\frac{28}{75}\end{aligned}$

Hence, the correct answer is option (2).

Question: Let the mean and the standard deviation of the observation $2,3,3,4,5,7, \mathrm{a}, \mathrm{b}$ be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is :

(1) 1

(2) $\frac{3}{4}$

(3) 2

(4) $\frac{1}{2}$

Solution: Mean Deviation About Mean

First find the mean, i.e.

$
\bar{x}=\frac{\sum_{i=1}^n x_i f_i}{\sum_{i=1}^n f_i}=\frac{1}{\mathrm{~N}} \sum_{i=1}^n x_i f_i
$

$N$ is the sum of all frequencies
Then, find the deviations of observations $x_i$ from the mean $\bar{x}$ and take their absolute values, i.e., $\left|x_i-\bar{x}\right|$ for all $i=1,2, \ldots, n$
After this, find the mean of the absolute values of the deviations
$\operatorname{M.D.}(\bar{x})=\frac{\sum_{i=1}^n f_i\left|x_i-\bar{x}\right|}{\sum_{i=1}^n f_i}=\frac{1}{N} \sum_{i=1}^n f_i\left|x_i-\bar{x}\right|$

Mean Deviation About any value 'a'

$
\text { M.D.(a) }=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left|x_i-\mathrm{a}\right|
$
Mean deviation for a grouped frequency distribution
The formula for mean deviation is the same as in the case of an ungrouped frequency distribution. Here, $x_i$ is the midpoint of each class.

Standard deviation

The standard deviation is a measure of the amount of variation or dispersion in a set of values. It quantifies how much the values in a data set deviate from the mean (average) value. A low standard deviation indicates that the data points tend to be close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range.

For a data set with values $x_1, x_2, \ldots, x_n$ and mean $\bar{x}$, the {population standard deviation is defined as

$\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2}$

If the data represents a sample rather than the entire population, the sample standard deviation is calculated as

$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$

where $n$ is the number of observations.

$\text { Mean is given by } \frac{2+3+3+4+5+7+a+b}{8}=4 \text {. }$

$\begin{aligned} & \frac{24+a+b}{8}=4 \\ & a+b=8\ldots(1)\end{aligned}$

The standard deviation is given by $\sqrt{\frac{\sum_{i=1}^8\left(x_i-4\right)^2}{8}}=\sqrt{2}$.
Squaring both sides:

$
\frac{(2-4)^2+(3-4)^2+(3-4)^2+(4-4)^2+(5-4)^2+(7-4)^2+(a-4)^2+(b-4)^2}{8}=2
$

$\begin{aligned} &2=\frac{4+1+1+0+1+9+(a-4)^2+(b-4)^2}{8} \\ & 16=48+a^2+b^2-8 a-8 b \\ & a^2+b^2=32 \\ & 32=2 a b \\ & a b=16 \\ & a=4, b=4 \\ & \text { mode }=4 \\ & \text { mean deviation }=\frac{2+1+1+0+1+3+0+0}{8}=1\end{aligned}$

Hence, the answer is option (1).

Question: Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=b, a>b$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n, n=1,2, \ldots \ldots \ldots 5$ is

(1) 17

(2) 16.4

(3) 17.4

(4) 16

Solution: Variance

The mean of the squares of the deviations from the mean is called the variance and is denoted by $\sigma^2$ (read as sigma square).
Variance is a quantity which leads to a proper measure of dispersion.
The variance of $n$ observations $x_1, x_2, \ldots, x_n$ is given by

$
\sigma^2 = \frac{\sum x_i^2}{n} - \left( \frac{\sum x_i}{n} \right)^2
$

Calculate the mean:

$5=\frac{1+3+a+7+b}{5}$

$\Rightarrow a+b=14$

$\frac{1+9+a^2+49-b^2}{5}-(5)^2=10$

$a^2+b^2=116$

$\Rightarrow a=10, b=4$

New observations: $2,5,13,11,9$

Var $=\frac{4+26+169+121+81}{5}-64$

Var $=80.2-64$

Var $\approx 16$
Hence, the answer is option (4).

Question: The variance of the numbers $8,21,34,47, \ldots, 320$, is____________.

Solution: The mean of the squares of the deviations from the mean is called the variance and is denoted by $\sigma^2$ (read as sigma square).
Variance is a quantity that leads to a proper measure of dispersion.
The variance of $n$ observations $x_1, x_2, \ldots, x_n$ is given by

$
\sigma^2=\frac{1}{n} \displaystyle\sum_{i=1}^n\left(x_i-\bar{x}\right)^2
$

Variance formula for an AP:

$\sigma^2=\frac{(n^2-1)d^2}{d-1}$

Given the arithmetic sequence: $ 8, 21, 34, 47, \ldots, 320 $

First term, $ a = 8 $, common difference, $ d = 13 $

Use the $ n^{th} $ term formula:
$
a_n = a + (n-1)d
$

Put $ a_n = 320 $:
$
320 = 8 + (n-1) \times 13 \\
312 = 13(n-1) \\
n-1 = \frac{312}{13} = 24 \\
n = 25
$

Number of terms, $ n = 25 $

Mean of the AP:
$
\bar{x} = \frac{a + l}{2} = \frac{8 + 320}{2} = 164
$

Variance of AP is given by:
$
\sigma^2 = \frac{(n^2 - 1)d^2}{12}
$

Substitute values:
$
\sigma^2 = \frac{(25^2 - 1) \times 13^2}{12} = \frac{624 \times 169}{12}
$

Calculate numerator:
$
624 \times 169 = 105456
$

Divide:
$
\sigma^2 = \frac{105456}{12} = 8788
$

Hence, the answer is 8788.

Question: For a statistical data $x_1, x_2, \ldots, x_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i=1}^{10} \mathrm{x}_{\mathrm{i}}^2=371$. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is

(1) 7

(2) 4

(3) 9

(4) 5

Solution: Variance

The mean of the squares of the deviations from the mean is called the variance and is denoted by $\sigma^2$ (read as sigma square).
Variance is a quantity which leads to a proper measure of dispersion.
The variance of $n$ observations $x_1, x_2, \ldots, x_n$ is given by

$
\sigma^2=\text { Variance }=\sigma^2=\frac{1}{n} \sum x_i^2-\bar{x}^2$

Given:

Number of data points: $n=10$
Incorrect mean: $\bar{x}_{\text {wrong }}=5.5$
$\sum_{i=1}^{10} x_i^2=371$ (sum of squares of the incorrect data)
Two values were recorded incorrectly as 4 and 5, but the correct values are 6 and 8

The incorrect sum of values using the mean:

$
\sum x_i=10 \times 5.5=55
$

Corrected sum $=55-(4+5)+(6+8)=55-9+14=60$

Corrected mean:

$
\bar{x}_{\text {correct }}=\frac{60}{10}=6
$

Original incorrect squared sum:

$
\sum x_i^2=371
$
Remove the incorrect squares, add the correct ones:

$
\begin{gathered}
\sum x_i^2(\text { correct })=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=371-(16+25)+(36+64) \\
=371-41+100=430
\end{gathered}
$

$\begin{aligned} & \text { Variance }=\frac{1}{n} \sum x_i^2-\bar{x}^2 \\ & =\frac{430}{10}-6^2=43-36=7\end{aligned}$

Hence, the correct answer is Option (1).

Why Probability is Important in JEE Mains

Probability is an important part of JEE Main Maths because questions from this chapter come every year in the exam. If the basics from this chapter are well prepared then it is an easy and scoring chapter that will help you improve your overall marks. Given below some point on why this chapter is important for JEE Mains:

1.This chapter is less time consuming and can be covered with proper efficiency and attention.

2. Every year this chapter has a significant amount of weightage associated with it. Probability JEE Mains questions with solutions can increase your rank too if studied properly.

3. It is linked with chapters like Permutation and combination, Sets, Relations. This can help you keep the flow and understand concepts better.

4. In JEE Advanced, it has core importance as this forms the base of mathematical reasoning.

5. Probability JEE Mains weightage is approximately 6.32% which cannot be ignored.

Download: JEE Main 2026 Important Formulas for Maths PDF

JEE Main Previous Year Maths Questions With Solutions

Solving previous year questions is one of the best ways to prepare for the JEE Main exam. These questions help you to understand the exam pattern and the types of questions asked. Solving these questions improves your problem solving speed and accuracy. To find the probability PYQs you can download the ebooks given below for practice anytime.

JEE Main Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters

Frequently Asked Questions (FAQs)

Q: Is Probability an important topic for JEE Main?
A:

Yes, Probability is an important topic in JEE Main Mathematics. It carries a good weightage every year and is considered a high-scoring chapter if the concepts and formulas are well understood. It also links with other chapters like Permutations & Combinations and Sets. 

Q: Is Probability difficult for JEE Main?
A:

Not really. It is considered moderate in difficulty and requires strong basics in formulas and logic.

Q: How many questions are asked from Probability in JEE Main?
A:

1-2 questions are asked constituting 4-8 marks.

Q: What are the main topics of Probability in JEE Main?
A:

The core concepts of probability are Bayes theorem, random variables, binomial distribution and basic terms/events.

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If you follow a proper plan and practice regularly, you can boost your chances of scoring well in JEE Mains.

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Currently, there is no official release of JEE Mains 2026 registration begin. But on an average, every year there will be 11 Lakh to 12.5 Lakh students register

Hello,

If you belong to the BC (Backward Class) category, you can choose OBC-NCL in JEE Mains only if your caste is included in the Central OBC list released by the Government of India.

Here’s how it works:

  • Central OBC list: Used for national-level exams like JEE, NEET, UPSC, etc.

  • State BC list: Used only for state-level admissions or jobs.

So, if your BC caste name is in the Central list, select OBC–NCL while filling the JEE form.
If your caste is not in the Central list, choose General category, because the state BC certificate won’t be valid for JEE.

Example:
If you’re BC in Tamil Nadu but your caste isn’t in India’s Central list, you’ll be counted as General for JEE.

Hope you understand.

Hello,

No, it’s not compulsory to fill the APAAR ID in the JEE Main application form right now.

If your APAAR ID shows a “credential mismatch” error even after entering the correct number, don’t worry, you can simply leave it blank and continue filling the form.

  • The APAAR ID field is optional for most students.

  • It’s mainly for linking your academic records under the Academic Bank of Credits (ABC) system.

  • The mismatch often happens if your Aadhaar details or school record name format differ.

So, you can submit your JEE form without the APAAR ID, and it won’t affect your application.


Hope you understand.