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

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:

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
$

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

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.