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

Importance of Probability JEE Mains Questions

1. Low time consuming: This chapter can be covered with proper efficiency and attention.

2. High scoring: 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. Conceptually link: It is linked with chapters like PnC, Sets, Relations. This can help you keep the flow and understand concepts better.

4. Application Based Topic: In JEE Advanced, it has core importance as this forms the base of mathematical reasoning. Probability JEE Mains questions 2026 is therefore significantly important.

5. Probability JEE Mains weightage is 6.32% which cannot be ignored in JEE Main Math.

Probability JEE Main Previous Year Questions With Solutions

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Main Topics of Probability JEE Mains questions 2026

The following are some main topics of Probability JEE Mains questions with solutions asked in previous years:

Probability JEE Mains Questions

1. 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

$\frac{11}{15}$ $\frac{28}{75}$ $\frac{2}{15}$ $\frac{3}{5}$

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).

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 $\frac{3}{4}$ 2 $\frac{1}{2}$

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).

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

17 16.4 17.4 16

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).

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

8788

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.

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

7 4 9 5

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 answer is Option (1).