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

Important Concepts of Probability for JEE Main 2027

Probability in JEE Main 2027 is a part of many concepts that require technical reasoning and knowledge of mathematics. These questions are designed to check how to apply formulas and solve questions. The following are some main topics of Probability JEE Main questions with solutions asked in previous years:

Probability Previous Year Questions JEE Main

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 are some previous year questions of JEE Main from Probability:

Question: A bag contains $(\mathrm{N}+1)$ coins - N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then N is equal to:

(1) 5

(2) 7

(3) 8

(4) 9

Solution:

Total ( $\mathrm{N}+1$ ) coins
$\mathrm{N} \Rightarrow$ fair coins
$1 \Rightarrow$ Head on both side

$\begin{aligned}
& P\left(1^{+}\right)=\left(\frac{N}{N+1}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{N+1}\right) 1 \\
& =\frac{N+2}{2(N+1)}=\frac{9}{16} \\
& 8 N+16=9 N+9 \\
& \Rightarrow N=7
\end{aligned}$

Question: If $A$ and $B$ are two events such that $\mathrm{P}(\mathrm{A})=0.7$, $\mathrm{P}(\mathrm{B})=0.4$ and $\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=0.5$, where $\overline{\mathrm{B}}$ denotes the complement of $B$, then $P(B \mid(A \cup \bar{B}))$ is equal:-

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

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

(3) $\frac{1}{6}$

(4) $\frac{1}{3}$

Solution:

$\begin{aligned} & P(A)=0.7 \\ & P(B)=0.4 \\ & P\left(A \cap B^C\right)=0.5\end{aligned}$

$\begin{aligned}
& P\left(B / A \cup B^C\right)=\frac{P\left(B \cap\left(A \cup B^C\right)\right)}{P\left(A \cup B^C\right)} \\
& =\frac{P(A \cap B)}{P\left(A \cup B^C\right)}
\end{aligned}$

Using Venn diagram

$\Rightarrow \quad \frac{P(A \cap B)}{P\left(A \cup B^C\right)}=\frac{0.2}{0.5+0.2+0.1}=\frac{2}{8}=\frac{1}{4}$

Hence, the answer is option (1).

Question: A card from a pack of 52 cards is lost. From the remaining 51 cards, $n$ cards are drawn and are found to be spades. If the probability of the lost card to be a spade is $\frac{11}{50}$, the n is equal to

(1) 1

(2) 2

(3) 4

(4) 5

Solution:

Let the event $L$ be: "Lost card is a spade"
Let the event $D$ be: "All $n$ drawn cards are spades"

We are given:

$
P(L \mid D) = \frac{11}{50}
$

We will use Bayes' Theorem:

$
P(L \mid D) = \frac{P(D \mid L) \cdot P(L)}{P(D)}
$

First, compute the prior probabilities:
Number of spades in a full deck = 13
Probability that the lost card is a spade:

$
P(L) = \frac{13}{52} = \frac{1}{4}
$

Probability that the lost card is not a spade:

$
P(L') = \frac{39}{52} = \frac{3}{4}
$

Now calculate: If the lost card is a spade, then the number of spades left = $13 - 1 = 12$, and the total cards left = 51.

So, $
P(D \mid L) = \frac{{^{12}C_n}}{{^{51}C_n}}
$

If the lost card is not a spade, then all 13 spades are still there.

So, $
P(D \mid L') = \frac{{^{13}C_n}}{{^{51}C_n}}
$

Now apply Bayes' Theorem:

$
\frac{11}{50} = \frac{\frac{{^{12}C_n}}{{^{51}C_n}} \cdot \frac{1}{4}}{\frac{{^{12}C_n}}{{^{51}C_n}} \cdot \frac{1}{4} + \frac{{^{13}C_n}}{{^{51}C_n}} \cdot \frac{3}{4}}
$

Cancel common factor ${^{51}C_n}$ in numerator and denominator:

$
\frac{11}{50} = \frac{{^{12}C_n} \cdot \frac{1}{4}}{{^{12}C_n} \cdot \frac{1}{4} + {^{13}C_n} \cdot \frac{3}{4}}
$

Multiply numerator and denominator by 4:

$
\frac{11}{50} = \frac{{^{12}C_n}}{{^{12}C_n} + 3 \cdot {^{13}C_n}}
$

Now, cross-multiply:

$
11 \left( {^{12}C_n} + 3 \cdot {^{13}C_n} \right) = 50 \cdot {^{12}C_n}
$

$
11 \cdot {^{12}C_n} + 33 \cdot {^{13}C_n} = 50 \cdot {^{12}C_n}
$

$
33 \cdot {^{13}C_n} = 39 \cdot {^{12}C_n}
$

Divide both sides by 3: $
11 \cdot {^{13}C_n} = 13 \cdot {^{12}C_n}
$

Use identity: $
{^{13}C_n} = \frac{13}{13 - n} \cdot {^{12}C_n}
$

So: $
11 \cdot \frac{13}{13 - n} \cdot {^{12}C_n} = 13 \cdot {^{12}C_n}
$

Cancel ${^{12}C_n}$: $
\frac{143}{13 - n} = 13
$

Now solve: $
143 = 13 \cdot (13 - n)
$

$
143 = 169 - 13n
$

$
13n = 169 - 143 = 26
$

$n = 2$

Hence, the answer is option (2).

Question: If the probability that the random variable X takes the value x is given by $\mathrm{P}(\mathrm{X}=\mathrm{x})=\mathrm{k}(\mathrm{x}+1) 3^{-x}$, $\mathrm{x}=0,1,2,3 \ldots \ldots$., where k is a constant, then $\mathrm{P}(\mathrm{X} \geq 3)$ is equal to

(1) $\frac{7}{27}$

(2) $\frac{4}{9}$

(3) $\frac{8}{27}$

(4) $\frac{1}{9}$

Solution:

$s = \frac{k}{3^0} + \frac{2k}{3} + \frac{3k}{3^2} + \ldots$

Dividing both sides by 3,

$\frac{s}{3} = \frac{k}{3} + \frac{2k}{3^2} + \frac{3k}{3^3} + \ldots$

Subtracting,

$s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \ldots$

The right side is a geometric series with sum

$k \times \frac{1}{1 - \frac{1}{3}} = \frac{3k}{2}$

Thus,

$\frac{2s}{3} = \frac{3k}{2} \implies s = \frac{9k}{4}$

Since total probability $s=1$,

$\frac{9k}{4} = 1 \implies k = \frac{4}{9}$

Now, calculate $P(x \geq 3)$:

$P(x \geq 3) = 1 - \left(P(x=0) + P(x=1) + P(x=2)\right)$

$1 - \left(k + \frac{2k}{3} + \frac{3k}{3^2}\right) = 1 - 2k$

$= 1 - 2 \times \frac{4}{9}$

$= \frac{1}{9}$

Hence, the correct answer is option (4).

Question: One die has two faces marked 1, two faces marked 2, one face marked 3, and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3, and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both dice are thrown together, is:

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

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

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

(4) $\frac{4}{9}$

Solution:

Dice 1 : 1, 1, 2, 2, 3, 4
Dice 2 : 1, 2, 2, 3, 3, 4

For Sum $=4$

$
\begin{aligned}
& (1,3) \Rightarrow 2 \times 2=4 \\
& (3,1) \Rightarrow 1 \times 1=1 \\
& (2,2) \Rightarrow 2 \times 2=4
\end{aligned}
$

Total ways $=9$

$
\text { Sum = } 5
$
$
\begin{aligned}
& (1,4) \Rightarrow 2 \times 1=2 \\
& (4,1) \Rightarrow 1 \times 1=1 \\
& (2,3) \Rightarrow 2 \times 2=4 \\
& (3,2) \Rightarrow 1 \times 2=2
\end{aligned}
$
Total ways $=9$

$
\text { Probability }=\frac{18}{36}=\frac{1}{2}
$

Hence, the correct answer is option (1).

Probability Weightage and Importance in JEE Main 2027

Probability is an important part of JEE Main math 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 are some points on why this chapter is important for the JEE Main exam:

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

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

How to Prepare Probability for JEE Main 2027

Learn the formulas first, but don't stop there — understand where each formula comes from. A formula you've only memorized will fail you the moment a question is slightly rephrased.

• Solve the previous 10 years' PYQs before anything else. Not after finishing the chapter while you're studying it. JEE Main repeats question types more often than it repeats questions, and spotting those patterns early changes how you prioritize your time.

• Don't stay in your comfort zone with easy problems. Moderate and difficult questions are where marks are actually won or lost. If you're only solving textbook exercises, you're not testing yourself against what the paper actually looks like.

• Take chapter-wise mock tests before moving on. Finishing the chapter isn't the same as being ready to answer questions from it under time pressure.

• Revise your short notes every week, not the night before the exam. Probability has enough interconnected concepts—conditional probability feeding into Bayes, random variables feeding into distributions—that weekly revision keeps the whole picture intact.