Aakash Repeater Courses
Take Aakash iACST and get instant scholarship on coaching programs.
The game starts with the dice roll followed by a coin flip which is allowed only when the dice roll produces an even number. If you already rolled an even number in the dice what are your odds to obtain a head outcome? Conditional probability applies perfectly to circumstances where an upcoming event requires the previous one to meet particular conditions. These daily scenarios where the chance of one event rests upon a preceding event are precisely.
This Story also Contains
This exercise covers the conditional probability formula and teaches how to solve dependent events alongside using the multiplication theorem of probability. In this exerise we discuss basic probability rules to check whether two events maintain independence with detailed NCERT solutions for Class 12 Maths Chapter 13 Exercise 13.2 solutions you will be able to master all these concepts and solve the questions step by step with confidence.
You can download the PDF file of the Chapter 13 Exercise 13.2 Class 12 Maths Solutions of NCERT that contains crucial Probability problems and practice them without difficulty. Download PDF
Question 1: If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$ find $P(A\cap B)$ if $A$ and $B$ are independent events.
Answer:
$P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5},$
Given : $A$ and $B$ are independent events.
So we have, $P(A\cap B)=P(A).P(B)$
$\Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{5}\times \frac{1}{5}$
$\Rightarrow \, \, \, \, P(A\cap B)=\frac{3}{25}$
Answer:
Two cards are drawn at random and without replacement from a pack of 52 playing cards.
There are 26 black cards in a pack of 52.
Let $P(A)$ be the probability that first cards is black.
Then, we have
$P(A)= \frac{26}{52}=\frac{1}{2}$
Let $P(B)$ be the probability that second cards is black.
Then, we have
$P(B)= \frac{25}{51}$
The probability that both the cards are black $=P(A).P(B)$
$=\frac{1}{2}\times \frac{25}{51}$
$=\frac{25}{102}$
Answer:
Total oranges = 15
Good oranges = 12
Bad oranges = 3
Let $P(A)$ be the probability that first orange is good.
The, we have
$P(A)= \frac{12}{15}=\frac{4}{5}$
Let $P(B)$ be the probability that second orange is good.
$P(B)=\frac{11}{14}$
Let $P(C)$ be the probability that third orange is good.
$P(C)=\frac{10}{13}$
The probability that a box will be approved for sale $=P(A).P(B).P(C)$
$=\frac{4}{5}.\frac{11}{14}.\frac{10}{13}$
$=\frac{44}{91}$
Answer:
A fair coin and an unbiased die are tossed,then total outputs are:
$= \left \{ (H1),(H2),(H3),(H4),(H5),(H6),(T1),(T2),(T3),(T4),(T5),(T6) \right \}$
$=12$
A is the event ‘head appears on the coin’ .
Total outcomes of A are : $= \left \{ (H1),(H2),(H3),(H4),(H5),(H6) \right \}$
$P(A)=\frac{6}{12}=\frac{1}{2}$
B is the event ‘3 on the die’.
Total outcomes of B are : $= \left \{ (T3),(H3)\right \}$
$P(B)=\frac{2}{12}=\frac{1}{6}$
$\therefore A\cap B = (H3)$
$P (A\cap B) = \frac{1}{12}$
Also, $P (A\cap B) = P(A).P(B)$
$P (A\cap B) = \frac{1}{2}\times \frac{1}{6}=\frac{1}{12}$
Hence, A and B are independent events.
Answer:
Total outcomes $=\left \{ 1,2,3,4,5,6 \right \}=6$.
$A$ is the event, ‘the number is even,’
Outcomes of A $=\left \{ 2,4,6 \right \}$
$n(A)=3.$
$P(A)=\frac{3}{6}=\frac{1}{2}$
$B$ is the event, ‘the number is red’.
Outcomes of B $=\left \{ 1,2,3 \right \}$
$n(B)=3.$
$P(B)=\frac{3}{6}=\frac{1}{2}$
$\therefore (A\cap B)=\left \{ 2 \right \}$
$n(A\cap B)=1$
$P(A\cap B)=\frac{1}{6}$
Also,
$P(A\cap B)=P(A).P(B)$
$P(A\cap B)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}\neq \frac{1}{6}$
Thus, both the events A and B are not independent.
Answer:
Given :
$P(E)=\frac{3}{5},P(F)=\frac{3}{10}$ and $P(E\cap F)=\frac{1}{5}.$
For events E and F to be independent , we need
$P(E\cap F)=P(E).P(F)$
$P(E\cap F)=\frac{3}{5}\times \frac{3}{10}=\frac{9}{50}\neq \frac{1}{5}$
Hence, E and F are not indepent events.
(i) mutually exclusive
Answer:
Given,
$P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}$
Also, A and B are mutually exclusive means $A\cap B=\phi$.
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$\frac{3}{5}=\frac{1}{2}+P(B)-0$
$P(B)=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$
(ii) independent
Answer:
Given,
$P(A)=\frac{1}{2},P(A\cup B)=\frac{3}{5}$
Also, A and B are independent events means
$P(A\cap B) = P(A).P(B)$. Also $P(B)=p.$
$P(A\cap B) = P(A).P(B)=\frac{p}{2}$
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$\frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}$
$\frac{p}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$
$p=\frac{2}{10}=\frac{1}{5}$
Question 8: Let A and B be independent events with $P(A)=0.3$ and $P(B)=0.4$ Find
(i) $P(A\cap B)$
Answer:
$P(A)=0.3$ and $P(B)=0.4$
Given : A and B be independent events
So, we have
$P(A\cap B)=P(A).P(B)$
$P(A\cap B)=0.3\times 0.4=0.12$
Question 8: Let$A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$ Find
(ii) $P(A\cup B)$
Answer:
$P(A)=0.3$ and $P(B)=0.4$
Given : A and B be independent events
So, we have
$P(A\cap B)=P(A).P(B)$
$P(A\cap B)=0.3\times 0.4=0.12$
We have, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$P(A\cup B)=0.3+0.4-0.12=0.58$
Question 8: Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$ Find
(iii) $P(A\mid B)$
Answer:
$P(A)=0.3$ and $P(B)=0.4$
Given : A and B be independent events
So, we have $P(A\cap B)=0.12$
$P(A|B)=\frac{P(A\cap B)}{P(B)}$
$P(A|B)=\frac{0.12}{0.4}= 0.3$
Question 8: Let A and B be independent events with $P(A)=0.3$ and $P(B)=0.4$ Find
(iv) $P(B\mid A)$
Answer:
$P(A)=0.3$ and $P(B)=0.4$
Given : A and B be independent events
So, we have $P(A\cap B)=0.12$
$P(B|A)=\frac{P(A\cap B)}{P(A)}$
$P(B|A)=\frac{0.12}{0.3}= 0.4$
Answer:
If $A$ and $B$ are two events such that $P(A)=\frac{1}{4},P(B)=\frac{1}{2}$ and $P(A\cap B)=\frac{1}{8},$
$P(not\; A\; and\; not\; B)= P(A'\cap B')$
$P(not\; A\; and\; not\; B)= P(A\cup B)'$ use, $(P(A'\cap B')= P(A\cup B)')$
$= 1-(P(A)+P(B)-P(A\cap B))$
$= 1-(\frac{1}{4}+\frac{1}{2}-\frac{1}{8})$
$= 1-(\frac{6}{8}-\frac{1}{8})$
$= 1-\frac{5}{8}$
$= \frac{3}{8}$
Answer:
If $A$ and $B$ are two events such that $P(A)=\frac{1}{2},P(B)=\frac{7}{12}$ and $P(not \; A \; or\; not\; B)=\frac{1}{4}.$
$P(A'\cup B')=\frac{1}{4}$
$P(A\cap B)'=\frac{1}{4}$ $(A'\cup B'=(A\cap B)')$
$\Rightarrow \, \, 1-P(A\cap B)=\frac{1}{4}$
$\Rightarrow \, \, \, P(A\cap B)=1-\frac{1}{4}=\frac{3}{4}$
$Also \, \, \, P(A\cap B)=P(A).P(B)$
$P(A\cap B)=\frac{1}{2}\times \frac{7}{12}=\frac{7}{24}$
As we can see $\frac{3}{4}\neq \frac{7}{24}$
Hence, A and B are not independent.
Question 11: Given two independent events $A$ and $B$ such that $P(A)=0.3,P(B)=0.6,$ Find
(i) $P(A \; and\; B)$
Answer:
$P(A)=0.3,P(B)=0.6,$
Given two independent events $A$ and $B$.
$P(A\cap B)=P(A).P(B)$
$P(A\cap B)=0.3\times 0.6=0.18$
Also , we know $P(A \, and \, B)=P(A\cap B)=0.18$
Question 11: Given two independent events A and B such that $P(A)=0.3,P(B)=0.6,$ Find
(ii) $P(A \; and \; not\; B)$
Answer:
$P(A)=0.3,P(B)=0.6,$
Given two independent events $A$ and $B$.
$P(A \; and \; not\; B)$$=P(A)-P(A\cap B)$
$=0.3-0.18=0.12$
Question 11: Given two independent events A and B such that $P(A)=0.3,P(B)0.6,$ Find
(iii) $P(A\; or \; B)$
Answer:
$P(A)=0.3,P(B)=0.6,$
$P(A\cap B)=0.18$
$P(A\; or \; B)=P(A\cup B)$
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$=0.3+0.6-0.18$
$=0.9-0.18$
$=0.72$
Question 11: Given two independent events $A$ and $B$ such that $P(A)=0.3,P(B)=0.6,$ Find
(iv) $P(neither\; A\; nor\; B)$
Answer:
$P(A)=0.3,P(B)=0.6,$
$P(A\cap B)=0.18$
$P(A\; or \; B)=P(A\cup B)$
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$=0.3+0.6-0.18$
$=0.9-0.18$
$=0.72$
$P(neither\; A\; nor\; B)$ $=P(A'\cap B')$
$= P((A\cup B)')$
$=1-P(A\cup B)$
$=1-0.72$
$=0.28$
Question 12: A die is tossed thrice. Find the probability of getting an odd number at least once.
Answer:
A die is tossed thrice.
Outcomes $=\left \{ 1,2,3,4,5,6 \right \}$
Odd numbers $=\left \{ 1,3,5 \right \}$
The probability of getting an odd number at first throw
$=\frac{3}{6}=\frac{1}{2}$
The probability of getting an even number
$=\frac{3}{6}=\frac{1}{2}$
Probability of getting even number three times
$=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$
The probability of getting an odd number at least once = 1 - the probability of getting an odd number in none of throw
= 1 - probability of getting even number three times
$=1-\frac{1}{8}$
$=\frac{7}{8}$
(i) both balls are red.
Answer:
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Total balls =18
Black balls = 10
Red balls = 8
The probability of getting a red ball in first draw
$=\frac{8}{18}=\frac{4}{9}$
The ball is repleced after drawing first ball.
The probability of getting a red ball in second draw
$=\frac{8}{18}=\frac{4}{9}$
the probability that both balls are red
$=\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}$
(ii) first ball is black and second is red.
Answer:
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Total balls =18
Black balls = 10
Red balls = 8
The probability of getting a black ball in the first draw
$=\frac{10}{18}=\frac{5}{9}$
The ball is replaced after drawing the first ball.
The probability of getting a red ball in the second draw
$=\frac{8}{18}=\frac{4}{9}$
the probability that the first ball is black and the second is red
$=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}$
(iii) one of them is black and other is red.
Answer:
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.
Total balls =18
Black balls = 10
Red balls = 8
Let the first ball is black and the second ball is red.
The probability of getting a black ball in the first draw
$=\frac{10}{18}=\frac{5}{9}$
The ball is replaced after drawing the first ball.
The probability of getting a red ball in the second draw
$=\frac{8}{18}=\frac{4}{9}$
the probability that the first ball is black and the second is red
$=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}$ $...........................1$
Let the first ball is red and the second ball is black.
The probability of getting a red ball in the first draw
$=\frac{8}{18}=\frac{4}{9}$
The probability of getting a black ball in the second draw
$=\frac{10}{18}=\frac{5}{9}$
the probability that the first ball is red and the second is black
$=\frac{4}{9}\times \frac{5}{9}=\frac{20}{81}$ $...........................2$
Thus,
The probability that one of them is black and the other is red = the probability that the first ball is black and the second is red + the probaility that the first ball is red and the second is black $=\frac{20}{81}+\frac{20}{81}=\frac{40}{81}$
(i) the problem is solved
Answer:
$P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$
Since, problem is solved independently by A and B,
$\therefore$ $P(A\cap B)=P(A).P(B)$
$P(A\cap B)=\frac{1}{2}\times \frac{1}{3}$
$P(A\cap B)=\frac{1}{6}$
probability that the problem is solved $= P(A\cup B)$
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$P(A\cup B)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}$
$P(A\cup B)=\frac{5}{6}-\frac{1}{6}$
$P(A\cup B)=\frac{4}{6}=\frac{2}{3}$
(ii) exactly one of them solves the problem
Answer:
$P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{3}$
$P(A')=1-P(A)$, $P(B')=1-P(B)$
$P(A')=1-\frac{1}{2}=\frac{1}{2}$ , $P(B')=1-\frac{1}{3}=\frac{2}{3}$
probability that exactly one of them solves the problem $=P(A\cap B') + P(A'\cap B)$
probability that exactly one of them solves the problem $=P(A).P(B')+P(A')P(B)$
$=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}$
$= \frac{2}{6}+\frac{1}{6}$
$= \frac{3}{6}=\frac{1}{2}$
(i) E : ‘the card drawn is a spade’
F : ‘the card drawn is an ace’
Answer:
One card is drawn at random from a well shuffled deck of $52$ cards
Total ace = 4
total spades =13
E : ‘the card drawn is a spade
F : ‘the card drawn is an ace’
$P(E)=\frac{13}{52}=\frac{1}{4}$
$P(F)=\frac{4}{52}=\frac{1}{13}$
$E\cap F :$ a card which is spade and ace = 1
$P(E\cap F)=\frac{1}{52}$
$P(E).P(F)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}$
$\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{52}$
Hence, E and F are indepentdent events .
(ii) E : ‘the card drawn is
F : ‘the card drawn is a king’
Answer:
One card is drawn at random from a well shuffled deck of $52$ cards
Total black card = 26
total king =4
E : ‘the card drawn is black’
F : ‘the card drawn is a king’
$P(E)=\frac{26}{52}=\frac{1}{2}$
$P(F)=\frac{4}{52}=\frac{1}{13}$
$E\cap F :$ a card which is black and king = 2
$P(E\cap F)=\frac{2}{52}=\frac{1}{26}$
$P(E).P(F)=\frac{1}{2}\times \frac{1}{13}=\frac{1}{26}$
$\Rightarrow P(E\cap F)=P(E).P(F)=\frac{1}{26}$
Hence, E and F are indepentdent events .
(iii) E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’.
Answer:
One card is drawn at random from a well shuffled deck of $52$ cards
Total king or queen = 8
total queen or jack = 8
E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’.
$P(E)=\frac{8}{52}=\frac{2}{13}$
$P(F)=\frac{8}{52}=\frac{2}{13}$
$E\cap F :$ a card which is queen = 4
$P(E\cap F)=\frac{4}{52}=\frac{1}{13}$
$P(E).P(F)=\frac{2}{13}\times \frac{2}{13}=\frac{4}{169}$
$\Rightarrow P(E\cap F)\neq P(E).P(F)$
Hence, E and F are not indepentdent events
(a) Find the probability that she reads neither Hindi nor English newspapers
Answer:
H : $60\; ^{o}/_{o}$ of the students read Hindi newspaper,
E : $40\; ^{o}/_{o}$ read English newspaper and
$H \cap E :$ $20\; ^{o}/_{o}$ read both Hindi and English newspapers.
$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$
$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$
$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$
the probability that she reads neither Hindi nor English newspapers $=1-P(H\cup E)$
$=1-(P(H)+P(E)-P(H\cap E))$
$=1-(\frac{3}{5}+\frac{2}{5}-\frac{1}{5})$
$=1-\frac{4}{5}$
$=\frac{1}{5}$
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
Answer:
H : $60\; ^{o}/_{o}$ of the students read Hindi newspaper,
E : $40\; ^{o}/_{o}$ read English newspaper and
$H \cap E :$ $20\; ^{o}/_{o}$ read both Hindi and English newspapers.
$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$
$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$
$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$
The probability that she reads English newspape if she reads Hindi newspaper $=P(E|H)$
$P(E|H)=\frac{P(E\cap H)}{P(H)}$
$P(E|H)=\frac{\frac{1}{5}}{\frac{3}{5}}$
$P(E|H)=\frac{1}{3}$
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Answer:
H : $60\; ^{o}/_{o}$ of the students read Hindi newspaper,
E : $40\; ^{o}/_{o}$ read English newspaper and
$H \cap E :$ $20\; ^{o}/_{o}$ read both Hindi and English newspapers.
$P(H)=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$
$P(E)=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$
$P(H\cap E)=\frac{20}{100}=\frac{2}{10}=\frac{1}{5}$
the probability that she reads Hindi newspaper if she reads English newspaper $= P(H |E)$
$P(H |E)=\frac{P(H\cap E)}{P(E)}$
$P(H |E)=\frac{\frac{1}{5}}{\frac{2}{5}}$
$P(H |E)=\frac{1}{2}$
Question 17: The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
(A) $0$
(B) $\frac{1}{3}$
(C) $\frac{1}{12}$
(D) $\frac{1}{36}$
Answer:
when a pair of dice is rolled, total outcomes $=6^2=36$
Even prime number $=\left \{ 2 \right \}$
$n(even \, \, prime\, \, number)=1$
The probability of obtaining an even prime number on each die $=P(E)$
$P(E)=\frac{1}{36}$
Option D is correct.
Question 18: Two events A and B will be independent, if
(A) $A$ and $B$ are mutually exclusive
(B) $P(A'B')=\left [ 1-P(A) \right ]\left [ 1-P(B) \right ]$
(C) $P(A)=P(B)$
(D) $P(A)+P(B)=1$
Answer:
Two events A and B will be independent, if
$P(A\cap B)=P(A).P(B)$
Or $P(A'\cap B')=P(A'B')=P(A').P(B')=(1-P(A)).(1-P(B))$
Option B is correct.
1. $
\text { Multiplication Theorem on Probability }
$
$
\begin{aligned}
&\begin{aligned}
\mathrm{P}(\mathrm{E} \cap \mathrm{~F}) & =\mathrm{P}(\mathrm{E}) \mathrm{P}(\mathrm{~F} \mid \mathrm{E}) \\
& =\mathrm{P}(\mathrm{~F}) \mathrm{P}(\mathrm{E} \mid \mathrm{F}) \text { provided } \mathrm{P}(\mathrm{E}) \neq 0 \text { and } \mathrm{P}(\mathrm{~F}) \neq 0 .
\end{aligned}\\
&\text { The above result is known as the multiplication rule of probability. }
\end{aligned}
$
2. Independent Events
Take Aakash iACST and get instant scholarship on coaching programs.
$
P(A \cap B)=P(A) \cdot P(B)
$
This condition must hold both ways:
$
P(A \mid B)=P(A), \quad \text { and } \quad P(B \mid A)=P(B)
$
Also Read
Also see-
Frequently Asked Questions (FAQs)
There are 18 questions in NCERT Class 12 Maths chapter 13 exercise 13.2.
If the occurrence of one event doesn’t affect the probability of the other event such events are called independent events.
The probability of getting a tail is 0.5 when we toss a fair coin.
You will get NCERT Exemplar Solutions for Class 12 Maths Probability by clicking on the given link.
An event is called an impossible event if the number of favorable outcomes is zero.
The probability of an impossible event is zero.
Click here to get NCERT Solutions for Class 12 Maths for other subjects.
The probability of getting head of this biased coin is 0.53.
On Question asked by student community
Hello
Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.
Hello Aspirant,
Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.
Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE