NCERT Solutions for Class 11 Maths Chapter 14 Exercise 14.2 Probability  PDF Download
JEE Mains & AdvancedNCERT solutions for class 11 maths chapter 14 exercise 14.2 probability is concerned with the axiomatic approach of finding probability. Ex 14.2 class 11 maths chapter 14 now combines sample space knowledge and events with axioms probability. This gives us a set of questions that not only help us practice but also help us gain a deeper understanding of the axiomatic method of probability.
Class 11 maths chapter 14 exercise 14.2 NCERT solutions consists of 21 questions. The first few questions will help you understand that the probability of all the elements in the sample area is always equal to one and next few questions that are based on calculating the probability of different events. Ex 14.2 class 11 maths solutions have been explained in a detailed manner by the academic team of mathematics at eSaral that will help you score good marks in exams. Class 11 maths chapter 14 exercise 14.2 NCERT solutions are also provided in downloadable PDF format which you can download for free from eSaral website and practice all questions included in ex 14.2.
Topics Covered in Exercise 14.2 Class 11 Mathematics Questions
Ex 14.2 class 11 maths ch 14 has included some essential topics such as axiomatic approach to probability, probability of an event, probabilities of equally likely outcomes, probability of the event ‘A or B’, and probability of event ‘not A’. Check below the detailed explanation of each topic to understand the concepts given by subject experts of eSaral.
1. 
Axiomatic Approach to Probability 


Axiomatic Approach to Probability
Another way to describe the probability of an event is to use an axiomatic approach. In this approach, some general principles or axioms are used to represent the probability.
Let S be the sample space of a random experiment. The probability P is a real
valued function whose domain is the power set of S and range is the interval [0,1]
satisfying the following axioms
(i) For any event E, P (E) ≥ 0
(ii) P(S) = 1
(iii) If E and F are mutually exclusive events, then P(E ∪ F) = P(E) + P(F).
It follows from (iii) that P($\phi$) = 0. To prove this, we take F = $\phi$ and note that E and φ are disjoint events.
Therefore, from axiom (iii), we get
P (E ∪ $\phi$ ) = P (E) + P ($\phi$) or P(E) = P(E) + P ($\phi$) i.e. P ($\phi$) = 0.
Let S be a sample space containing outcomes ω_{1}, ω_{2},.....,ω_{n}, i.e.,
S = {ω_{1} , ω_{2} , ..., ω_{n} }
It follows from the axiomatic definition of probability that
(i) 0 ≤ P (ω_{i} ) ≤ 1 for each ω_{i} ∈ S
(ii) P (ω_{1} ) + P (ω_{2} ) + ... + P (ω_{n }) = 1
(iii) For any event A, P(A) = ∑ P(ω_{i} ), ω_{i} ∈ A.

Probability of an event  For a finite sample space with equally likely outcomes Probability of an event P(A) = $\frac{n(A)}{n(S)}$, where
n(A) = number of elements in the set A
n(S) = number of elements in the set S
If A and B are any two events, then
P(A or B) = P(A) + P(B) – P(A and B)
equivalently, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
If A and B are mutually exclusive, then P(A or B) = P(A) + P(B)
If A is any event, then P(not A) = 1 – P(A)

Probabilities of equally likely outcomes  Let S be a sample space and E be an event, such that n(S) = n and n(E) = m. If each outcome is equally likely, then it follows that
P(E) = $\frac{m}{n}=\frac{\text { number of outcomes favourable to } E}{\text { Total possible outcomes }}$

Probability of the event ‘A or B’  P(A∪B) = P(A) + P(B)  P(A∩B)

Probability of event ‘not A’  P( A′ ) = P(not A) = 1  P(A)
Tips for Solving Exercise 14.2 Class 11 Chapter 14 Probability
Ex 14.2 class 11 maths chapter 14 NCERT solutions Probability is a comprehensive introduction to axiomatic probability. To solve questions of ex 14.2 class 11 maths you need to go through these tips provided below.

Ex 14.2 class 11 maths chapter 14 is all about axiomatic probability which requires clear understanding of its concepts. Students must study these concepts before solving questions.

It is important to read the questions carefully as each one introduces a new problem and you have to think about the possible answers. Therefore, it is best to concentrate on the question wordings and note down the points that help define the sample area.

Students should solve questions provided in NCERT solutions class 11 maths ex 14.2 to get a better understanding of concepts.
Importance of Solving Ex 14.2 Class 11 Maths Chapter 14 Probability
There are numerous benefits of solving questions of ex 14.2 class 11 maths chapter 14 probability. You can go through these benefits to get a better understanding of concepts and questions.

NCERT solutions class 11 maths chapter 14 ex 14.2 has questions that are based on axiomatic probability, you must read the topics and comprehend the conceptual understanding of probability to get the nature of the questions.

Ex 14.2 class 11 maths NCERT solutions have questions that are solved precisely and in step by step manner by our expert teachers of eSaral.

In ex 14.2 class 11 maths chapter 14, you will also learn about venn diagrams which is an essential concept to solve questions related to probability.

In NCERT solutions class 11 maths chapter 14, you will get to solve examples and questions associated with probability which will help you achieve high marks in examinations.
Frequently Asked Questions
Question 1. What is the probability of equally likely events?
Answer 1. When the results of an experiment have the equal probability of occurrence, they are referred to as equally likely events. It can be denoted by
P(E) = $\frac{m}{n}=\frac{\text { number of outcomes favourable to } E}{\text { Total possible outcomes }}$
Question 2. Where can I download NCERT solutions for class 11 maths chapter 14 ex 14.2?
Answer 2. You can download NCERT solutions for class 11 maths chapter 14 ex 14.2 from the official website of eSaral. You should not only read the solutions but also practice the questions included in these solutions to get a clear understanding of concepts.
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