# NCERT Solutions for Class 11 Maths Chapter 16 Probability (Ex 16.2) Exercise 16.2

Both public and private schools receive secondary education from the Central Board of Secondary Education, which is headquartered in New Delhi. CBSE is governed and administered by the Indian government. Since 1961, the National Council of Educational Research and Training has been working to improve the quality of school education in the country. This organisation provides advisory and assistance services to the State Governments and the Central Government. In addition to model textbooks, supplementary materials, newsletters, journals, educational kits, and multimedia digital products, they also create and publish educational materials. For the students studying in Class 11, a minimum score of 33% must be achieved on both the theory and practical examinations in order to be promoted to Class 12.

The study of Mathematics is an important part of the academic curriculum for students who have chosen to study it. As part of their course curriculum, all Commerce and Science stream students are required to take Mathematics. The Exercises Question And Answers For Class 11 Mathematics are designed to simplify all the problems found in the CBSE textbooks for Class 11. All the exercises, questions, and answers in NCERT Class 11 Mathematics are organised systematically and cover the entire curriculum. Along with Probability, other chapters include Sets, the Principle of Mathematical Induction, Linear Inequalities, Trigonometric Functions, the Binomial Theorem, Limits and Derivatives, among others.

Class 11 Mathematics covers Probability in Chapter 16. Probability had been studied previously by the students in preceding classes. Therefore, they are familiar with the concepts discussed in the chapter. The frequency with which unexpected events occur can be defined as Probability. Exercise 16.2 in the textbook focuses on the occurrence of events, the types of events, and event algebra, among other topics. There are numerous examples of each type of event provided in the textbook. For students to be adequately prepared, it is essential that they complete exercise 16.2. Probability is one of the most fascinating and rewarding chapters of statistics. Through the use of Extramarks’ NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2, students can better prepare for the exam.

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NCERT Solutions are also included in the CBSE exam preparation study materialsBefore taking the final examination, students should review the NCERT Solutions Class 11 Mathematics at least twice.Extramarks offers solutions not only for Mathematics but also for other chapters and classes. Using Extramarks’ website, students in Class 1 through Class 12 have easy access to these solutions. For all subjects, the solutions have been curated by experts and are considered to be an invaluable resource. For various classes, students can access the NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions Class 9, NCERT Solutions Class 10, and NCERT Solutions Class 12 on the Extramarks’ website and learning platform.

## NCERT Solutions for Class 11 Maths Chapter 16 Probability (Ex 16.2) Exercise 16.2

On the Extramarks website, the students can easily find NCERT Solutions Class 11 Maths Chapter 16 Exercise 16.2. During the preparation for the final Mathematics Class 11 exam, students should pay particular attention to Class 11 Maths Chapter 16 Exercise 16.2 e.g. questions. Extramarks’ website and mobile app provide students with access to a variety of study materials. The website provides a variety of study resources to help students prepare well, which include past years’ papers, revision notes, sample papers, and extra questions. The NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2 can be downloaded from Extramarks and used to assist students in preparing for their exams.

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### NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.2

For answers to questions in the Class 11 Mathematics textbook, students should refer to the NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2. It is also imperative that students understand the concepts presented in this chapter. Because the NCERT book frequently contains exam questions, practising exercises from it is generally beneficial. There are a number of questions contained in this chapter that may be challenging for students. The NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2 are available at Extramarks. To prepare for the examination, students should refer to authentic and original solutions. In order to access reliable study materials like NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2 should utilise Extramarks’ learning platform. Through the use of NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2, students can accelerate their learning process and improve their academic performance. These solutions help them gain a deeper understanding of the problems. The Extramarks website provides students with the opportunity to solve practise modules in order to improve their understanding of the concepts.

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### Exercise 16.2

The topic of probability is covered in Chapter 16 of Class 11 Mathematics. The students had previously studied probability in previous classes. This has resulted in their familiarity with the concept of the chapter. A measure of the frequency with which unanticipated events occur. There are several topics discussed in Exercise 16.2 in the textbook. These topics include the occurrence of events, the types of events, event algebra, and others. Throughout the textbook, a variety of examples are provided for each type of event. In exercise 16.2 of Class 11 Mathematics, there are a total of seven questions divided into sub-parts. It is advised that students study diligently and refer to the NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2 in order to cross-check their answers. It is imperative that students complete the exercise in order to be adequately prepared for the chapter. The study of probability is an enthralling and rewarding experience. In this way, students can prepare for their exams more effectively by using NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2.

### NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.2

For a deeper understanding of the concepts presented in Class 11 Mathematics NCERT Solutions, students should refer to the examples provided in Class 11 Maths Exercise 16.2. The answers to each exercise are provided in detail, so students can follow the steps in order to complete them. It is evident that some questions and patterns are repeated among them. The NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2 have been prepared by highly qualified professionals to meet the needs of students preparing for CBSE examinations. The Extramarks’ NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2 helps students achieve high levels of effectiveness because of this. It is critical to thoroughly understand the concepts involved in NCERT exercise 16.2 in order to be able to solve it. In order to facilitate students’ understanding of how to solve complex problems, each step of the method is explained in a simple manner. According to the CBSE examination mark distribution, the NCERT Solutions For Class 11 Maths Chapter 16 Exercise 16.2 provide step-by-step solutions. This chapter has been updated according to the most recent CBSE guidelines. In order to achieve a higher grade, students will be required to solve a variety of problems in this chapter.

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**Q.1** A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?

**Ans.**

Here, E = {4}, F = {2, 4, 6}

E ∩ F = {4}

Since, E ∩ F ≠

$\mathrm{\varphi}$

So, E and F are not mutually exclusive events.

**Q.2 ** A die is thrown. Describe the following events:

(i) A: a number less than 7

(ii) B: a number greater than 7

(iii) C: a multiple of 3

(iv) D: a number less than 4

(v) E: an even number greater than 4

(vi) F: a number not less than 3

Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F′, F′

**Ans.**

Here,

(i) A = {1, 2, 3, 4, 5, 6}

(ii) B = {} = Φ

(iii) C = {3, 6}

(iv) D = {1, 2, 3}

(v) E = {6}

(vi) F = {3, 4, 5, 6}

A ∪ B = {1, 2, 3, 4, 5, 6},

A ∩ B = {} = Φ

B ∪ C = {3, 6}

E ∩ F = {6}

D ∩ E = {} = Φ

A – C = {1, 2, 4, 5}

D – E = {1, 2, 3}

E ∩ F’ = {6} ∩ [{1, 2, 3, 4, 5, 6} – {3, 4, 5, 6}]

= {6} ∩ {1, 2}

= {} = Φ

F’ = {1, 2}

**Q.3** An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

A: the sum is greater than 8,

B: 2 occurs on either die

C: the sum is at least 7 and a multiple of 3.

Which pairs of these events are mutually exclusive?

**Ans.**

$\mathrm{S}=\left\{\begin{array}{l}(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)\\ (2,1)(2,2)(2,3)(2,4)(2,5)(2,6)\\ (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)\\ (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)\\ (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)\\ (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)\end{array}\right\}$

A = {(3, 6), (4, 5), (4, 6),(5, 4), (5, 5),(5, 6), (6, 3),(6, 4), (6, 5),(6, 6)}

B = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}

C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}

Since, A ∩ B =

$\mathrm{\varphi}$

. So, A and B are mutually exclusive.

And B ∩ C =

$\mathrm{\varphi}$

. So, B and C are mutually exclusive.

**Q.4 ** Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event ‘a head shows on the first coin”.

Which events are (i) mutually exclusive? (ii) simple? (iii) Compound?

**Ans.**

$\begin{array}{l}\mathrm{S}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right),\left(\mathrm{TTT}\right)\}\\ \mathrm{A}=\left\{\left(\mathrm{HHH}\right)\right\}\\ \mathrm{B}=\{\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right)\}\\ \mathrm{C}=\left\{\left(\mathrm{TTT}\right)\right\}\\ \mathrm{D}=\{\left(\mathrm{HTT}\right),\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{HHT}\right)\}\\ \left(\mathrm{i}\right)\mathrm{Since},\text{ A}\cap \text{B =}\mathrm{\varphi},\text{so, A and B are Mutually exclusive.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Since},\text{ A}\cap \text{C =}\mathrm{\varphi}\text{so, A and C are Mutually exclusive.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Since},\text{ B}\cap \text{C =}\mathrm{\varphi}\text{so, B and C are Mutually exclusive.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Since},\text{ C}\cap \text{D =}\mathrm{\varphi}\text{so, C and D are Mutually exclusive.}\\ \left(\mathrm{ii}\right)\text{A and C are simple events because they have single element.}\\ \left(\mathrm{iii}\right)\mathrm{Compound}\text{events are B and D.}\end{array}$

**Q.5 ** Three coins are tossed. Describe

(i) Two events which are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive.

(iii) Two events, which are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive.

(v) Three events which are mutually exclusive but not exhaustive.

**Ans.**

$\begin{array}{l}\mathrm{S}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right),\left(\mathrm{TTT}\right)\}\\ \left(\mathrm{i}\right)\text{Two mutually exclusive events are:}\\ \text{}\left\{\left(\mathrm{HHH}\right)\right\}\text{and}\left\{\left(\mathrm{TTT}\right)\right\}.\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}Three events which are mutually exclusive and\hspace{0.17em}}\\ \text{exhaustive:\hspace{0.17em}}\\ \text{A}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right)\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=\{\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right)\}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{C}=\text{\hspace{0.17em}\hspace{0.17em}}\{\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right),\left(\mathrm{TTT}\right)\}\\ \because \mathrm{A}\cap \mathrm{B}=\mathrm{\varphi},\mathrm{B}\cap \mathrm{C}=\mathrm{\varphi},\mathrm{A}\cap \mathrm{C}=\mathrm{\varphi}\text{and A}\cup \text{B}\cup \mathrm{C}=\text{S}\\ \left(\mathrm{iii}\right)\mathrm{D}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right)\}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{E}=\{\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\left(\mathrm{TTH}\right)\}\\ \because \mathrm{D}\cap \mathrm{E}\ne \mathrm{\varphi},\text{so, D and E are not mutually exclusive.}\\ \left(\mathrm{iv}\right)\mathrm{F}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right),\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right),\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right),\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{G}=\left\{\left(\mathrm{TTH}\right)\right\}\\ \mathrm{Since},\text{\hspace{0.17em}}\mathrm{F}\cap \mathrm{G}=\mathrm{\varphi}\text{but F}\cup \text{G}=\text{S}\\ \text{So, F and G are exclusive but not exhaustive.}\\ \left(\mathrm{v}\right)\text{H}=\{\left(\mathrm{HHH}\right),\left(\mathrm{HHT}\right)\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}K}=\{\left(\mathrm{HTH}\right),\left(\mathrm{THH}\right)\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}L}=\{\left(\mathrm{HTT}\right),\left(\mathrm{THT}\right)\}\\ \because \mathrm{H}\cap \mathrm{K}=\mathrm{\varphi},\text{\hspace{0.17em}}\mathrm{H}\cap \mathrm{L}=\mathrm{\varphi}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{K}\cap \mathrm{L}=\mathrm{\varphi}\text{and H}\cup \text{K}\cup \mathrm{L}\ne \text{S}\\ \text{So, H, K and L are mutually exclusive but not exhaustive.}\end{array}$

**Q.6 ** Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5.

Describe the events

(i) A′ (ii) not B ,

(iii) A or B (iv) A and B

(v) A but not C (vi) B or C

(vii) B and C (viii) A ∩ B′∩C′

**Ans.**

$\begin{array}{l}\mathrm{S}=\left\{\begin{array}{l}(1,1),\text{\hspace{0.17em}\hspace{0.17em}}(1,2),\text{\hspace{0.17em}\hspace{0.17em}}(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \text{A}=\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \text{B}=\left\{\begin{array}{l}(1,1),\text{\hspace{0.17em}\hspace{0.17em}}(1,2),\text{\hspace{0.17em}\hspace{0.17em}}(1,3),(1,4),(1,5),(1,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\end{array}\right\}\\ \mathrm{C}=\left\{\begin{array}{l}(1,1),\text{\hspace{0.17em}\hspace{0.17em}}(1,2),\text{\hspace{0.17em}\hspace{0.17em}}(1,3),(1,4),(2,1),\\ (2,2),(2,3),(3,1),(3,2),(4,1)\end{array}\right\}\end{array}$

$\begin{array}{l}\left(\mathrm{i}\right)\text{A\u2019}=\mathrm{S}-\mathrm{A}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(1,1),\text{\hspace{0.17em}\hspace{0.17em}}(1,2),\text{\hspace{0.17em}\hspace{0.17em}}(1,3),(1,4),(1,5),(1,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\end{array}\right\}=\mathrm{B}\\ \left(\mathrm{ii}\right)\text{not B}=\mathrm{S}-\mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}=\mathrm{A}\\ \left(\mathrm{iii}\right)\mathrm{A}\text{or B}=\mathrm{A}\cup \mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(1,1),\text{\hspace{0.17em}\hspace{0.17em}}(1,2),\text{\hspace{0.17em}\hspace{0.17em}}(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}=\mathrm{S}\\ \left(\mathrm{iv}\right)\text{A and B}=\mathrm{A}\cap \mathrm{B}\text{\hspace{0.17em}\hspace{0.17em}}=\mathrm{\varphi}\\ \left(\mathrm{v}\right)\mathrm{A}\text{but not C}=\mathrm{A}-\mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ -\left\{\begin{array}{l}(1,1),\text{\hspace{0.17em}\hspace{0.17em}}(1,2),\text{\hspace{0.17em}\hspace{0.17em}}(1,3),(1,4),(2,1),\\ (2,2),(2,3),(3,1),(3,2),(4,1)\end{array}\right\}\end{array}$

\begin{array}{l}\end{array} $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),\\ (4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \left(\mathrm{vi}\right)\text{B or C}=\mathrm{B}\cup \mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(1,1),\text{\hspace{0.17em}\hspace{0.17em}}(1,2),\text{\hspace{0.17em}\hspace{0.17em}}(1,3),(1,4),(1,5),(1,6),(2,1),\\ (2,2),(2,3),(3,1),(3,2),(3,3),(3,4),(3,5),\\ (3,6),(4,1),(5,1),(5,2),(5,3),(5,4),(5,5),\\ (5,6)\end{array}\right\}\\ \left(\mathrm{vii}\right)\mathrm{B}\text{}\mathrm{and}\text{}\mathrm{C}=\mathrm{B}\cap \mathrm{C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\{(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)\}\\ \left(\mathrm{viii}\right)\mathrm{A}\cap \mathrm{B}\u2018=\mathrm{A}\cap (\mathrm{S}-\mathrm{B})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{A}\cap \mathrm{A}\left[\mathrm{From}\text{\hspace{0.17em}}\left(\mathrm{ii}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{A}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{C}\u2018=(\mathrm{S}-\mathrm{C})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{\begin{array}{l}(1,5),(1,6),(2,4),(2,5),(2,6),(3,3),(3,4),\\ (3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),\\ (6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\cap \mathrm{B}\u2018\cap \mathrm{C}\u2018=\left\{\begin{array}{l}(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),\\ (4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}\end{array}$

**Q.7** Refer to question 6 above, state true or false: (Give reason for your answer)

(i) A and B are mutually exclusive

(ii) A and B are mutually exclusive and exhaustive

(iii) A = B′

(iv) A and C are mutually exclusive

(v) A and B′ are mutually exclusive.

(vi) A′, B′, C are mutually exclusive and exhaustive.

**Ans.**

1. True, because A ∩ B = Φ

2. True, because A ∩ B = Φ and A È B = S.

3. True, because B’= S – B = A.

4. False, because A ∩ C ≠ Φ

5. False, because A ∩ B’ = A ∩ A ≠ Φ [Here, B’ = A]

6. False, because A’ ∩ B’ = Φ, B’ ∩ C ≠ Φ

## FAQs (Frequently Asked Questions)

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