NCERT Solutions For Class 11 Maths Chapter 16 Probability (Ex 16.3) Exercise 16.3

The primary textbooks for CBSE students are NCERT books, and with proper utilisation, students can be well-prepared for their new endeavours. NCERT books are highly recommended for students and offer a variety of benefits for enhancing their foundational knowledge. NCERT books support the development of a thorough understanding of the essential concepts in every topic. Students will be able to adequately comprehend the ideas by carefully reading the NCERT textbooks. Once students have the appropriate conceptual clarity on topics, they will be able to solve problems successfully and efficiently. Students are advised to understand the concepts rather than memorise them, which will be beneficial in future competitive examinations as well. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will help students understand the complex questions in the NCERT Mathematics book for the particular chapter. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 are specially designed for the students who are planning to appear in competitive examinations like Joint Entrance Exam (JEE) or Central Universities Entrance Test (CUET). Students must know NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 is crucial for their annual examinations as well.

Mathematics is offered as an ‘instrumental subject’ in some curricula to supplement the study of other school subjects, while in others, integrated courses that combine Mathematics and other fields are offered.   Mathematics concepts are also used in physics, chemistry, economics, engineering, and other fields.Mathematics can be a complex academic discipline at times, but students must become aware that Mathematics includes the contributions of great mathematicians from all over the world, compiled over centuries. NCERT books are written with the idea of what students’ minds can comprehend at a given age in mind, and are based on that age.The NCERT textbooks’ goal is for students to grow intellectually. Extramarks is an online learning platform that has assembled NCERT Solutions to support students in their academic endeavours while keeping in mind the base of NCERT books. Mathematics is required to fully comprehend some topics in other subjects as well. Mathematics is present everywhere in one form or another. The subject matter can be found in different shapes, sizes, times, degrees, and others. Mathematics is an exacting subject, and students must remember to solve questions with precise steps and to obtain precise results.Hence, Extramarks is also all about accuracy and precision, so, students should feel free to get help from NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 as it is trustworthy and reliable.

The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 is  key material for any student who is considering qualifying for different competitive examinations. NCERT books assist students in clearly understanding concepts so that they can easily solve questions that may be asked in examinations.Therefore, Extramarks focuses on the question pattern of the NCERT book and provides solutions for every exercise. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will help students so that they will face no complexities while solving questions.  NCERT books strictly adhere to CBSE rules and regulations.NCERT is the supreme body for management and research in India. Students who study under the CBSE board must know that CBSE examinations can be passed with high scores if students focus on NCERT books. Students can easily access NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 from the Extramarks website or mobile application. They can also access every subject’s NCERT solutions and more from Extramarks.

NCERT Solutions for Class 11 Maths Chapter 16 Probability (Ex 16.3) Exercise 16.3

Students in CBSE Class 11 can quickly download the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 from the Extramarks website or mobile application, so they can see and revise the answers to Exercise 16.3 in Class 11 Maths NCERT Solutions even when they are not online. Additionally, Class 11 Maths Exercise 16.3 will aid in their understanding of the questions.

Students who diligently study from NCERT textbooks perform well in exams and have a strong conceptual understanding.Therefore, learning material like the NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3 is essential for students when they encounter challenges. While trying to solve the questions, it can get complicated when students get stuck on a question. Extramarks has been a great help to students in overcoming all of these problems and has improved their conceptual grasp. When students complete the exercises, they will quickly comprehend the questions, since the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 among other exercises, are written in an approachable and straightforward manner. Extramarks has left no stone unturned to make students understand each topic of every subject easily. Class 11 Mathematics topics are broad but simple, and a deeper understanding of the topics will help students  solve questions in the annual Mathematics examination.

Students need to be aware that Extramarks is a platform where they will be supported and encouraged by qualified instructors in particular subject areas. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will be helpful in a quick review of the chapter. Students can simply use NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 so that they can depend on Extramarks for their queries and concerns. Students should use the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 since it was framed with the assistance of skilled Mathematics professionals and studying it thoroughly with such ease will help students decide what career path to choose after Class 12. The Science stream could be challenging at times, students should become familiar with Class 11 early on. Students are recommended to consistently study and maintain focus on their areas of weakness. They must continue to utilise the Extramarks NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 because it will assist them during exam preparations.

What is Exercise 16.3 of Class 11 Maths Chapter 16 Probability About?

The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 has been solved by the experts of Mathematics and has been uploaded on the Extramarks website and mobile application. Probability is covered extensively in the Class 11 Mathematics curriculum and is crucial for both the Class 11 examinations and other engineering examinations like the JEE as well. Students may have studied the idea of probability as a concept of the degree of uncertainty surrounding certain phenomena in prior classes. Students must download the  NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 to know clearly about Class 11 Maths Exercise 16.3.

The following significant areas of Probability will be covered in depth by students in Class 11 Maths Chapter 16 Exercise 16.3.

  • Axiomatic Approach to Probability
  • Probability of an event
  • Probabilities of equally likely outcomes
  • Probability of the event ‘A or B’
  • Probability of event ‘not A’

To answer questions in this exercise, students must be able to remember the following three key ideas from the axiomatic theory of probability:

  • The probability of any event is either greater or equal to zero
  • The probability of sample space is equal to 1
  • For two mutually exclusive events A and B, P (A OR B) = P(A) + P(B)

Questions now incorporate events and sample space knowledge with axioms of probability. This gives students a set of problems that they can practise while also strengthening their knowledge of this axiomatic method of probability. To practise probability, students should consult NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3.

Access NCERT Solutions for Class 11 Chapter 16 – Probability

Extramarks’ website and mobile application give students access to the NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3. Students who prepare for examinations like the Joint Entrance Exam (JEE) can use this application very easily. This gives learners who aspire to pursue their studies at the Indian Institute of Technology (IITs) and National Institute of Technology (NITs) a fundamental understanding with the help of NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3. Students can consider careers outside of engineering as well by applying for competitive tests like the Central Universities Entrance Test (CUET). The questions asked in such competitive examinations are mostly Multiple Choice Questions (MCQs), NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will be beneficial for them. Mathematics is the most important subject for qualifying examinations like JEE.

All the exercises of Chapter 16- Probability are available with easy solutions on the Extramarks website and mobile applications. There are three exercises and a miscellaneous exercise for students to practise so that they will get a grasp of all the topics that have been discussed in the given chapter.

Class 11 is a decisive year in a student’s life, and they must have a clear concept and know the plus and minus of the stream they will be choosing. If students choose Mathematics, they must seek help from NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3. NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will help students to solve Mathematics problems in a simple and easy manner. The class 10 and class 12 boards are the two most difficult stages in a student’s life.In between the two classes, Class 11 is the transition period in students’ lives. Class 11 sometimes gets challenging for students, and NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will be helpful for them. Class 10 gives a general idea of every subject, and afterwards, students must decide which subject interests them and choose their stream accordingly. Mathematics has been significant in everyone’s day-to-day life, and understanding the application of Mathematics is even more important. Therefore, students who choose Mathematics must know that it will be helpful for a lifetime, even ifthey decide to change their stream after Class 12. Hence, students must decide on the stream that will be beneficial for them in the long run. NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will help students to have a clear mindset while studying. Those students who choose Science must know that it will help them grow forever. NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will help students  do their best in examinations. Students are advised to solve the questions in Mathematics again and again to understand the concept of the  question. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 are helpful in the revision of the exercise.

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Exercise 16.3

Students must go through the exercises in Chapter 16- Probability and try to analyse the questions. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will help to understand the concepts of chapter 16- Probability. Students can download NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 from the Extramarks website and mobile application in PDF format so that they can study offline as well.

There are a total of 21 questions in the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3. Students must follow the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 to know the solutions of the exercise. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 are solved by the experts of Mathematics. The experts solved the NCERT Solutions for Class 11 Maths, Chapter 16: Probability Exercise 16.3 with precision and accuracy, so students will not face any difficulties while completing the Class 11 Maths Exercise 16.3.

There are a total of three Exercises, namely 16.1, 16.2 and 16.3 and a Miscellaneous Exercise of Chapter 16. All these exercises are solved by experts at Extramarks. Students must know that the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 are solved step by step. Students can follow the question-solving methods from NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3. They should know the importance of step marking in Mathematics and how to solve questions systematically. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 are solved with the easiest method that can be easily understandable by the students of Class 11.

Students must remember to understand the topics and concepts first before diving into solving and practising the exercise questions in NCERT books. The exercise in the NCERT book may be difficult for students to solve on their own.The NCERT Solutions for Class 11 Maths, Chapter 16, Probability Exercise 16.3, will come in handy when students need them.Hence, students must stay focused on understanding the concepts rather than memorising them.

NCERT Solutions for Class 11 Maths Chapters

Every chapter of the Class 11 solutions has been solved by Extramarks experts, students can also search for other chapters, and they will find them very useful. Students must click on the sections below to learn more about other chapters in Class 11. Students must download the Extramarks application or visit the Extramarks website to learn more about other subjects and classes.They can download NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 from there.

Every NCERT Solution of Class 11 is available on the Extramarks website or mobile application in chapter-wise sequence, with the exact numbering of the pages.

NCERT Solutions Class 11 Chapter 1- Sets

NCERT Solutions Class 11 Chapter 2 – Relations and Functions

NCERT Solutions Class 11 Chapter 3- Trigonometric Functions

NCERT Solutions Class 11 Chapter 4- Principle of Mathematical Induction

NCERT Solutions Class 11 Chapter 5- Complex Numbers and Quadratic Equations

NCERT Solutions Class 11 Chapter 6- Linear Inequalities

NCERT Solutions Class 11 Chapter 7- Permutations and Combinations

NCERT Solutions Class 11 Chapter 8- Binomial Theorem

NCERT Solutions Class 11 Chapter 9- Sequences and Series

NCERT Solutions Class 11 Chapter 10- Straight Lines

NCERT Solutions Class 11 Chapter 11- Conic Sections

NCERT Solutions Class 11 Chapter 12- Introduction to Three Dimensional Geometry

NCERT Solutions Class 11 Chapter 13- Limits and Derivatives

NCERT Solutions Class 11 Chapter 14- Mathematical Reasoning

NCERT Solutions Class 11 Chapter 15- Statistics

NCERT Solutions Class 11 Chapter 16- Probability

Students must complete the NCERT book exercise to ensure that their concepts are clear.To know the solutions to those exercises, students must follow NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 and others.

NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3

Students must know that the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 are of a total of 21 questions and each question has been solved with attention and clarity. They must solve each question to understand the basic concepts of the exercise. Each question of the exercise has been solved easily, simply and correctly. Students should always seek help from NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 whenever they feel stuck on any question. Students must try to solve questions on their own after using NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3. They must assess their understanding of the NCERT Solutions for Class 11 Maths Chapter 16 Probability Exercise 16.3.Students always try to understand the concepts of questions so that they will be able to solve them even in examinations. Students who understand question concepts will be able to solve complex questions based on the same concept.Therefore, students should always practise the exercises in NCERT books, as most of the questions asked in examinations are from NCERT books only. NCERT’s exercises could be complex at times, and NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will help students remove those complexities.

NCERT Solution Class 11 Maths of Chapter 16 All Exercises

Apart from NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 there are other NCERT Solutions as well on Extramarks website or mobile application. There are 3 exercises and a miscellaneous exercise in chapter 16- Probability. The first exercise is 16.1 which has 16 questions, then the second exercise is 16.2 which has 7 questions, then the third exercise which has 21 questions and lastly, a miscellaneous exercise which has 10 questions. All the exercises of Chapter 16- Probability have been solved in an easy and understandable way. Students must seek help for other exercises apart from NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.1, NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.2, NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 and NCERT Solutions For Class 11 Maths Chapter 16 Probability Miscellaneous Exercise. Students must solve all the questions from every given exercise in the NCERT book.

Students must remember to practise Mathematics problems to learn the formulas and derivations of the questions. They are advised to download the previous year’s question papers and sample papers that are available on Extramarks’ website and mobile application. Students should solve those sample papers and previous years question papers to evaluate their knowledge and not  miss out on any questions. Further, they must time themselves while solving each paper so that they will not run out of time in the examination hall. Hence, practising is the key to understanding every question in the examination. The NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3 will also be significant while solving questions.

With the proper exposure provided by these NCERT Solutions, students are better able to assess their aptitude and develop their skills. The experts at Extramarks have simplified these solutions so that students can comprehend the topics more quickly and retain them for longer.

The major topics and subtopics covered in Chapter 16 of NCERT Solutions for Class 11 Maths are –

16.1 – Introduction

16.2 – Random Experiments

16.2.1 – Outcomes and Sample Space

16.3 – Event

16.3.1 – Occurrence of an event

16.3.2 – Types of events

16.3.3 – Algebra of events

16.3.4 – Mutually exclusive events

16.3.5 – Exhaustive events

16.4 – Axiomatic Approach to Probability

16.4.1- Probability of an event

16.4.2 – Probabilities of equally likely outcomes

16.4.3 – Probability of the event ‘A or B’

16.4.4 – Probability of event ‘not A’.

Every exercise has concept-based questions, and students must understand the concepts before checking the NCERT Solutions For Class 11 Maths Chapter 16 Probability Exercise 16.3. They must remember that solving the questions will help them qualify for competitive examinations easily, just as practise is the key to Mathematics.

Q.1

Which of the following cannot be valid assignment of probabilitiesfor outcomes of sample Space S = {ω1, ω2, ω3, ω4, ω5, ω6, ω7}Assignmentω1ω2ω3ω4ω5ω6ω7(a)0.10.010.050.030.010.20.6(b)17171717171717(c)0.10.20.30.40.50.60.7(d)−0.10.20.30.4−0.20.10.3(e)1142143144145146141514

Ans.

a ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1Yes, it is a valid assignment of probabilities.b ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 17 + 17 + 17 + 17 + 17 + 17 + 17 = 77 = 1Yes, it is a valid assignment of probabilities.c ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8No, it is not a valid assignment of probabilities. Because sum of probabilities is greater than 1.d ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = - 0.1 + 0.2 + 0.3 + 0.4 + – 0.2 +  0.1 + 0.3No, it is not a valid assignment of probabilities. Because probability can not be negative.e ω1 + ω2 + ω3 + ω4 + ω5 + ω6 + ω7 = 114 +  214 + 314 + 414 + 514 + 614 + 1514 = 1 + 2 + 3 + 4 + 5 + 6 + 1514

Q.2 A coin is tossed twice, what is the probability that at least one tail occurs?

Ans.
Sample space when a coin is tossed twice,
S(E) = {HH, HT, TH, TT}
Favorable events of getting at least one tail
E = {HT, TH, TT}

P(E)=n(E)S(E) =34Thus, the probability of at least one tail is 34.

Q.3 A die is thrown; find the probability of following events:

(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.

Ans.
Sample space of a thrown die is:
S(E) = {1,2,3,4,5,6}
(i) Prime number (E) = {2, 3, 5}

P(Prime number)=36=12

(ii) Events of getting a number greater than or equal to 3
= {3, 4, 5, 6}

P(E)=46=23Thus, the required probability is 23.

(iii) Event of getting a number less than or equal to one
= {1}

P(E)=16Thus, the required probability is 13.

(iv) Event of getting a number more than 6
= {}

P(E)=06=0Thus, the required probability is 0.

(v) Event of getting a number less than 6
= {1, 2, 3, 4, 5}

P(E)=56Thus, the required probability is 56.

Q.4 A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is
(i) an ace (ii) black card.

Ans.
(a) There are 52 points in the sample space.
(b) Number of ace of spades = 1

P(ace of spades)=152Thus, the probability of an ace of spades is 152.

(c) (i) Number of aces in a deck of 52 cards = 4

P(ace)=452=113Thus, the probability of an ace card is 113.

(ii) Number of black cards in a deck of 52 cards

P(ace)=2652=12Thus, the probability of an ace card is 12.

Q.5 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed, find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

Ans.

Sample events when one coin(faced 1 and 6) and a dice: S={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}n(S)=12(i) Events in which sum is 3: E={(1,2)} P(E)=112

Thus, the required probability is 112.(ii) Events in which sum is 12: E={(6,6)} P(E)=112Thus, the required probability is 112.

Q.6 There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Ans.
Number of people in a council = 10
Number of women in a council = 6

 P(woman)=610=35Thus, the required probability is 35.

Q.7 A fair coin is tossed four times, and a person win ₹ 1 for each head and lose ₹1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Ans.
Sample space when a coin is tossed four times:
{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT,THHT, THTH, TTHH, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}

A person gain on getting each head = ₹ 1
A person loses on getting each tail = ₹ 1.50
Gain on getting HHHH = ₹ 4
Gain on getting HHHT = ₹ 1.50
Gain on getting HHTH = ₹ 1.50
Gain on getting HTHH = ₹ 1.50
Gain on getting THHH = ₹ 1.50
Loss on getting HHTT = ₹ 1.00
Loss on getting HTHT = ₹ 1.00
Loss on getting THHT = ₹ 1.00
Loss on getting THTH = ₹ 1.00
Loss on getting TTHH = ₹ 1.00
Loss on getting HTTH = ₹ 1.00
Loss on getting TTTH = ₹ 3.50
Loss on getting TTHT = ₹ 3.50
Loss on getting THTT = ₹ 3.50
Loss on getting HTTT = ₹ 3.50
Loss on getting TTTT = Rs. 6

       P(Winning ₹4)=116P(Winning ₹1.50)=416   =14   P(Losing ₹1.00)=616   =38   P(Losing ₹3.50)=416  =14

   P(Losing ₹6.00)=116

Q.8 Three coins are tossed once. Find the probability of getting

(i) 3 heads
(ii) 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head (vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails

Ans.
Sample space of tossing three coins is:
{HHH, HHT, HTH, THH, HTT, THT, HTT, TTT}

i P3 heads=18ii P2 heads=38iii Pat least 2 heads=48=12iv Pat most 2 heads=78v PNo head=18vi P3 tails=18vii Pexactly 2 tails=38viii PNo tail=18ix Pat most 2 tails=78

Q.9

If ​ 211 is the probability of an event, what is the probability of the event ‘not A’.

Ans.

Let probability of an event be P(A).Then,       P(A)=211P(not A)=1−P(A)          =1−211          =11−211          =911Thus, the probability of the event ‘not A’ is 911.

Q.10 A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is
(i) a vowel (ii) a consonant

Ans.

Number of letters in the word “ASSASSINATION“=13Number of vowels in the word “ASSASSINATION“=6Number of consonants in the word “ASSASSINATION“=13−6

=7(i)            P(a vowel)=613 [∵P(E)=n(E)S(E)](ii) P(a consonant)=713

Q.11 In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

Ans.
Total numbers from 1 to 20 = 20
Numbers to be selected = 6

Number of ways to select 6 numbers from 20 numbers = 20C6 =20!6!(20−6)! =20×19×18×17×16×15×14!6×5×4×3×2×1×14! =38760Number of ways to fix 6 numbers for winning lottery = 1   P(Winning the prize)=138760

Q.12 Check whether the following probabilities P(A) and P(B) are consistently defined

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

Ans.

(i) Here,​ P(A∩B)>P(A) [∵P(A∩B)=0.6 and P(A)=0.5]    So, the probabilities P(A) and P(B) are not consistently defined.(ii) Here, P(A∪B)>P(A) and P(A∪B)>P(B).    So, the probabilities P(A) and P(B) are consistently defined.

Q.13

Fill in the blanks in following table:P(A)P(B)P(A∩B)P(A∪B)(i)1315115.……(ii)0.35.……0.250.6(iii)0.50.35.……0.7

Ans.

i PA∪B=PA+PB−PA∩B                   =13+15−115                   =5+3−115                   =715ii PA∪B=PA+PB−PA∩B               0.6=0.35+PB−0.25        0.6− 0.1 =PB⇒               PB=0.5iii  PA∪B=PA+PB−PA∩B                         0.7 =0.5+0.35−PA∩B          PA∩B=0.85−0.7                                 =0.15

Q.14

Given P(A) = 35  and P(B) = 15. Find P(A or B), if A and B are mutually exclusive events.

Ans.

Since, A and B are mutually exclusive events. So,A∩B=ϕ⇒P(A∩B)=0∵P(A∪B)=P(A)+P(B)−P(A∩B)     =35+15−0     =45Therefore, P(A or B) is 45.

Q.15

If E and F are events such that P(E) = 14, P(F) = 12 and P(E and F) = 18, find (i)P(E or F), (ii)P(not E and not F).

Ans.

Given:  P(E)=14,  P(F)=12 and P(E∩F)=18(i) ∵ P(E∪F)=P(E)+P(F)−P(E∩F)          =14+12−18          =2+4−18          =58Therefore, P(E or F) is 58.(ii) P(not E and not F)=P(E‘ ∩F‘)        =P(E∪F)‘[By Demorgan’s Law]        =1−P(E∩F)        =1−58        =8−58        =38Therefore, P(not E and not F) is equal to 38.

Q.16 Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.

Ans.

∵P(not E or not F) =0.25⇒                    P(E‘ ∪F‘)=0.25⇒                    P(E∩F)‘=0.25[By Demorgarn’s law]⇒             1−P(E∩F)=0.25[∵P(E‘)=1−P(E)]⇒     P(E∩F)=1−0.25⇒     P(E∩F)=0.75⇒            E∩F≠0So, E and F are not mutually exclusive.

Q.17 A and B are events such that P(A) = 0.42,
P(B) = 0.48 and P(A and B) = 0.16.
Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)

Ans.

Given:P(A)=0.42, P(B)=0.48 and P(A∩B)=0.16 (i)    P(not A)=P(A‘)         =1−P(A)         =1−0.42         =0.58(ii)    P(not B)=P(B‘)       =1−P(B)       =1−0.48       =0.52(iii) P(A or B)=P(A∪B) =P(A)+P(B)−P(A∩B) =0.42+0.48−0.16 =0.74

Q.18 In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Ans.

PMathematics​ = 40%               PM​​​ = 0.40PBiology​ = 30%         PB ​= 0.30

PMathematics and Biology=10% PM∩B =0.10 PMathematics or Biology=PM∪B     =PM+PB−PM∩B      =0.40+0.30−0.10      =0.60Therefore, probability that student wiall be studying Mathematics or Biology is 0.60.

Q.19 In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?

Ans.

Let probability of passing in both examinations  be  P(A) and P(B) respectively.Then, we haveP(A)=0.8, P(B)=0.7 and P(A∪B)=0.95∵ P(A∩B)=P(A)+P(B)−P(A∪B)      =0.8+0.7−0.95      =0.55Thus, the probability of passing in both the examination is 0.55.

Q.20 The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

Ans.

Let the probability that a student will pass the final examination in both English and Hindi =P(E∩H)⇒ P(E∩H)=0.5Let the probability that a student will not pass the final examination in neither English nor Hindi =P(E‘ ∩H‘)⇒        P(E‘ ∩H‘)=0.1⇒         P(E ∪H)‘=0.1[By Demorgan’s law]⇒   1−  P(E ∪H)=0.1⇒           P(E ∪H)= 1− 0.1⇒           P(E ∪H)= 0.9The probability of passing the English examination        =P(E)        =0.75∵                                   P(E∪H)=P(E)+P(H)−P(E∩H)⇒         0.9 =0.75+P(H)−0.5⇒       P(H)=0.9−0.75+0.5        =0.65Therefore, the probability of passing the Hindi examination is 0.65.

Q.21 In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.

Ans.

               Number of students in a class=60                     Number of students in NCC=30                     Number of students in NSS=32Number of students in NCC and NSS=24∴      P(NCC)=3060=12      P(NSS)=3260=815 P(NCC∩NSS)=2460=615(i)P(NCC or NSS)=P(NCC∪NSS)      =P(NCC)+P(NSS)−P(NCC∩NSS)      =12+815−615      =15+16−1230      =1930(ii) P(Neither ​NCC nor NSS)      =P{(NCC)‘ ∪ (NSS)‘}      =P(NCC∩NSS)‘      =1−P(NCC∪NSS)      =1−1930      =1130(iii) P(NSS but not NCC)=P{(NSS)∩(NCC)‘}   =P(NSS)−P(NSS∩NCC)   =815−615   =215

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5. What is conditional probability?

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