NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.3) Exercise 13.3

The Central Board of Secondary Education CBSE is one of the biggest and most efficient boards in India. It is a central board and has many schools affiliated with it spread out all over India.  CBSE deploys a very objective approach regarding the structure of its academic curriculum. They mainly focus on application-based learning, where students learn the applications of the theory they have learned in class. They focus on comprehensive learning,where a student can answer any question if the concept explored in a chapter is well understood by the student. Students are made to refrain from memorising everything that they are studying and are rather encouraged to understand the concept and try to answer the questions analytically. Students, therefore, have to have a very deep understanding of the concepts and paradigms to be able to score well in exams.

Teachers all over the country are consulted by Extramarks, and upon discussing ways for a student to score well in their CBSE Exams, some common pointers are observed. Teachers have reiterated the importance of paying attention to the lessons taught in class and developing the habit of revising everything astudent has learned and doing great reference work. Teachers emphasise the importance of solving problems and answering questions as opposed to the common trend of just recapitulating what a student has learned. According to teachers, students should be able to answer application-based questions during the exam. It is only logical that they practise more similar questions and questions that have already appeared in the exams.

Countrywide students who come from different cultural, academic, and geographical backgrounds associate CBSE with its enormous syllabus. The syllabus of a CBSE board might be primarily objective, but their syllabus is vast and enormous. With the advanced level of difficulty that comes with the transition from a 10 level to a 10+2 level, coupled with the enormity of the CBSE syllabus.

The National Council for Educational Research and Training (NCERT), is a resource-based organisation that takes control over matters that are related to the education of the country. NCERT is an autonomous organisation that was set up in 1961 and was built so that it could assist the Central Government and all the other State Governments. CBSE and NCERT walk hand in hand, and CBSE strictly follows and adheres to all the guidelines laid out by NCERT. NCERT formulates the syllabus, and it has a significant executive function in the examination procedures. NCERT publishes books based on the syllabus laid out by them, and all CBSE students should treat their NCERT textbooks as their main textbooks. NCERT has single-handedly contributed enormously to the qualitative improvement of the Indian Education System. They strive to promote and provide assistance and aim to help all different kind of students. NCERT tries to help the students instead of making false promises. In recent years, NCERT has expanded its reach by coordinating areas of research for its students, providing model textbooks, and providing other participatory allotments in its newsletter and Given the stature of NCERT and CBSE, it is understandable that they expect excellence and accuracy from their students.students. The CBSE board suits the students who are preparing for all the different kinds of competitive exams that are held in India. These competitive exams ensure admissions into programmes like engineering, medicine, and law. These exams have an entirely objective question pattern that includes multiple choice questions. While some students who were part of the CBSE board prior to their Class 10 boards continue their education in the same board, a large number of students switch from other boards to CBSE.Students who have been a part of the CBSE before Class 11 are very familiar with the vastness of the CBSE board syllabus, and therefore the only new thing that they need to get accustomed to is the advanced level of difficulty of the exams. Although it is a very different reality for the students who transitioned from a different board, they are adjusting to two different major shifts. Learners must practise their Mathematics numerical regularly to not get confused during the exams. Students have been found to benefit greatly from regular revision. Every time a student comes across doubts, they are advised to refer to the NCERT Solutions provided by Extramarks. For students to deal with the difficulty level of these exams, they must solve as many problems and exercises as they possibly can because the more a student practises the more different kinds of problems they encounter, and therefore it prepares them for unfamiliar problems in an exam.

A student is expected to complete the entire syllabus in the time allotted to them. With the syllabus being massive, the concepts that are being discussed are extremely advanced. All that a student needs to do to address these issues is to regularly practise and revise the problems discussed in the NCERT textbooks. A CBSE Class 12 student has an enormous syllabus to attend to as well as an enormous amount of reference work to do. As a result, given India’s diverse demographic, the resources made available are certain to benefit as many people as possible.Students come up with doubts that arise and are answered when the student meets the teacher next. Extramarks, therefore, have found the perfect solution to this problem.Extramarks offers a wide range of NCERT Solutions. These solutions are made available for every course that CBSE offers in all the subjects that are available for students under the CBSE curriculum. The solutions contain answers to every unsolved question that the NCERT books contain. The best part of these solutions is the process of acquiring them, which is an extensive process that makes sure the answers provided are all correct. When students use these solutions, they have a solved reference at their expense all the time.

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Mathematics is one of the core subjects required of students who pursue a career in science after completing their Class 10 exams.Every student in India who has passed their Class 10 board exam has to select a stream that would help them with their future career. The streams available to the students are Science, Commerce and Humanities. Every student who chooses science as their field of study has mathematics as their core subject.Mathematics is particularly tricky and difficult concerning the syllabus of other subjects in the CBSE Class 12 curriculum. Not only is the syllabus of Mathematics vast and massive, but students are also made accustomed to a whole new set of mathematical vocabulary, which is very difficult to grasp for every student who learns these chapters for the first time.

The new learning curve in Mathematics includes memorising all the various formulas and concepts and solving as many problems as possible. Mathematics is highly interdisciplinary and if some ideas are not well understood in Mathematics, then they might create hurdles in other subjects like Physics and Chemistry. The primary reason for this is Mathematics is the core of all Science, and they surface in other subjects all the time. Students are always reminded to maintain a brisk pace and move ahead with the syllabus by keeping up with the pace at which their school is moving ahead. When students keep up with the pace of their schools and revise everything they learn on a regular basis, they will have a firm grasp on the topics covered in the syllabus.In regard to this, students and teachers alike have commented on the diverse use of the Extramarks NCERT Solutions. Extramarks has released solutions for every CBSE subject in every class. Extramarks offers various other tools to help students with their preparations for board examinations.NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex. 13.3) Exercise 13.3

Chapter 13 of Class 12 Mathematics Syllabus CBSE is Probability. Exercise 13.3 Class 12th Maths is particularly tricky. Probability marks one of the biggest transformations that a student observes in their shift from the class 10 level to the 10 +2 level. Students in their Class 10 exams had an extremely brief introduction to the concept of probability. In the 10+2 level, the chapter on Probability becomes a section in itself. The CBSE, as well as competitive exams, ask a plethora of questions from this chapter.This can be seen in Ex 13.3 Class 12.

Chapter 13 as the name suggests, divulges mathematical concepts that help calculate the chance of the occurrence of events. Students learn about different kinds of events and the different conditions based on which the calculation would be done. There are different theories and different case studies. Sometimes there are multiple independent events which take place at a time when they are all connected to the other events, on such occasions, the events are called mutually exclusive events. During such events, the calculations are done in one way. There have been other observations in Chapter 13 where there are multiple events but none of the individual events influence each other. These are called independent events, and the calculations based on these events are completely different from the calculations done when the events are mutually dependent.

Mathematics Class 12 CBSE Chapter 13 Exercise 13.3 is specifically based on three specific ideas that a student must know if they aim at solving Probability questions with ease. The concepts are as follows –

  1. Partition of a sample space.
  2. The theorem of Total Probability.
  3. Baye’s Theorem.

Mathematics is a subject that demands regular practice. The Mathematics syllabus is particularly vast and advanced. Chapter 12 in the CBSE Mathematics syllabus is Probability. The ideas discussed in this chapter are referenced widely in subjects like Physics, Chemistry and Statistics. Extramarks has released Exercise 13.3 Class 12 Maths NCERT Solutions on its website. These solutions are compiled in accordance with the latest CBSE guidelines. Students are implored to follow through with the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 for reference. Students get acclimated with the way CBSE expects its students to answer. CBSE values a lucid step-by-step deduction of the problems and the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 are created similarly. A student has to be really observant in order to understand exactly the type of questions that the CBSE asks in their exam. When a student is accustomed to the type of questions that appear in the board exams, itbecomes easier for the student to score good marks in the exams. Students, therefore, have found the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 extremely helpful.

Probability includes deceptively simple chapters, but when students tend to solve these exercises, it has been observed that they come across many doubts and complications. Therefore, Extramarks has released the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 for the third exercise of the chapter. NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 provides all the correct answers to all the questions in the NCERT book. All of these solutions are ensured to be completely accurate and explained in a detailed manner that clearly explains each step of the problemNCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 have explanations for all the complicated steps that are involved in a particular problem.

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Teachers have found new and easier techniques to solve a problem with the help of the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 include elaborate solutions to all the problems in the third exercise of Chapter 13.

Exercise 13.3 is based on the following topics:

Probability has many subtopics. NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 deals with all the different concepts very tactfully. NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 provides extremely comprehensible explanations. Probability uses several concepts from set theory and therefore NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 has a few pages dedicated to a revision of the set theory. NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 even explains how and why set theory and probability merge and how the solutions in NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 can be easily achieved using them.

As seen in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 there are a variety of concepts discussed in this chapter. NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 deals with every concept discussed in every exercise whose solutions are provided. Probability seems difficult,but if a student comprehends every case study, then the solutions in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 can be understood using basic cognition. The solutions in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 direct the students to solve the problems using the fundamental concepts discussed instead of complicating them. Students have often contacted Extramarks just to complement the perfect explanations provided in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 for every problem.

The various topics covered in Chapter 13 for which the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 provides solutions are –

  • Introduction To Probability
  • Conditional Probability
  • Properties of Conditional Probability
  • Multiplication Theorem on Probability
  • Independent Events
  • Bayes Theorem
  • Partition of Sample Space
  • The theorem of Total Probability
  • Random Variables and its Probability Distribution
  • Probability Distribution of Random Variable
  • Mean of Random Variables
  • The variance of Random Variables
  • Binomial Distribution
  • Miscellaneous Exercise

 

The solutions provided in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 are completely based on the concepts discussed here in the list. NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 have all correct answers. The answers in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 have been matched with the solutions provided in the NCERT textbook.

Access NCERT Solutions for Class 12 Maths Chapter 13 – Probability

The Extramarks’ website provides multiple resources for students to follow. A Class 12 student preparing for boards is generally under a lot of pressure, and therefore, NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 helps the student  get a good grasp over the subject, one chapter at a time. There are additional resources that are available on the Extramarks’ website, like recorded classes, live classes, doubt clearing sessions, and test quizzes with detailed analysis reports, besides the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3. Extramarks provide solutions for every exercise in Chapter 12 CBSE Mathematics, Probability. The Extramarks’ website is therefore a one-stop destination for students to access simple and accurate answers to all the questions in their NCERT textbook.

The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 is one of many in this particular series. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 can be easily accessed through the link that is provided under the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 heading. The link directly forwards the students from where they can easily refer to and download the solutions. Students spend an immense amount of time looking for a source that provides them with correct NCERT solutions with all the steps of a problem discussed in precise detail. When a student is given the entire calculation process of a particular problem, they can locate their own mistakes. When a student discovers their mistakes, they can actively address them, ensuring that they never make the same mistake again.NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 does exactly that. Extramarks provides solutions for students to not only understand how a problem is solved, but also to know what mistakes they should never make.Importance of Probability

Probability is one of the branches of Mathematics that equips students to make a decision when they do not have perfect information for an event or a situation. As observed through the solutions in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 which provide solutions to every NCERT word problem, it also gives an idea of the nature of the questions in the exercise. Probability prepares individuals to make decisions in situations in which there is no degree of certainty but there is an observable pattern. NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 aims to help students solve these problems easily.

NCERT Solutions for Class 12 Maths PDF Download

NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 are carefully compiled making sure that there are no mistakes and inaccuracies in them. Mathematics exercises comprise a wide range of problems. Among the problems, some are easy while some are extremely difficult for the student to solve. When a student goes through the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 they can understand clearly and definitively the reason why they could not solve a problem. When students learn the reason beyond their mistakes, they learn from them. Experts, therefore, recommend that students revise what they have learned. When a student revises,  they can check if they are making the same mistakes again.  If they make the same mistake again, they will be aware of the areas that need to be addressed.The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 helps out students to do exactly that.

NCERT Solution Class 12 Maths of Chapter 13 All Exercises

NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 provides the solutions for the third exercise of Chapter 13. Like the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 there are other solutions provided by Extramarks for every chapter and all its exercises from all classes. Since the question comes in two languages – Hindi and English the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 also has a Hindi version of it as well for anyone to follow. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 makes sure that the information is easily accessible and well understood by the students.

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.3

Students who are working on the probability chapter and referring to the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 should keep in mind that if they have any questions, they can visit the Extramarks website.Students in Class 12 CBSE are often preparing for competitive exams like JEE Mains, JEE Advanced, NEET, etc. These examinations put a student under a lot of stress, and NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 helps reduce stress. With the use of the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 students are constantly under the guidance of experts. The concepts discussed in this chapter can be difficult, but students find great excitement in learning about them. Although to score good marks, a student has to go beyond just understanding the concepts. They must regularly revise, and they are recommended to reference the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 to keep in check their progress.

Scoring well in exams depends on multiple factors. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 provides very systemic solutions to all the NCERT questions, and therefore it is one of the most important resources that a student can use. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3 helps students work out problems on their own. With every practice, the students brush up on the concepts that they have learned. When a student is able to solve problems on their own, it amplifies their confidence. This helps students answer the questions with great ease. Although to reach this level of expertise, learners are advised to practice sums that they solved earlier again. Every time they come across a doubt they must refer to the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.3.

Q.1 An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Ans

    Number of red balls in urn=5 Number of black balls in urn=5 Let a red ball be drawn in the first attempt.∴P(red ball)=510=12If two red balls are added to the urn, then    Number of red balls in urn=7      Total balls in urn=12     P(drawing a red ball)=712Let a black ball be drawn in the first attempt.∴  P (black ball)=510=12Probability of red ball in second attempt,       P(red ball)=712Therefore, probability of drawing second ball as red=12×712+12×512=12(712+512)=12×1=12

Q.2 A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Ans

            Number of red balls in first bag=4       Number of black balls in first bag=4    Number of red balls in second bag=2Number of black balls in second bag=6Let E1 and E2 be the events of selecting first bag and second bag respectively.∴P(E1)=12; P(E2)=12Let A be the event of getting a red ball.        P(A|E1)=P(red ball from first bag)=48=12P(A|E2)=P(red ball from second bag)=28=14The probability of drawing a ball from the first bag, given that it is red  =P(E1|A)   =P(E1).P(A|E1)P(E1). P(A|E1)+P(E2). P(A|E2)[By using Bayes‘ theorem]  =12.1212.12+12.14 =1414+18=1438=14×83=23Therefore, the probability of drawing a ball from the first bag, given that it is red is 23.

Q.3 Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Ans

Let E1=The student is a hostler          E2=The student is a day scholar and   A=The chosen student gets grade A. ∴P(E1)=60%=60100=0.6    P(E2)=40%=40100=0.4P(A|E1)=P(student getting an A grade is a hostler)        =30%=0.3P(A|E2)=P(student getting an A grade is a day scholar)        =20%=0.2The probability that a randomly chosen student is a hostler, given that he has an A grade        =P(E1|A)        =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)[By Bayes‘ Theorem]        =0.6×0.30.6×0.3+0.4×0.2        =0.180.18+0.08        =0.180.26        =1826        =913Therefore, the required probability is 913.

Q.4 In answering a question on a multiple choice test, a student either knows the answer or guesses. Let (3/4) be the probability that he knows the answer and (1/4) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability (1/4). What is the probability that the student knows the answer given that he answered it correctly?

Ans

Let   E1=The student knows the answerand E2=The student guesses the answerLet    A=The answer is correct∴P(E1)=34    P(E2)=14The probability that the student answered correctly, given that he knows the answer,P(A|E1)=1P(A|E2)=P(student answered correctly, given that he guessed)        =14The probability that the student knows the answer, given that he answered it correctly        =P(E1|A)        =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)[By Bayes‘ Theorem]        =34. 134. 1+14. 14        =3434+116        =3412+116=1213Therefore, the required probability is 1213.

Q.5 A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Ans

Let E1=A person has disease.        E2=A person has no disease.Since E1 and E2 are events complimentary to each other, ∴ P (E1) + P (E2)=1     P (E2)=1−P (E1)         =1−0.1%         =1−0.1100         =1−0.001         =0.999Let A be the event that the blood test result is positive.      P (E1)=0.1%         =0.001P(A|E1)=P(result is positive given the person has disease)          =99%=0.99P(A|E2)=P(result is positive given the person has no disease)          =0.5%=0.005P(E1|A)=Probability that a person has a disease, given that his    test result is positive         =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)×P(A|E2)  [By Bayes’ Theorem]         =0.000990.005985         =9905985         =1981197Therefore, the required probability is 1981197.

Q.6 There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Ans

Let    E1=choosing a two headed coin          E2=choosing a biased coin          E3=choosing an unbiased coin∴P(E1)=13,  P(E2)=13 and P(E3)=13Let    A=the coin shows headsA two–headed coin will always show heads. P(A|E1)=1   P(A|E2)=75%   =75100 =34∵The third coin is unbiased, the probability that it shows heads   =12∴P(A|E3)=12The probability that the coin is two–headed, given that it shows heads,   =P(E1|A)   =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)×P(A|E2)+P(E3)×P(A|E3)  [By Bayes‘ Theorem]   =13. 113. 1+13. 34+13. 12   =1313+14+16   =134+3+212   =13912=49Thus, the required probability is 49.

Q.7 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Ans

Let    E1=The driver is a scooter driver E2=The driver is a  car driverE3=The driver is a truck driver          A=The person meets with an accident. There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers. Total number of drivers = 2000 + 4000 + 6000 = 12000       P(E1)=200012000=16       P(E2)=400012000=13       P(E3)=600012000=12P(A|E1)=P(scooter driver met with an accident)        =0.01=1100P(A|E2)=P(car driver met with an accident)=0.03=3100P(A|E3)=P(truck driver met with an accident)=0.15=15100The probability that the driver is a scooter driver, given that he met with an accident, =P(E1|A)=P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)×P(A|E2)+P(E3)×P(A|E3) [By Bayes‘ Theorem]=16×110016×1100+13×3100+12×15100=1616+33+152=161+6+456=152Therefore, the required probability is 152.

Q.8 A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Ans

Let    E1=Items produced by machines A          E2= Items produced by machines B.            A= The Item was found to be defective∴P(E1)=60%=0.6    P(E2)=40%=0.4Probability that machine A produced defective items,P(A|E1)=2%=0.02Probability that machine B produced defective items,P(A|E2)=1%=0.01The probability that the randomly selected item was from machine B, given that it is defective,P(E2|A)=P(E2)P(A|E2)P(E1)P(A|E1)+P(E2)×P(A|E2)  [By Bayes‘ Theorem]        =0.4×0.010.6×0.02+0.4×0.01        =0.0040.012+0.004        =0.0040.016        =416=14Therefore, the probability that the randomly selected item was from machine B, given that it is defective, is 14.

Q.9 Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Ans

Let      E1=The first group win the competition   E2=The second group win the competition             A=The event of introducing a new product        P(E1)=Probability that the first group wins the competition

=0.6
 P( E 2 )=Probability that the second group wins the competition =0.4 P (A| E 1 )=Probability of introducing a new product if the first    group wins  
= 0.7 P (A| E 2 )=Probability of introducing a new product if the second group wins =0.3 The probability that the new product is introduced by the second group is given by           
= P(E 2 |A) P( E 2 |A )= P( E 2 )P( A| E 2 ) P( E 1 )P( A| E 1 )+P( E 2 )×P( A| E 2 )   [ By Bayes’ Theorem ]          
 = 0.4×0.3 0.6×0.7+0.4×0.3           
= 0.12 0.42+0.12          
= 0.12 0.54 = 2 9 Therefore, the required probability is 2 9 .

Q.10 Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Ans

  Let E1=The outcome on the die is 5 or 6and E2=The outcome on the die is 1, 2, 3, or 4. ∴P(E1)=26=13   P(E2)=46=23Let   A=The event of getting exactly one head.P (A|E1)=Probability of getting exactly one head by tossing the coin three times if she gets 5 or 6         =38P (A|E2)=Probability of getting exactly one head in a single throw of coin if she gets 1,2, 3, or 4         =12The probability that the girl threw 1, 2, 3, or 4 with the die, if she obtained exactly one head,         =P (E2|A)         =P(E2)P(A|E2)P(E1)P(A|E1)+P(E2)P(A|E2)  [By Bayes‘ Theorem]         =23×1238×13+23×12         =1318+13         =133+824         =131124         =13×2411=811Therefore, the required probability is 811.

Q.11 A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Ans

  Let E1=the time consumed by machines A for the job    E2=the time consumed by machines B for the joband E3=the time consumed by machines C for the job   P(E1)=50%=0.5   P(E2)=30%=0.3   P(E3)=20%=0.2 A= The event of producing defective itemsP(A|E1)=1%=0.01P(A|E2)=5%=0.05P(A|E3)=7%=0.07The probability that the defective item was produced by A,=P (E1|A)=P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)+P(E3)P(A|E3)  [By Bayes‘ Theorem]=0.5×0.010.5×0.01+0.3×0.05+0.2×0.07=0.0050.005+0.015+0.014=0.0050.034=534Therefore, the required probability is 534.

Q.12 A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Ans

Let E1=events of choosing a diamond card E2=events of choosing a card which is not diamondand A = the lost cardOut of 52 cards, 13 cards are diamond and 39 cards are not diamond.∴P(E1)=1352=14   P(E2)=3952=34When one diamond card is lost, there are 12 diamond cards outof 51 cards. Two cards can be drawn out of 12 diamond cards in12C2 ways.Similarly, 2 diamond cards can be drawn out of 51 cards in51C2 ways. The probability of getting two cards, when one diamond card is   lost=P(A|E1)  =12C251C2=12!2!10!×2!49!51!  =12×111×151×50  =12×1151×50  =661275When the lost card is not a diamond, there are 13 diamond cardsout of 51 cards. Two cards can be drawn out of 13 diamond cards in 13C2 ways whereas 2 cards can be drawn out of 51 cards in 51C2 ways.The probability of getting two cards, when one card is lost which is not diamond  =P(A|E2)   =13C251C2  =13!2! ×11!×2! ×49!51!  =13×121×151×50  =26425The probability that the lost card is diamond  =P (E1|A)  =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)  [By Bayes‘ Theorem]  =14×2242514×22425+34×26425  =1150Therefore, the required probability is 1150.

Q.13 Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is
(A) 4/5 (B) 1/2 (C) 1/5 (D) 2/5

Ans

Let E1 and E2 be the events such that E1: A speaks truth E2: A speaks falseLet A be the event that a head appears. P(E1)=45P(E2)=1−45[P(E2)=1−P(E1)]  =15If a coin is tossed, then it may result in either head (H) or tail (T).The probability of getting a head is 12 whether A speaks truthor not.∴P(A|E1)=12 and P(A|E2)=12The probability that there is actually a head=P(E1|A)  =P(E1)P(A|E1)P(E1)P(A|E1)+P(E2)P(A|E2)  [By Bayes‘ Theorem]  =45×1245×12+15×12=45Therefore, the required probability is 45.Thus, correct option is A.

Q.14

If A and B are two events such that A⊂B and P(B)≠0,thenwhich of the following is correct?(A)P(A|B)=P(B)P(A) (B)P(A|B)<P(A)(C)P(A|B)≥P(A) (D)None of these

Ans

If A⊂B, then A∩B=A⇒ P (A∩B)=P(A)...(i)(A)P(A|B)=P(A∩B)P(B)     =P(A)P(B)[From equation(i)]     ≠P(B)P(A)Therefore, option A is not correct.(B)P(A|B)=P(A∩B)P(B)    =P(A)P(B)[From equation(i)]⇒P(A|B)<P(A)[∵P(A)<P(B)]Therefore, option B is not correct.(C)P(A|B)=P(A∩B)P(B)⇒P(A|B)=P(A)P(B)[From equation(i)]⇒P(A|B)≥P(A)[∵P(B)≤1, so 1P(B)≥1]Therefore, option C is correct.

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