# NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.5) Exercise 13.5

The importance of Science and its various sub-disciplines has been recognised and upheld since the beginning of civilisation. Physics, Chemistry, and Biology have been studied and explored in various forms and through a variety of intriguing methods,, in various civilisations across the globe. It can be said without a doubt that Mathematics is one of the most prominent and vital scientific disciplines. The relative importance of Mathematics as compared to other sciences can be a subject of intense debate.  Mathematics and knowledge of the basic working principles and tenets that are a part of Mathematics and its operations are crucial to attaining a holistic and comprehensive understanding of subjects like Physics, Chemistry etc. Physics and Mathematics are often perceived as sister disciplines, and it can be comfortably pointed out that both of these disciplines are quite interdisciplinary concerning the importance of calculations, quantities, and other nuances.

Mathematics comprises a multitude of underlying themes and topics. These are quite distinct from each other and are simultaneously interconnected in a myriad of ways. Intricate calculations, comprehensive formulae and their derivatives, intriguing theorems, and thought-provoking postulates make up the entirety of the curriculum in Mathematics. The history of a major proportion of this versatile curriculum can be traced back to thousands of years of research and exploration. As a result, Mathematics can appear to be quite a complex, dynamic, and sophisticated subject at first glance. However, it is also noteworthy that it is a fundamental academic discipline.

The knowledge of Mathematics is vital for the smooth and unhindered functioning of human activities as well as of man-made machines and operations. Mathematical operations are also indispensable towards ensuring the efficient organizsation of daily life. Mathematics, as a discipline, has also made extensive contributions towards uncovering many intriguing mysteries of the world and the workings of various natures and natural phenomena. Mathematics has a vital role to play in almost all professional sectors as well as in creative occupations like Art, Literature, etc. A variety of modern art involves the use and practical application of concepts of Mathematics like Geometry and Proportions, etc.

Therefore, Mathematics is also a vital and indispensable academic discipline. The knowledge of Mathematics is imparted to students from the earliest stages of education and gradually grows more sophisticated and is classified as a rise in grades. Beginning with simple and easy-to-understand concepts like Addition, Subtraction, Multiplication, Division etc, students proceed to learn about complex themes such as Mensuration, Percentage, Trigonometry, Calculus, Geometry, Probability, etc. All of these themes are inseparable parts of the prescribed NCERT academic curriculum for Mathematics for the students of various classes.The entire NCERT academic curriculum for school students in mathematics has been carefully designed to lay a solid foundation for the students in order to prepare them for their future endeavors.This is precisely because Mathematics  is a vital subject within a huge number of professions and occupations like Engineering, Architecture, Technology, Astronomy etc. Additionally, Mathematics as an academic discipline with a long history also has immense and unfathomable potential for research. Therefore, curious students may find Mathematics their ideal choice of subject for pursuing higher education and scientific research. Mathematics is also a crucial part of the prescribed NCERT academic curriculum for Class 12.

The prescribed NCERT academic syllabus for Mathematics in Class 12 is composed of thirteen distinct chapters. These chapters include themes such as Differential Equations, Integrals, Theorems of Calculus, Probability, etc. All of these themes are dynamic and vast academic topics that have been the subject of great interest over the years in various fields of academic research and study. Therefore, these themes have been comprehensively covered in the NCERT prescribed academic syllabus for Class 12.

The academic content on these themes includes various challenging and diverse assessments associated with different chapters in the form of exercises. Probability is a theme which is the focus of Chapter 13. Ex 13.5 Class 12 is one of the exercises that is  part of Chapter 13. It is very crucial for students to be able to solve this exercise comprehensively to be adequately prepared for examinations. Reliable and authentic NCERT Solutions like the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 can prove to be important reference materials in this context.

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## NCERT Solutions for Class 12 Maths Chapter 13 Probability (Ex 13.5) Exercise 13.5

A diverse number of constituting topics and themes make up the composition of Mathematics as an academic discipline. While these underlying themes are indeed interlinked and associated with each other in many interesting ways, they are also very different from each other at the same time. The academic curriculum of Mathematics comprises many elements such as complex and minute calculations, detailed formulae and their respective derivatives, sophisticated theorems, and conventionally established postulates, among others. A significant chunk of this dynamic composition can be put under a historical lens. This can lead to the discovery that these themes have sparked a huge amount of research, invention, and experimentation over the ages.

Accordingly, Mathematics has taken on the appearance of a complex, versatile, and elite subject and appears  mysterious at first glance. The indispensability of Mathematics as an academic discipline, however, cannot be overlooked. Continuous and unbarred operation of a large proportion of human activities along with the smooth functioning of a great number of man-made crafts prerequisites basic knowledge of Mathematics. Mathematical knowledge holds great importance in humans’ daily lives and ensures that everything is in perfect, workable order. As an academic discipline, Mathematics has also been crucial in discovering answers to many secrets of the world, nature and natural phenomena, and the working of the universe itself.

Thirteen different chapters make up the entirety of the academic curriculum of Mathematics prescribed for Class 12 by the NCERT. Themes like Differential Equations, Integrals, Theorems of Calculus, Probability etc.Each one of these themes has been the subject of vast academic studyand research and has sparked great interest in the minds of intellectuals across the world. Therefore, the academic syllabus for Mathematics for Class 12 prescribed by NCERT has been compiled to include the nuances and fundamental tenets of these themes. This has been done to ensure that students are appropriately knowledgeable concerning these themes in time for their future academic and professional ventures.

Along with the academic elements, various comprehensive and competitive assessments have been included within the chapters covering various themes.The last chapter covers the theme of Probability. Exercise 13.5 is part of this chapter. To excel in this topic in the examinations, regular practise and revision of Exercise 13.5 with the help of the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 is crucial. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 have been designed to serve this purpose. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 consists of comprehensive and easy-to-understand solutions for Exercise 13.5.  Important sub-topics related to the broader theme of probability are also covered in the NCERT Solutions for Class 12 Maths Chapter 13 Exercise 13.5.

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### NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.5

Within the prescribed academic curriculum for Mathematics in Class 12 by the NCERT, Probability is a major theme. It is also quite a complex topic. It consists of many individual formulae and their particular derivatives. These formulae, including the symbols and equations used in their composition, have to be adequately comprehended and understood. This is important because, without appropriate comprehension and logical understanding of the specific structure of the given formulae, students would find it difficult to apply the formulae practically for problem-solving. However, merely by retaining formulae and figuring out their practical application, one cannot excel in Probability. To ace the mathematical concept of Probability, students would also need to familiarise themselves with the variety of theorems, functions, and uses of Probability associated with the practical world. Probability becomes a part of the academic syllabus of Mathematics in very early grades, whichis a testimony to its importance as a concept. Therefore, while it is essential to attain knowledge of the practical application of various formulae, it is equally vital to acquire theoretical knowledge of the relevance of Probability and the variety of facts that accompany this knowledge. The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 can undoubtedly cater to this purpose.

Classroom teaching is a lengthy and complex process. There could be many reasons for this. The primary reason could be that each student is an individual in their own right.By virtue of this fact, students have their own methods of grasping things, expressing themselves, and learning or comprehending academic content that is delivered to them. However, all of these different individuals are brought together in a classroom setting for the convenience of imparting education. The time allotted for the teaching of each subject and topic is limited, and teachers have to also comply with a strict schedule within which they are expected to complete the prescribed syllabus.

In such a situation, it may become cumbersome and difficult to accommodate the individuality of students and their apparent diversity. Teachers may often find it difficult to accommodate all doubts and queries raised by students during school hours and may not be able to respond to all of them adequately. Students may be left with many unclarified doubts or unresolved queries. The accumulation of such doubts and queries can take a toll on the conceptual clarity of students. However, conceptual clarity is an imperative resource for a conceptually rich academic discipline like Mathematics. Inadequate conceptual clarity can turn into a major obstacle during examinations. ConsequentlyAs a result, the NCERT Solutions for Class 12 Maths, Chapter 13, Exercise 13.5, can be particularly useful in this regard.

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Probability is an inseparable part of the academic curriculum of Mathematics prescribed for Class 12. It is a very interesting and thought-provoking concept. It is also a very versatile academic theme. Therefore, it has to be read, comprehended, and analysed at many levels. While on the one hand, students are expected to acquire knowledge of and retain important formulae, on the other hand, they also have to be made aware of the practical applications and uses of concepts of probability in daily life.This latter type of knowledge is critical in determining the context of many critical questions and appropriately solving them.The NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 cover both of these aspects of the concept of Probability. Remarkably, the concepts which constitute the theme of Probability have multiple uses in our daily lives. Some of them can be listed as follows – flipping a coin, choosing a card from the deck, throwing a die, winning a lottery, predicting the outcome of a cricket match, etc. All of these applications and practical questions involving them have been covered in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5.

Probability can be defined as the likelihood of the occurrence of any event. The earliest known form of Probability and Statistics can be found in the work of a Central Asian mathematician, called Al-Khalil.  He authored a book titled Cryptographic Messages, which involved the first use of permutation and combination to list all the Arabic words with or without vowels. Probability is thatProbability is the branch of mathematics that deals with calculating the likelihood of an event occurring.he occurrence of any event could be a number between 0 and 1. Here, 0 indicates the impossibility of an event and 1 indicates certainty. There are a multitude of areas in which Probability theory and concepts are widely used. A few notable examples of fields of studies such as Statistics, Finance, Gambling, Artificial Intelligence (AI), Computer Science, Machine Learning, Game Theory and Philosophy. The Probability concept is also used intensively in fields such as Weather Forecasting, Sports Strategising, Insurance, etc. Therefore, it is very evident that Probability as a mathematical concept has unparalleled relevance. Its wide uses in daily life also make it a fascinating area of study. It is highly recommended that students  use the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 to gain a good understanding of the concept of Probability and its applications.

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Q.1 A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?

Ans

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{die}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{denotes}\\ \mathrm{the}\mathrm{number}\mathrm{of}\mathrm{successes}\mathrm{of}\mathrm{getting}\mathrm{odd}\mathrm{numbers}\mathrm{in}\mathrm{an}\\ \mathrm{experiment}\mathrm{of}6\mathrm{trials}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{an}\mathrm{odd}\mathrm{number}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{a}\mathrm{die}\mathrm{is},\\ \mathrm{p}=\frac{3}{6}=\frac{1}{2}\\ \therefore \mathrm{q}=1-\mathrm{p}\\ \mathrm{ }=1-\frac{1}{2}=\frac{1}{2}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{.}\\ \mathrm{Therefore},\\ \mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{\mathrm{x}}{\left(\frac{1}{2}\right)}^{6-\mathrm{x}}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \left(\mathrm{i}\right)\mathrm{P}\left(5\mathrm{success}\right)=\mathrm{P}\left(\mathrm{X}=5\right)\\ \mathrm{ }=\mathrm{C}_{5}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=6.{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=\frac{6}{64}\\ \mathrm{ }=\frac{3}{32}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{at}\mathrm{least}5\mathrm{successes}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}=5\right)+\mathrm{P}\left(\mathrm{X}=6\right)\\ \mathrm{ }=\mathrm{C}_{5}^{6}{\left(\frac{1}{2}\right)}^{6}+\mathrm{C}_{6}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=6.\frac{1}{64}+1.\frac{1}{64}\\ \mathrm{ }=\frac{7}{64}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{at}\mathrm{most}5\mathrm{successes}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}\le 5\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}>5\right)\\ =1-\mathrm{P}\left(\mathrm{X}=6\right)\\ \mathrm{ }=1-\mathrm{C}_{6}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{ }=1-\frac{1}{64}\\ \mathrm{ }=\frac{63}{64}\end{array}$

Q.2 A pair of dice is thrown 4 times. If getting a doublet is considered a success, Find the probability of two successes.

Ans

$\begin{array}{l}\mathrm{Total}\mathrm{outcomes}\mathrm{of}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{dice}=6×6\\ =36\\ \mathrm{Number}\mathrm{of}\mathrm{doublets}\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\right\}\\ =6\\ \mathrm{The}\mathrm{repeated}\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{pair}\mathrm{of}\mathrm{dice}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\\ \mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{doublets}\mathrm{in}\mathrm{an}\\ \mathrm{experiment}\mathrm{of}\mathrm{throwing}\mathrm{two}\mathrm{dice}\mathrm{simultaneously}\mathrm{four}\mathrm{times}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{doublets}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{pair}\mathrm{of}\\ \mathrm{dice}\mathrm{is} \mathrm{p}=\frac{6}{36}=\frac{1}{6}\\ \mathrm{and} \mathrm{ }\mathrm{q}=1-\mathrm{p}\\ =1-\frac{1}{6}\\ =\frac{5}{6}\\ \mathrm{Here},\mathrm{X}\mathrm{has}\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=4,\mathrm{p}=\frac{1}{6}\mathrm{and}\mathrm{q}=\frac{5}{6}\\ \therefore \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{4}{\left(\frac{1}{6}\right)}^{\mathrm{x}}{\left(\frac{5}{6}\right)}^{4-\mathrm{x}}\\ \mathrm{So}, \mathrm{ }\mathrm{P}\left(\mathrm{X}=2\right)=\mathrm{C}_{2}^{4}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{4-2}\\ =\frac{4!}{2!2!}×\frac{1}{36}×\frac{25}{36}\\ =6×\frac{1}{36}×\frac{25}{36}\\ =\frac{25}{216}\end{array}$

Q.3 There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{defective}\mathrm{items}\mathrm{in}\mathrm{a}\mathrm{sample}\mathrm{of}10\\ \mathrm{items}\mathrm{drawn}\mathrm{successively}\mathrm{.}\\ \mathrm{Since}\mathrm{the}\mathrm{drawing}\mathrm{is}\mathrm{done}\mathrm{with}\mathrm{replacement},\mathrm{the}\mathrm{trials}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{P}\left(\mathrm{defective}\mathrm{item}\right),\mathrm{p}=\frac{5}{100}\\ =\frac{1}{20}\\ \mathrm{q}=1-\mathrm{p}\\ =1-\frac{1}{20}\\ =\frac{19}{20}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 10\mathrm{and}\mathrm{p}=\frac{1}{20}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{10}{\left(\frac{1}{20}\right)}^{\mathrm{x}}{\left(\frac{19}{20}\right)}^{10-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{not}\mathrm{more}\mathrm{than}1\mathrm{defective}\mathrm{item}\right)\\ =\mathrm{P}\left(\mathrm{X}\le 1\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)\\ =\mathrm{C}_{0}^{10}{\left(\frac{1}{20}\right)}^{0}{\left(\frac{19}{20}\right)}^{10-0}+\mathrm{C}_{1}^{10}{\left(\frac{1}{20}\right)}^{1}{\left(\frac{19}{20}\right)}^{10-1}\\ ={\left(\frac{19}{20}\right)}^{10}+10.\left(\frac{1}{20}\right){\left(\frac{19}{20}\right)}^{9}\\ ={\left(\frac{19}{20}\right)}^{9}\left(\frac{19}{20}+\frac{10}{20}\right)\\ ={\left(\frac{19}{20}\right)}^{9}\left(\frac{29}{20}\right)\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}{\left(\frac{19}{20}\right)}^{9}\left(\frac{29}{20}\right).\end{array}$

Q.4 Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade?

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}=\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{spade}\mathrm{cards}\mathrm{among}\mathrm{the}\mathrm{five}\mathrm{cards}\mathrm{drawn}\\ \mathrm{Since}\mathrm{the}\mathrm{drawing}\mathrm{of}\mathrm{card}\mathrm{is}\mathrm{with}\mathrm{replacement},\mathrm{the}\mathrm{trials}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{In}\mathrm{a}\mathrm{well}\mathrm{shuffled}\mathrm{deck}\mathrm{of}52\mathrm{cards},\mathrm{there}\mathrm{are}13\mathrm{spade}\mathrm{cards}\mathrm{.}\\ \therefore \mathrm{p}=\frac{13}{52}=\frac{1}{4}\\ \mathrm{and} \mathrm{q}=1-\frac{1}{4}=\frac{3}{4}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 5\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{4}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{1}{4}\right)}^{\mathrm{x}}{\left(\frac{3}{4}\right)}^{5-\mathrm{x}}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{all}\mathrm{}5\mathrm{cards}\mathrm{are}\mathrm{spades}\right)\\ =\mathrm{C}_{5}^{5}{\left(\frac{1}{4}\right)}^{5}{\left(\frac{3}{4}\right)}^{5-5}\\ =1.{\left(\frac{1}{4}\right)}^{5}{\left(\frac{3}{4}\right)}^{0}\\ =\frac{1}{{4}^{5}}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{only}3\mathrm{cards}\mathrm{are}\mathrm{spades}\right)=\mathrm{P}\left(\mathrm{X}=3\right)\\ =\mathrm{C}_{3}^{5}{\left(\frac{1}{4}\right)}^{3}{\left(\frac{3}{4}\right)}^{5-3}\\ =\mathrm{C}_{3}^{5}{\left(\frac{1}{4}\right)}^{3}{\left(\frac{3}{4}\right)}^{2}\\ =10.\frac{1}{64}.\frac{9}{16}\\ =\frac{45}{512}\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{none}\mathrm{is}\mathrm{spade}\right)=\mathrm{P}\left(\mathrm{X}=0\right)\\ =\mathrm{C}_{0}^{5}{\left(\frac{1}{4}\right)}^{0}{\left(\frac{3}{4}\right)}^{5-0}\\ =1.1.\frac{243}{1024}\\ =\frac{243}{1024}\end{array}$

Q.5 The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{bulbs}\mathrm{that}\mathrm{will}\mathrm{fuse}\mathrm{after}150\\ \mathrm{days}\mathrm{of}\mathrm{use}\mathrm{in}\mathrm{an}\mathrm{experiment}\mathrm{of}5\mathrm{trials}.\mathrm{The}\mathrm{trials}\mathrm{are}\mathrm{Bernoulli}\\ \mathrm{trials}\mathrm{.}\\ \mathrm{Given},\mathrm{p}=\mathrm{0}.05\mathrm{and}\\ \mathrm{ }\mathrm{q}=1-0.05=0.95\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 5\mathrm{and}\mathrm{ }\mathrm{p}=\mathrm{0}.05\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\mathrm{0}.05\right)}^{\mathrm{x}}{\left(0.95\right)}^{5-\mathrm{x}}\\ \left(\mathrm{i}\right)\mathrm{P}\left(\mathrm{none}\right)=\mathrm{P}\left(\mathrm{X}=0\right)\\ =\mathrm{C}_{0}^{5}{\left(\mathrm{0}.05\right)}^{0}{\left(0.95\right)}^{5-0}\\ =1.1.{\left(0.95\right)}^{5}\\ ={\left(0.95\right)}^{5}\\ \left(\mathrm{ii}\right)\mathrm{P}\left(\mathrm{not}\mathrm{more}\mathrm{than}\mathrm{one}\right)\\ =\mathrm{P}\left(\mathrm{X}\le 1\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)\\ =\mathrm{C}_{0}^{5}{\left(\mathrm{0}.05\right)}^{0}{\left(0.95\right)}^{5-0}+\mathrm{C}_{1}^{5}{\left(\mathrm{0}.05\right)}^{1}{\left(0.95\right)}^{5-1}\\ =1.1.{\left(0.95\right)}^{5}+5\left(0.05\right){\left(0.95\right)}^{4}\\ ={\left(0.95\right)}^{4}\left(0.95+0.25\right)\\ ={\left(0.95\right)}^{4}×\left(1.2\right)\\ \left(\mathrm{iii}\right)\mathrm{P}\left(\mathrm{more}\mathrm{than}\mathrm{one}\right)\\ =\mathrm{P}\left(\mathrm{X}>1\right)\\ =1-\mathrm{P}\left(\mathrm{X}\le 1\right)\\ =1-{\left(0.95\right)}^{4}×\left(1.2\right)\\ \left(\mathrm{iv}\right)\mathrm{P}\left(\mathrm{at}\mathrm{least}\mathrm{one}\right)\\ =\mathrm{P}\left(\mathrm{X}\ge 1\right)\\ =1-\mathrm{P}\left(\mathrm{X}<1\right)\\ =1-\mathrm{P}\left(\mathrm{X}=0\right)\\ =1-\mathrm{C}_{0}^{5}{\left(\mathrm{0}.05\right)}^{0}{\left(0.95\right)}^{5-0}\\ =1-{\left(0.95\right)}^{5}\end{array}$

Q.6 A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{balls}\mathrm{marked}\mathrm{with}\mathrm{the}\mathrm{digit}0\mathrm{among}\\ \mathrm{the}4\mathrm{balls}\mathrm{drawn}\mathrm{.}\\ \mathrm{Since}\mathrm{the}\mathrm{balls}\mathrm{are}\mathrm{drawn}\mathrm{with}\mathrm{replacement},\mathrm{the}\mathrm{trials}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=4\mathrm{and}\mathrm{p}=\frac{1}{10}\\ \therefore \mathrm{q}=1-\frac{1}{10}=\frac{9}{10}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 4\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{10}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{4}{\left(\frac{1}{10}\right)}^{\mathrm{x}}{\left(\frac{9}{10}\right)}^{4-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{none}\mathrm{marked}\mathrm{with}0\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)\\ =\mathrm{C}_{0}^{4}{\left(\frac{1}{10}\right)}^{0}{\left(\frac{9}{10}\right)}^{4-0}\\ ={\left(\frac{9}{10}\right)}^{4}\end{array}$

Q.7 In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{correctly}\mathrm{answered}\mathrm{questions}\mathrm{out}\\ \mathrm{of}20\mathrm{questions}.\mathrm{The}\mathrm{repeated}\mathrm{tosses}\mathrm{of}\mathrm{a}\mathrm{coin}\mathrm{are}\mathrm{Bernoulli}\mathrm{trails}.\\ \mathrm{Since}\mathrm{head}\mathrm{on}\mathrm{a}\mathrm{coin}\mathrm{represents}\mathrm{the}\mathrm{true}\mathrm{answer}\mathrm{and}\mathrm{tail}\\ \mathrm{represents}\mathrm{the}\mathrm{false}\mathrm{answer},\mathrm{the}\mathrm{correctly}\mathrm{answered}\mathrm{questions}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{So},\mathrm{p}=\frac{1}{2}\mathrm{and}\mathrm{q}=1-\frac{1}{2}=\frac{1}{2}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=20\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{10}\\ \mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{20}{\left(\frac{1}{2}\right)}^{\mathrm{x}}{\left(\frac{1}{2}\right)}^{20-\mathrm{x}}\\ =\mathrm{C}_{\mathrm{x}}^{20}{\left(\frac{1}{2}\right)}^{20}\\ \mathrm{P}\left(\mathrm{at}\mathrm{least}12\mathrm{questions}\mathrm{answered}\mathrm{correctly}\right)\\ =\mathrm{P}\left(\mathrm{X}\ge 12\right)\\ =\mathrm{P}\left(\mathrm{X}=12\right)+\mathrm{P}\left(\mathrm{X}=13\right)+...+\mathrm{P}\left(\mathrm{X}=20\right)\\ =\mathrm{C}_{12}^{20}{\left(\frac{1}{2}\right)}^{20}+\mathrm{C}_{13}^{20}{\left(\frac{1}{2}\right)}^{20}+...+\mathrm{C}_{\mathrm{x}}^{20}{\left(\frac{1}{2}\right)}^{20}\\ ={\left(\frac{1}{2}\right)}^{20}\left(\mathrm{C}_{12}^{20}+\mathrm{C}_{13}^{20}+...+\mathrm{C}_{20}^{20}\right)\end{array}$

Q.8 Suppose X has a binomial distribution B (6, 1/2). Show that X = 3 is the most likely outcome.
(Hint: P(X = 3) is the maximum among all P (xi), xi = 0, 1, 2, 3, 4, 5, 6)

Ans

$\begin{array}{l}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{random}\mathrm{variable}\mathrm{whose}\mathrm{binomial}\mathrm{distribution}\mathrm{is}\left(6, \frac{1}{2}\right)\mathrm{.}\\ \mathrm{Therefore},\mathrm{n}=6\mathrm{and}\mathrm{p}=\frac{1}{2}\\ \therefore \mathrm{ }\mathrm{q}=1-\frac{1}{2}=\frac{1}{2}\\ \mathrm{Then},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{\mathrm{x}}{\left(\frac{1}{2}\right)}^{6-\mathrm{x}}\\ =\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{2}\right)}^{6}\\ \mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)\mathrm{will}\mathrm{be}\mathrm{maximum}\mathrm{if}\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{is}\mathrm{maximum}\mathrm{.}\\ \mathrm{Then},\mathrm{}\mathrm{C}_{0}^{6}=\mathrm{C}_{6}^{6}=1, \mathrm{ }\mathrm{C}_{1}^{6}=\mathrm{C}_{5}^{6}=6\\ \mathrm{C}_{2}^{6}=\mathrm{C}_{4}^{6}=15, \mathrm{C}_{3}^{6}=20\\ \mathrm{The}\mathrm{value}\mathrm{of}\mathrm{C}_{3}^{6}\mathrm{ }\mathrm{is}\mathrm{maximum}.\mathrm{Therefore},\mathrm{for}\mathrm{x}= 3,\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)\\ \mathrm{is}\mathrm{maximum}\mathrm{.}\\ \mathrm{Thus},\mathrm{X}= 3\mathrm{is}\mathrm{the}\mathrm{most}\mathrm{likely}\mathrm{outcome}\mathrm{.}\end{array}$

Q.9 On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Ans

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{guessing}\mathrm{of}\mathrm{correct}\mathrm{answers}\mathrm{from}\mathrm{multiple}\mathrm{choice}\\ \mathrm{questions}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\\ \mathrm{correct}\mathrm{answers}\mathrm{by}\mathrm{guessing}\mathrm{in}\mathrm{the}\mathrm{set}\mathrm{of}5\mathrm{multiple}\mathrm{choice}\\ \mathrm{questions}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{correct}\mathrm{answer}\mathrm{is},\\ \mathrm{p}=\frac{1}{3}\\ \therefore \mathrm{q}=1-\frac{1}{3}=\frac{2}{3}\\ \mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}= 5\mathrm{and}\mathrm{ }\mathrm{p}=\frac{1}{3}\\ \mathrm{Then},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{1}{3}\right)}^{\mathrm{x}}{\left(\frac{2}{3}\right)}^{5-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{guessing}\mathrm{more}\mathrm{than}4\mathrm{correct}\mathrm{answers}\right)\\ =\mathrm{P}\left(\mathrm{X}\ge 4\right)\\ =\mathrm{P}\left(\mathrm{X}=4\right)+\mathrm{P}\left(\mathrm{X}=5\right)\\ =\mathrm{C}_{4}^{5}{\left(\frac{1}{3}\right)}^{4}{\left(\frac{2}{3}\right)}^{5-4}+\mathrm{C}_{5}^{5}{\left(\frac{1}{3}\right)}^{5}{\left(\frac{2}{3}\right)}^{5-5}\\ =5.\left(\frac{1}{81}\right)\left(\frac{2}{3}\right)+1.\left(\frac{1}{243}\right).1\\ =\frac{10}{243}+\frac{1}{243}\\ =\frac{11}{243}\end{array}$

Q.10 A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100. What is the probability that he will win a prize
(a) at least once (b) exactly once (c) at least twice?

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{winning}\mathrm{prizes}\mathrm{in}50\mathrm{lotteries}.\\ \mathrm{The}\mathrm{trials}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}\mathrm{.}\\ \mathrm{So},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=50\mathrm{and}\\ \mathrm{p}=\frac{1}{100}\\ \therefore \mathrm{q}=1-\frac{1}{100}=\frac{99}{100}\\ \therefore \mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ \mathrm{ }=\mathrm{C}_{\mathrm{x}}^{50}{\left(\frac{1}{100}\right)}^{\mathrm{x}}{\left(\frac{99}{100}\right)}^{50-\mathrm{x}}\\ \left(\mathrm{a}\right)\mathrm{P}\left(\mathrm{winning}\mathrm{at}\mathrm{least}\mathrm{once}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}\ge 1\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}<1\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}=0\right)\\ \mathrm{ }=1-\mathrm{C}_{0}^{50}{\left(\frac{1}{100}\right)}^{0}{\left(\frac{99}{100}\right)}^{50-0}\\ \mathrm{ }=1-{\left(\frac{99}{100}\right)}^{50}\\ \left(\mathrm{b}\right)\mathrm{P}\left(\mathrm{winning}\mathrm{exactly}\mathrm{once}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}=1\right)\\ \mathrm{ }=\mathrm{C}_{1}^{50}{\left(\frac{1}{100}\right)}^{1}{\left(\frac{99}{100}\right)}^{50-1}\\ \mathrm{ }=50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}\\ \mathrm{ }=\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}\\ \left(\mathrm{c}\right)\mathrm{P}\left(\mathrm{at}\mathrm{least}\mathrm{twice}\right)\\ \mathrm{ }=\mathrm{P}\left(\mathrm{X}\ge 2\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}<2\right)\\ \mathrm{ }=1-\mathrm{P}\left(\mathrm{X}=1\right)-\mathrm{P}\left(\mathrm{X}=0\right)\\ \mathrm{ }=1-\mathrm{C}_{1}^{50}{\left(\frac{1}{100}\right)}^{1}{\left(\frac{99}{100}\right)}^{50-1}-\mathrm{C}_{0}^{50}{\left(\frac{1}{100}\right)}^{0}{\left(\frac{99}{100}\right)}^{50-0}\\ \mathrm{ }=1-\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}-{\left(\frac{99}{100}\right)}^{50}\\ \mathrm{ }=1-{\left(\frac{99}{100}\right)}^{49}\left(\frac{1}{2}+\frac{99}{100}\right)\\ \mathrm{ }=1-{\left(\frac{99}{100}\right)}^{49}\left(\frac{149}{100}\right)\end{array}$

Q.11 Find the probability of getting 5 exactly twice in 7 throws of a die.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{tossing}\mathrm{of}\mathrm{a}\mathrm{die}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\\ \mathrm{Let}\mathrm{X}\mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}5\mathrm{in}7\mathrm{throws}\\ \mathrm{of}\mathrm{the}\mathrm{die}.\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}5\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{the}\mathrm{die},\\ \mathrm{p}=\frac{1}{6}⇒\mathrm{q}=1-\frac{1}{6}=\frac{5}{6}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{the}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=7\mathrm{}\mathrm{and}\mathrm{}\mathrm{p}=\frac{1}{6}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{7}{\left(\frac{1}{6}\right)}^{\mathrm{x}}{\left(\frac{5}{6}\right)}^{7-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{getting}5\mathrm{exactly}\mathrm{twice}\right)\\ =\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{C}_{2}^{7}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{7-2}\\ =21.\left(\frac{1}{36}\right){\left(\frac{5}{6}\right)}^{5}\\ =\left(\frac{7}{12}\right){\left(\frac{5}{6}\right)}^{5}\end{array}$

Q.12 Find the probability of throwing at most 2 sixes in 6 throws of a single die.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{tossing}\mathrm{of}\mathrm{the}\mathrm{die}\mathrm{are}\mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\\ \mathrm{represent}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{sixes}\mathrm{in}6\mathrm{throws}\\ \mathrm{of}\mathrm{the}\mathrm{die}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{six}\mathrm{in}\mathrm{a}\mathrm{single}\mathrm{throw}\mathrm{of}\mathrm{die},\\ \mathrm{p}=\frac{1}{6}⇒\mathrm{q}=1-\frac{1}{6}=\frac{5}{6}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=6\mathrm{and}\mathrm{p}=\frac{1}{6}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{6}{\left(\frac{1}{6}\right)}^{\mathrm{x}}{\left(\frac{5}{6}\right)}^{6-\mathrm{x}}\\ \mathrm{ }\mathrm{P}\left(\mathrm{at}\mathrm{most}2\mathrm{sixes}\right)=\mathrm{P}\left(\mathrm{X}\le 2\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{X}=2\right)\\ =\mathrm{C}_{0}^{6}{\left(\frac{1}{6}\right)}^{0}{\left(\frac{5}{6}\right)}^{6-0}+\mathrm{C}_{1}^{6}{\left(\frac{1}{6}\right)}^{1}{\left(\frac{5}{6}\right)}^{6-1}\\ +\mathrm{C}_{2}^{6}{\left(\frac{1}{6}\right)}^{2}{\left(\frac{5}{6}\right)}^{6-2}\\ ={\left(\frac{5}{6}\right)}^{6}+6.\left(\frac{1}{6}\right){\left(\frac{5}{6}\right)}^{5}+15.\left(\frac{1}{36}\right){\left(\frac{5}{6}\right)}^{4}\\ ={\left(\frac{5}{6}\right)}^{4}\left(\frac{25}{36}+\frac{5}{6}+\frac{15}{36}\right)\\ ={\left(\frac{5}{6}\right)}^{4}\left(\frac{25+30+15}{36}\right)\\ ={\left(\frac{5}{6}\right)}^{4}\left(\frac{70}{36}\right)\\ =\frac{35}{18}{\left(\frac{5}{6}\right)}^{4}\end{array}$

Q.13 It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?

Ans

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{selections}\mathrm{of}\mathrm{articles}\mathrm{in}\mathrm{a}\mathrm{random}\mathrm{sample}\mathrm{space}\\ \mathrm{are}\mathrm{Bernoulli}\mathrm{trails}.\\ \mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{selecting}\mathrm{defective}\mathrm{articles}\\ \mathrm{in}\mathrm{a}\mathrm{random}\mathrm{sample}\mathrm{space}\mathrm{of}12\mathrm{articles}\mathrm{.}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=12\mathrm{and}\\ \mathrm{p}=10%=\frac{1}{10}\\ \therefore \mathrm{q}=1-\frac{1}{10}=\frac{9}{10}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=12\mathrm{and}\mathrm{p}=\frac{1}{10}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{12}{\left(\frac{1}{10}\right)}^{\mathrm{x}}{\left(\frac{9}{10}\right)}^{12-\mathrm{x}}\\ \mathrm{P}\left(9\mathrm{defective}\mathrm{articles}\right)\\ =\mathrm{P}\left(\mathrm{X}=9\right)\\ =\mathrm{C}_{9}^{12}{\left(\frac{1}{10}\right)}^{9}{\left(\frac{9}{10}\right)}^{12-9}\\ =220{\left(\frac{1}{10}\right)}^{9}{\left(\frac{9}{10}\right)}^{3}\\ =220\left(\frac{{9}^{3}}{{10}^{12}}\right)\\ =\frac{22×{9}^{3}}{{10}^{11}}\end{array}$

Q.14

$\begin{array}{l}\text{I}\text{n}\text{}\text{a}\text{}\text{b}\text{o}\text{x}\text{}\text{c}\text{o}\text{n}\text{t}\text{a}\text{i}\text{n}\text{i}\text{n}\text{g}\text{}100\text{}\text{b}\text{u}\text{l}\text{b}\text{s},\text{}10\text{}\text{a}\text{r}\text{e}\text{}\text{d}\text{e}\text{f}\text{e}\text{c}\text{t}\text{i}\text{v}\text{e}.\text{}\text{T}\text{h}\text{e}\text{}\text{p}\text{r}\text{o}\text{b}\text{a}\text{b}\text{i}\text{l}\text{i}\text{t}\text{y}\text{}\text{t}\text{h}\text{a}\text{t}\text{}\text{o}\text{u}\text{t}\text{}\text{o}\text{f}\text{}\text{a}\text{}\text{s}\text{a}\text{m}\text{p}\text{l}\text{e}\text{}\text{o}\text{f}\text{}5\text{}\text{b}\text{u}\text{l}\text{b}\text{s},\text{}\text{n}\text{o}\text{n}\text{e}\text{}\text{i}\text{s}\text{}\text{d}\text{e}\text{f}\text{e}\text{c}\text{t}\text{i}\text{v}\text{e}\\ \text{}\text{i}\text{s}\\ \left(\text{A}\right)1{0}^{–1}\text{}\text{}\left(\text{B}\right){\left(\frac{1}{2}\right)}^{5}\text{}\text{}\left(\text{C}\right){\left(\frac{9}{10}\right)}^{5}\text{}\text{}\left(\text{D}\right)\frac{9}{10}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{selections}\mathrm{of}\mathrm{defective}\mathrm{bulbs}\mathrm{from}\mathrm{a}\mathrm{box}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{defective}\mathrm{bulbs}\mathrm{out}\\ \mathrm{of}\mathrm{a}\mathrm{sample}\mathrm{of}5\mathrm{bulbs}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{defective}\mathrm{bulb},\\ \mathrm{p}=\frac{10}{100}=\frac{1}{10}⇒\mathrm{q}=1-\frac{1}{10}=\frac{9}{10}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=5\mathrm{and}\mathrm{p}=\frac{1}{10}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{1}{10}\right)}^{\mathrm{x}}{\left(\frac{9}{10}\right)}^{5-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{None}\mathrm{of}\mathrm{the}\mathrm{bulbs}\mathrm{is}\mathrm{defective}\right)\\ =\mathrm{P}\left(\mathrm{X}=0\right)\\ ={\left(\frac{9}{10}\right)}^{5}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{C}\mathrm{.}\end{array}$

Q.15

$\begin{array}{l}\mathrm{The}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{that}\mathrm{}\mathrm{a}\mathrm{}\mathrm{student}\mathrm{}\mathrm{is}\mathrm{}\mathrm{not}\mathrm{}\mathrm{a}\mathrm{}\mathrm{swimmer}\mathrm{}\mathrm{is}\frac{1}{5}.\mathrm{Then}\mathrm{}\mathrm{the}\mathrm{}\mathrm{probability}\mathrm{}\mathrm{that}\mathrm{}\mathrm{out}\mathrm{}\mathrm{of}\mathrm{}\mathrm{five}\mathrm{}\mathrm{students},\mathrm{four}\mathrm{}\mathrm{are}\\ \mathrm{swimmers}\mathrm{}\mathrm{is}\\ \left(\mathrm{A}\right)\mathrm{C}_{4}^{5}{\left(\frac{4}{5}\right)}^{4}\frac{1}{5}\mathrm{}\left(\mathrm{B}\right){\left(\frac{4}{5}\right)}^{4}\frac{1}{5}\\ \left(\mathrm{C}\right)\mathrm{C}_{4}^{5}\frac{1}{5}{\left(\frac{4}{5}\right)}^{4}\mathrm{}\left(\mathrm{D}\right)\mathrm{None}\mathrm{}\mathrm{of}\mathrm{}\mathrm{these}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{repeated}\mathrm{selection}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{are}\mathrm{swimmers}\mathrm{are}\\ \mathrm{Bernoulli}\mathrm{trials}.\mathrm{Let}\mathrm{X}\mathrm{denote}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{students},\mathrm{out}\mathrm{of}5\\ \mathrm{students},\mathrm{who}\mathrm{are}\mathrm{swimmers}\mathrm{.}\\ \mathrm{Probability}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{are}\mathrm{not}\mathrm{swimmers},\\ \mathrm{q}=\frac{1}{5}⇒\mathrm{p}=1-\frac{1}{5}=\frac{4}{5}\\ \mathrm{Clearly},\mathrm{X}\mathrm{has}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{n}=5\mathrm{and}\mathrm{p}=\frac{4}{5}\\ \mathrm{Therefore},\mathrm{ }\mathrm{P}\left(\mathrm{X}=\mathrm{x}\right)=\mathrm{C}_{\mathrm{x}}^{\mathrm{n}}\mathrm{ }{\mathrm{p}}^{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}},\mathrm{where}\mathrm{n}=0,1,2,3\dots \mathrm{ }\mathrm{n}\\ =\mathrm{C}_{\mathrm{x}}^{5}{\left(\frac{4}{5}\right)}^{\mathrm{x}}{\left(\frac{1}{5}\right)}^{5-\mathrm{x}}\\ \mathrm{P}\left(\mathrm{four}\mathrm{students}\mathrm{are}\mathrm{swimmers}\right)\\ =\mathrm{P}\left(\mathrm{X}=4\right)\\ =\mathrm{C}_{4}^{5}{\left(\frac{4}{5}\right)}^{4}{\left(\frac{1}{5}\right)}^{5-4}\\ =\mathrm{C}_{0}^{5}{\left(\frac{1}{10}\right)}^{0}{\left(\frac{9}{10}\right)}^{5-0}\\ =\mathrm{C}_{4}^{5}\left(\frac{1}{5}\right){\left(\frac{4}{5}\right)}^{4}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

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### 3. Is Probability a complex and difficult topic in the NCERT prescribed curriculum for Class 12 for Mathematics?

Probability is a vital portion of the academic syllabus of Mathematics prescribed by the NCERT for the students of Class 12. It is a very intriguing and multidimensional concept. It is thus a very dynamic topic. This makes it evident that it ought to be understood, learned and absorbed step-by-step. Students should start with reading and attempting to retain basic formulae and their derivatives along with other fundamental knowledge like theorems. Then, students should aim at acquiring proficiency in the practical application of formulae and theorems through regular practice and revision of comprehensive exercises. Students should also read in detail about the practical uses and relevance of the concept of Probability in daily life. Knowing the purposes of theorems of Probability for human activities would help students to comprehend the problems presented to them more appropriately and efficiently. Formulae as well as the practical uses of Probability have been descriptively covered in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5. The concept of Probability has a myriad of uses in daily life. Predictions involved in phenomena like flipping a coin, choosing a card from the deck, throwing a die, the outcome of a cricket match etc. are all based on theorems of Probability. All of these applications and practical questions involving them have been covered in the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5.

Probability as a mathematical concept has great relevance. Its multiple uses in daily life also make it a particularly practical and interesting area of study. It is greatly recommended that students should use the NCERT Solutions Class 12 Maths Chapter 13 Exercise 13.5 to gain a good understanding of the concept of Probability and its applications.