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Class 10 maths case study based questions chapter 15 probability cbse board term 1 with answer key.

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Case Study Class 10 Maths Questions

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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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CBSE Case Study Questions for Class 10 Maths Probability Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Probability  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Probability PDF

Checkout our case study questions for other chapters.

• Chapter 11: Construction Case Study Questions
• Chapter 12: Area Related to Circles Case Study Questions
• Chapter 13: Surface Area and Volumes Case Study Questions
• Chapter 14: Statistics Case Study Questions

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Sit through chapter wise FULLY INVIGILATED TESTS

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Practice Assertion Reason & Case Study Based Questions

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28. Case study: Probability

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Class 10 Maths Case Study Questions Chapter 15 Probability

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Case study Questions in the Class 10 Mathematics Chapter 15  are very important to solve for your exam. Class 10 Maths Chapter 15 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 10 Maths Case Study Questions  Chapter 15  Probability

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Probability Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 15 Probability

Case Study/Passage-Based Questions

Question 1:

Answer: (a) 18

(ii) If the probability of distributing dark chocolates is 4/9, then the number of dark chocolates Rohit has, is

Answer: (c) 24

(iii) The probability of distributing white chocolates is

Answer: (d) 2/9

(iv) The probability of distributing both milk and white chocolates is

Answer: (b) 5/9

(v) The probability of distributing all the chocolates is

Answer: (b) 1

Question 2:

Rahul and Ravi planned to play Business ( board game) in which they were supposed to use two dice.

1. Ravi got first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8?

Answer: b) 5/36

2. Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 13?

Answer: d) 0

3. Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12?

Answer: a) 1

4. Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal to 7?

Answer: c) 1/6

5. Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is greater than 8?

Answer: d) 5/18

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 15 Probability with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Probability Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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Chapter 14 Class 10 Probability

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Updated for new NCERT Books - for 2024 Boards.

Get NCERT Solutions for Chapter 14 Class 10 free at teachoo. Solutions to all exercise questions, examples and optional is available with detailed explanations. 

In Class 9 , we studied about Empirical or Experimental Probability.

In this chapter, we will study

• Theoretical Probability , that is, P(E) = Number of outcomes with E / Total possible outcomes
• Probability of complementary event, i.e., P(not E)
• Probability of Impossible and sure events
• Probability of questions where die is thrown twice
• Probability of card questions
• Finding Probability using distance and area

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CBSE Class 10 Maths Case Study Questions for Chapter 15 - Probability (Published by CBSE)

Cbse class 10 maths case study questions for chapter 15 - probability are published by the cbse board itself to help students understand the new format of questions. these questions are important for the cbse class 10 maths 2022 board exam preparations..

CBSE Class 10 Maths Case Study Questions for Chapter 15 - Probability are provided here. These questions are quite useful to understand how real-life problems can be put in the form of questions. These case study questions are published by the CBSE board. Solve all these questions to prepare for your exams and score the desired marks.

Case Study Questions for Class 10 Maths Chapter 15 - Probability:

CASE STUDY 1:

On a weekend Rani was playing cards with her family. The deck has 52 cards. If her brother drew one card.

1. Find the probability of getting a king of red colour.

Answer: a) 1/26

2. Find the probability of getting a face card.

Answer: d) 3/13

3. Find the probability of getting a jack of hearts.

4. Find the probability of getting a jack of hearts.

Answer: a) 3/13

5. Find the probability of getting a jack of hearts.

Answer: d) 1/4

CASE STUDY 2:

Rahul and Ravi planned to play Business ( board game) in which they were supposed to use two dice.

1. Ravi got first chance to roll the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 8?

Answer: b) 5/36

2. Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is 13?

Answer: d) 0

3. Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is less than or equal to 12?

Answer: a) 1

4. Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is equal to 7?

Answer: c) 1/6

5. Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the top face of the dice is greater than 8?

Answer: d) 5/18

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

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CBSE Class 10 Maths Case Study

CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus. 

These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.

Class 10 Maths (Formula, Case Based, MCQ, Assertion Reason Question with Solutions)

In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.

Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too. 

Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.

For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.

CBSE Class 10th Case study Questions

Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.

Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.

CBSE Class 10th Assertion Reasoning Questions

These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement? 

Moreover, to solve the problem they need to look at the given options and then answer them.

CBSE Class 10 Maths Case Based MCQ

CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.

Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.

Class 10th Mathematics Multiple Choice Questions

Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.

Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.

The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.

To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.

In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.

CBSE Class 10 Exams Finish, When Can You Expect Results? Details Here

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CBSE Class 10 Information Technology Exam 2024 : Most Important Questions Answers for Last-Minute Revision

CBSE Class 10 Computer Applications Exam 2024 : Most Important Questions Answers for Last-Minute Revision

CBSE Board Class 10 Maths Answer Key 2024 and Question Papers, Download PDF All SETs

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CBSE 10th Standard Maths Subject Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

Based on the above information, answer the following questions. (i) The upper limit of a class and lower limit of its succeeding class is differ by

(ii) The median class is

(iii) The cumulative frequency of the class preceding the median class is

(iv) The median of the distance travelled is

(v) If the mode of the distance travelled is 223.78 km, then mean of the distance travelled by the bus is

An electric scooter manufacturing company wants to declare the mileage of their electric scooters. For this, they recorded the mileage (km/ charge) of 50 scooters of the same model. Details of which are given in the following table.

(ii) The modal value of the given data is

(ill) The median value of the given data is

(iv) Assumed mean method is useful in determining the

(v) The manufacturer can claim that the mileage for his scooter is

Household income in India was drastically impacted due to the COVID-19 loekdown. Most of the companies decided to bring down the salaries of the employees by 50%. The following table shows the salaries (in percent) received by 25 employees during loekdown.

(ii) Total number of persons whose salary is reduced by atmost 40%, is

(iii) The modal class is

(iv) The median class of the given data is

(v) The empirical relationship between mean, median and mode is

A petrol pump owner wants to analyse the daily need of diesel at the pump. For this he collected the data of vehicles visited in 1 hr. The following frequency distribution table shows the classification of the number of vehicles and quantity of diesel filled in them.

(ii) Average diesel required for a vehicle is

(iii) If approximately 2000 vehicles comes daily at the petrol pump, then how much litres of diesel the pump should have?

(iv) The sum of upper and lower limit of median class is

(v) If the median of given data is 8litres, then mode will be equal to

(ii) \(\begin{equation} ₹ \end{equation} \)  20 coin

(iii) not a   \(\begin{equation} ₹ \end{equation} \)  10 coin

(iv) of denomination of atleast   \(\begin{equation} ₹ \end{equation} \) 10. 

(v) of denomination of atmost \(\begin{equation} ₹ \end{equation} \)  5.

(ii) If the probability of distributing dark chocolates is 4/9, then the number of dark chocolates Rohit has, is

(iii) The probability of distributing white chocolates is

(iv) The probability of distributing both milk and white chocolates is

(v) The probability of distributing all the chocolates is

(ii) not of yellow colour

(iii) of green colour

(iv) of yellow colour 

(v) not of blue colour 

(ii) The probability of selecting a flowerhorn fish is

(iii) The probability of not selecting a koi fish is

(iv) The probability of selecting neither angle fish nor flowerhorn fish is 

(v) The probability of selecting a guppy fish is 

 (ii) The probability of getting the sum of numbers on two dice is 16, is

(iii) The probability that both the numbers are prime numbers, is

(iv) The probability that product of two numbers is odd, is|

(v) The probability that difference between numbers is zero, is 

(ii) The probability of getting a black diamond is

(iii) The probability of getting a face card is

(iv) The probability of getting a club is

(v) The probability of getting a red card is

(ii) If Anita spins the wheel, then the probability of getting no prize is

(iii) Anshu spins the wheel, the probability that the wheel stops at soccer ball is

(iv) The probability that one customer wins 15% discount is

(v) The probability of getting a free spin is 

(ii) A football is selected at random, the probability of selecting a non-defective football is

(iii) The total number of defective footballs produced in one day is

(iv) The total number of non-defective footballs produced in one day is 

(v) If the probability of selecting a defective football is , then the number of non-defective footballs produced in one day, if everyday same number of footballs produced in the factory, is 

(ii) The probability that Gupta's has atleast 1 boy is

(iii) The probability that Gupta's has atmost 1 girl is

(iv) The probability that Singhal's has no boy is

(v) The sum of probabilities that both families have exactly two girls is

(ii) The probability that the selected carton is not of orange juice is

(iii) The probability of selecting a carton of guava juice is

(iv) Vishal buys 4 cartons of apple juice, 3 cartons of orange juice and 3 cartons of guava juice. A customer comes to Vishal's shop and picks a tetrapack of juice at random. The probability that the customer picks a guava juice, if each carton has 10 tetrapacks of juice, is 

(v) If the storekeeper bought 14 more cartons of apple juice, then the probability of selecting a tetrapack of apple juice from the store is 

In the month of May, the weather forecast department gives the prediction of weather for the month of June, The given table shows the probabilities of forecast of different days:

(ii) If the number of cloudy days in June is 5, then x =

(iii) The probability that the day is not rainy is

(iv) If the sum of x < and y is  \(\begin{equation} \frac{3}{10} \end{equation}\) , then the number of rainy days in June is

(v) Find the number of partially cloudy days

*****************************************

Cbse 10th standard maths subject case study questions 2021 answer keys.

(i) (b): The upper limit of a class and the lower class of its succeeding class differ by 1. (ii) (d) : Here, class intervals are in inclusive form. So, we first convert them in exclusive form. The frequency distribution table in exclusive form is as follows:

\(\text { Here, } \Sigma f_{i} \text { i.e., } N=60 \) \(\Rightarrow \frac{N}{2}=30\)

Now, the class interval whose cumulative frequency is just greater than 30 is 219.5 - 229.5. \(\therefore\) Median class is 219.5 - 229.5. (iii) (b): Clearly, the cumulative frequency of the class preceding the median class is 18 (iv) (d) : Median  \(=l+\left[\frac{\frac{N}{2}-c . f .}{f}\right] \times h\) \(=219.5+\left(\frac{30-18}{26}\right) \times 10 \) \(=219.5+\frac{12 \times 10}{26}=219.5+4.62=224.12\) \(\therefore\) Median of the distance travelled is 224.12 km (v) (d) : We know, Mode = 3 Median - 2 Mean \(\therefore \quad \text { Mean }=\frac{1}{2}(3 \text { Median }-\text { Mode }) \) \(=\frac{1}{2}(672.36-223.78)=224.29 \mathrm{~km}\)

Given frequency distribution table can be drawn as:

(i) (d): Clearly, average mileage \(=\frac{7240}{50}=144.8 \mathrm{~km} / \text { charge }\) (ii) (b) : Since, highest frequency is IS, therefore, modal class is 140-160. Here, l = 140,f 1 = 18,f 0  = 12,f 2 = 13, h = 20 \(\therefore \quad \text { Mode }=140+\frac{18-12}{36-12-13} \times 20=140+\frac{6}{11} \times 20 \) \(=140+\frac{120}{11}=140+10.91=150.91\) (iii) (b) : Here  \(\frac{N}{2}=\frac{50}{2}=25\)  and the corresponding class whose cumulative frequency is just greater than 25 is 140-160. Here, l = 140, c.f = 19, h = 20 and f= 18 \(\therefore \quad \text { Median }=l+\left(\frac{\frac{N}{2}-c . f .}{f}\right) \times h\) \(=140+\frac{25-19}{18} \times 20=140+\frac{60}{9}=146.67\) (iv) (a) : Assumed mean method is useful in determining the mean. (v) (a): Since, Mean = 144.S, Mode = 150.91 and Median = 146.67 and minimum of which is 144 approx, therefore manufacturer can claim the mileage for his scooter 144 km/charge.

(i) (d): Required number of persons = 9 + 6 = 15 (ii) (c) : Required number of persons = 6 + 8 + 2 = 16 (iii) (a) : 50-60 is the modal class as the maximum frequency is 9. (iv) (b) : The cumulative frequency distribution table for the given data can be drawn as :

\(\text { Here, } \frac{N}{2}=\frac{25}{2}=12.5\) The cumulative frequency just greater than 12.5 lies in the interval 60-70. Hence, the median class is 60-70. (v) (a): We know, Mode = 3 Median - 2 Mean \(\therefore\)   3 Median = Mode + 2 Mean.

(i) (a): If f i  and x i are very small, then direct method is appropriate method for calculating mean. (ii) (a) : The frequency distribution table from the given data can be drawn as :

\(\therefore \quad \text { Mean }=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{326}{40}=8.15 \text { litres }\)

(iii) (b) : If 2000 vehicles comes daily and average quantity of diesel required for a vehicle is 8.15 litres, then total quantity of diesel required = 2000 x 8.15 = 16300 litres (iv) (c) : Here  \(N=40 \text { and } \frac{N}{2}=20\)  c.f for the distribution are 5, 15,25,32,40 Now, c.f just greater than 20 is 25 which is corresponding to the class interval 7-9. So median class is 7-9. \(\therefore\) Required sum of upper limit and lower limit = 7 + 9 = 16 (v) (b): We know, Mode = 3 Median -2 Mean = 3(8) - 2(8.15) = 24 - 16.3 = 7.7

Total number of coins = 50 + 48 + 36 + 28 + 8 = 170 (i) (c): Number of  \(\begin{equation} ₹ \end{equation} \)  5 coins = 36 Required probability =  \(\begin{equation} \frac{36}{70}=\frac{18}{85} \end{equation}\) (ii) (b): Number of \(\begin{equation} ₹ \end{equation} \) 20 coins = 8 Required probability =  \(\begin{equation} \frac{8}{170}=\frac{4}{85} \end{equation}\) (iii) (d): Number of \(\begin{equation} ₹ \end{equation} \) 10 coins = 28 Probability (coin is of \(\begin{equation} ₹ \end{equation} \) 10) =  \(\begin{equation} \frac{28}{170} \end{equation}\) Required probability = 1 - P(coin is of 10) \(\begin{equation} =1-\frac{28}{170}=\frac{142}{170}=\frac{71}{85} \end{equation}\) (iv) (a) : Total number of coins of  \(\begin{equation} ₹ \end{equation} \) 10 and \(\begin{equation} ₹ \end{equation} \)  20 = 28 + 8 = 36 Required probability =  \(\begin{equation} \frac{36}{170}=\frac{18}{85} \end{equation}\) (v) (a): Total number of coins of \(\begin{equation} ₹ \end{equation} \) 1, \(\begin{equation} ₹ \end{equation} \) 2 and \(\begin{equation} ₹ \end{equation} \)  5  = 50 + 48 + 36 = 134 Required probability =  \(\begin{equation} \frac{134}{170}=\frac{67}{85} \end{equation}\)

Since, every student get one chocolate. So, number of chocolates Rohit has is equal to the number of students in the class. (i) (a): Let number of milk chocolates Rohit has = x Probability of distributing milk chocolates =  \(\begin{equation} \frac{1}{3} \end{equation}\) \(\begin{equation} \Rightarrow \frac{x}{54}=\frac{1}{3} \Rightarrow x=\frac{54}{3}=18 \end{equation}\) (ii) (c): Let number of dark chocolates Rohit has = y Probability of distributing dark chocolates =  \(\begin{equation} \frac{4}{9} \end{equation}\) \(\begin{equation} \Rightarrow \frac{y}{54}=\frac{4}{9} \Rightarrow y=\frac{4 \times 54}{9}=24 \end{equation}\) (iii) (d) : Number of white chocolates Rohit has = 54 -(18 + 24) = 12 Required probability =  \(\begin{equation} \frac{12}{54}=\frac{2}{9} \end{equation}\) (iv) (b) : Total number of milk and white chocolates = 18 + 12 = 30 Required probability =  \(\begin{equation} \frac{30}{54}=\frac{5}{9} \end{equation}\) (v) (b): Since all students gets one chocolate. So, total number of chocolates distributed = 54 \(\begin{equation} \text { Required probability }=\frac{54}{54}=1 \end{equation}\)

Total number of blocks in the kit = 120 Number of red blocks = 40 Number of blue blocks = 25 Number of green blocks = 30 Number of yellow blocks = 120 - (40 + 25 + 30) = 120 - 95 = 25 (i) (d): P(block is red)  \(\begin{equation} =\frac{40}{120}=\frac{1}{3} \end{equation}\) (ii) (c): P(block is not yellow) = 1 - P(block is yellow) \(\begin{equation} =1-\frac{25}{120}=1-\frac{5}{24}=\frac{19}{24} \end{equation}\) (iii) (c) : P(block is green) =  \(\begin{equation} \frac{30}{120}=\frac{1}{4} \end{equation}\) (iv) (b) : P(block is yellow) =  \(\begin{equation} \frac{5}{24} \end{equation}\) (v) (b): P(block is not blue) = 1 - P(block is blue) \(\begin{equation} =1-\frac{25}{120}=1-\frac{5}{24}=\frac{19}{24} \end{equation}\)

Total number of fish in the aquarium = 13 + 18 + 12 + 11 = 54 (i) (b): Number of male fish in the aquarium = 36 Number of female fish in the aquarium = 54 - 36 = 18 So, required probability =  \(\begin{equation} \frac{18}{54}=\frac{1}{3} \end{equation}\) (ii) (b): Required probability =  \(\begin{equation} \frac{18}{54}=\frac{1}{3} \end{equation}\) (iii) (d) : P(selecting a koi fish) =  \(\begin{equation} \frac{12}{54}=\frac{2}{9} \end{equation}\) P(not selecting a koi fish) = 1 - P(selecting a koi fish)  \(\begin{equation} =1-\frac{2}{9}=\frac{7}{9} \end{equation}\) (iv) (b) : Total number of guppy fish and koi fish = 13 + 12 = 25 P(selecting neither angel fish nor flowerhorn fish =25/54. (v) (c) : P(selecting a guppy fish) = 13/54.

(i) (b): Total number of possible outcomes on throwing two dice simultaneously = 6 x 6 = 36 (ii) (c): As we know, that maximum sum of numbers on two dice = 6 + 6 = 12 It is an impossible event. So, required probability = 0 (iii) (c) : Let A be the event that both the numbers on dice are prime.  A = {(2, 2), (2, 3), (2, 5), (3,2), (3, 3), (3, 5), (5, 2), (5,3), (5, 5)} \(\Rightarrow\) n(A) = 9 \(\begin{equation} P(A)=\frac{9}{36}=\frac{1}{4} \end{equation}\) (iv) (c) : Let B the event that product of two numbers is odd. B = {(1, 1), (1, 3), (1, 5), (3,1), (3, 3), (3, 5), (5,1), (5,3), (5, 5)} \(\Rightarrow\) n(B) = 9 \(\begin{equation} P(B)=\frac{9}{36}=\frac{1}{4} \end{equation}\) (v) (c) : Let C be the event that difference of two numbers is zero. C = {(1, 1), (2,2), (3, 3), (4,4), (5, 5), (6, 6)} \(\Rightarrow\) n(C) = 6 \(\begin{equation} P(C)=\frac{6}{36}=\frac{1}{6} \end{equation}\)

Total number of cards left = 52 - (4 + 4 + 4) = 52 - 12 = 40 There are four 2's, four 5's and four Jacks in a deck (i) (c): P(getting 6 of spade) =  \(\begin{equation} \frac{1}{40} \end{equation}\) (ii) (a): As there is no black diamond in the deck of cards. P(black diamond) = 0 (iii) (b) : Total number of face cards in a deck = 12 But all Jacks are removed Number offace cards left = 12 - 4 = 8 So, P(getting a face card) =  \(\begin{equation} \frac{8}{40}=\frac{1}{5} \end{equation}\) (iv) (d) : Number of cards of club left = 13 - 3 = 10 3 cards are removed of each kind P(getting a club) =  \(\begin{equation} \frac{10}{40}=\frac{1}{4} \end{equation}\) (v) (c): Number of red cards left = 26 - 6 = 20 P(getting a red card) =  \(\begin{equation} \frac{20}{40}=\frac{1}{2} \end{equation}\)

Total number of possible outcomes = 10 (i) (b): P(getting 100% discount) =  \(\begin{equation} \frac{1}{10} \end{equation}\) (ii) (c): P(getting no prize) =  \(\begin{equation} \frac{3}{10} \end{equation}\) (iii) (b) : P(getting a soccer ball) =  \(\begin{equation} \frac{2}{10}=\frac{1}{5} \end{equation}\) (iv) (b) : P(getting 15% discount) =  \(\begin{equation} \frac{2}{10}=\frac{1}{5} \end{equation}\) (v) (a): P(getting a free spin) =  \(\begin{equation} \frac{1}{10} \end{equation}\)

Total number of footballs produced in one day = 220000 (i) (c): P(selecting a defective football) =  \(\begin{equation} \frac{15}{100}=\frac{3}{20} \end{equation}\) (ii) (d): P(selecting a non-defective football) = 1- P(selecting a defective football) =  \(\begin{equation} 1-\frac{3}{20}=\frac{17}{20} \end{equation}\) (iii) (b) : Total number of defective footballs produced in 1 day =  \(\begin{equation} \frac{3}{20} \times 22000=3300 \end{equation}\) (iv) (a) : Total number of non-defective footballs produced in 1 day = 22000 - 3300 = 18700 (v) (d): P(selecting defective football)  \(\begin{equation} =\frac{8}{25} \end{equation}\) P (selecting non-defective football) =  \(\begin{equation} 1-\frac{8}{25}=\frac{17}{25} \end{equation}\) Number of non-defective footballs produced  \(\begin{equation} =\frac{17}{25} \times 22000=14960 \end{equation}\)

Let Band G denotes boy and girl respectively. Then, Possible outcomes for Gupta's family having two kids is {BB, BG, GG} Similarly, possible outcomes for Singhal's family having 3 kids is {BBB, BBG, BGG, GGG} (i) (b): Required probability =  \(\begin{equation} \frac{1}{4} \end{equation}\) (ii) (b): Required probability =  \(\begin{equation} \frac{2}{3} \end{equation}\) (iii) (b): Required probability =  \(\begin{equation} \frac{2}{3} \end{equation}\) (iv) (b) : Required probability =  \(\begin{equation} \frac{1}{4} \end{equation}\) (v) (c): Required probability =  \(\begin{equation} \frac{1}{3}+\frac{1}{4}=\frac{7}{12} \end{equation}\)

Total number of cartons in the store = 80 + 90 + 38 + 42 = 250 (i) (d) : P(choosing an apple juice carton) =  \(\begin{equation} \frac{90}{250}=\frac{9}{25} \end{equation}\) (ii) (c) : P( choosing an orange juice carton ) =  \(\begin{equation} \frac{80}{250}=\frac{8}{25} \end{equation}\) P(choosing not an orange juice carton) =  \(\begin{equation} =1-\frac{8}{25}=\frac{17}{25} \end{equation}\) (Iii) (d): P(choosing a guava juice carton) =  \(\begin{equation} \frac{42}{250}=\frac{21}{125} \end{equation}\) (iv) (c) : Total number of cartons Vishal bought =4+3+3=10 Number of tetrapacks in 1 carton = 10 Total number of tetrapacks Vishal has = 100 So, P( customer picks a guava juice tetrapack)  \(\begin{equation} =\frac{3 \times 10}{100}=\frac{3}{10} \end{equation}\) (v) (b): Number of cartons left with storekeeper = 250 - 10 = 240 Number of cartons he bought = 14 Total number of cartons are with storekeeper now = 240 + 14 = 254 So, P(selecting a tetrapack of apple juice from store)  \(\begin{equation} =\frac{(90-4+14) \times 10}{254 \times 10}=\frac{100}{254}=\frac{50}{127} \end{equation}\)

Total number of days in June = 30 (i) (c): Number of sunny days = PCsunny day) x 30 =  \(\begin{equation} \frac{1}{2} \times 30=15 \end{equation}\) (ii) (b) : Number of cloudy days in June = 5 \(\begin{equation} x=\frac{5}{30}=\frac{1}{6} \end{equation}\) (iii) (a): Required probability =  \(\begin{equation} \frac{1}{2}+\frac{1}{6}+\frac{1}{5}=\frac{13}{15} \end{equation}\) (iv) (d) : We have,  \(\begin{equation} x+y=\frac{3}{10} \Rightarrow y=\frac{3}{10}-\frac{1}{6}=\frac{2}{15} \end{equation}\) So, number of rainy days =  \(\begin{equation} \frac{2}{15} \times 30=4 \end{equation}\) (v) (c): Number of partially cloudy days = P(partially cloudy days) x 30  \(\begin{equation} =\frac{1}{5} \times 30=6 \end{equation}\)

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• NCERT Solutions
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• Chapter 15: Probability

NCERT Solutions for Class 10 Maths Chapter 15 Probability

Ncert solutions for class 10 maths chapter 15 – cbse free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 14.

NCERT Solutions for Class 10 Maths Chapter 15 Probability is undoubtedly an essential study material for the students studying in CBSE Class 10. NCERT Solutions , along with the downloadable PDF provided in this article, can help students prepare effectively for the CBSE Board exams. The chapter is a continuation of what was taught in Chapter Probability in Class 9 and further explains the different concepts related to it.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 15 Probability

Download most important questions for class 10 maths chapter – 15 probability.

The subject experts prepared the NCERT Solutions for Class 10 Maths to assist students in preparing for their board examinations. Every step and concept used in solving a solution is explained clearly in the answers provided at BYJU’S, leaving no question unsolved. Along with exam preparation, these solutions can be used to check if the answers given to the exercise questions by the students are correct while doing their homework and assignments. So, it is advised to all the students to go through these NCERT Solutions regularly to excel in the CBSE Class 10 Maths examination.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics

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Access Answers of Maths NCERT Class 10 Chapter 15 – Probability

Exercise: 15.1 (Page No: 308)

1. Complete the following statements:

(i) Probability of an event E + Probability of the event ‘not E’ = ___________.

(ii) The probability of an event that cannot happen is __________. Such an event is called ________.

(iii) The probability of an event that is certain to happen is _________. Such an event is called _________.

(iv) The sum of the probabilities of all the elementary events of an experiment is __________.

(v) The probability of an event is greater than or equal to ___ and less than or equal to __________.

(i) Probability of an event E + Probability of the event ‘not E’ = 1 .

(ii) The probability of an event that cannot happen is 0 . Such an event is called an impossible event .

(iii) The probability of an event that is certain to happen is 1 . Such an event is called a sure or certain event .

(iv) The sum of the probabilities of all the elementary events of an experiment is 1 .

(v) The probability of an event is greater than or equal to 0 and less than or equal to 1 .

2. Which of the following experiments have equally likely outcomes? Explain.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to a solution: a true-false question. The solution is right or wrong.

(iv) A baby is born. It is a boy or a girl.

(i) This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.

(ii) This statement does not have equally likely outcomes as the player may shoot or miss the shot.

(iii) This statement has equally likely outcomes as it is known that the solution is either right or wrong.

(iv) This statement also has equally likely outcomes as it is known that the newly born baby can either be a boy or a girl.

3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

The tossing of a coin is a fair way of deciding because the number of possible outcomes is only 2, i.e. either head or tail. Since these two outcomes are equally likely outcomes, tossing is unpredictable and is considered to be completely unbiased.

4. Which of the following cannot be the probability of an event?

(A) 2/3, (B) -1.5, (C) 15%, (D) 0.7

The probability of any event (E) always lies between 0 and 1, i.e. 0 ≤ P(E) ≤ 1. So, from the above options, option (B) -1.5 cannot be the probability of an event.

5. If P(E) = 0.05, what is the probability of ‘not E’?

We know that,

P(E)+P(not E) = 1

It is given that, P(E) = 0.05

So, P(not E) = 1-P(E)

Or, P(not E) = 1-0.05

∴ P(not E) = 0.95

6. A bag contains lemon-flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she will take out

(i) an orange-flavoured candy?

(ii) a lemon-flavoured candy?

(i) We know that the bag only contains lemon-flavoured candies.

So, The number of orange-flavoured candies = 0

∴ The probability of taking out orange-flavoured candies = 0/1 = 0

(ii) As there are only lemon-flavoured candies, P(lemon-flavoured candies) = 1 (or 100%)

7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Let the event wherein 2 students having the same birthday be E

Given, P(E) = 0.992

Or, P(not E) = 1–0.992 = 0.008

∴ The probability that the 2 students have the same birthday is 0.008

8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

(ii) not red?

The total number of balls = No. of red balls + No. of black balls

So, the total number of balls = 5+3 = 8

We know that the probability of an event is the ratio between the number of favourable outcomes and the total number of outcomes.

P(E) = (Number of favourable outcomes/Total number of outcomes)

(i) Probability of drawing red balls = P (red balls) = (no. of red balls/total no. of balls) = 3/8

(ii) Probability of drawing black balls = P (black balls) = (no. of black balls/total no. of balls) = 5/8

9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

(ii) white?

(iii) not green?

The Total no. of balls = 5+8+4 = 17

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of red balls = 5

P (red ball) = 5/17 = 0.29

(ii) Total number of white balls = 8

P (white ball) = 8/17 = 0.47

(iii) Total number of green balls = 4

P (green ball) = 4/17 = 0.23

∴ P (not green) = 1-P(green ball) = 1-(4/7) = 0.77

10. A piggy bank contains hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin

(i) will be a 50 p coin?

(ii) will not be a ₹5 coin?

Total no. of coins = 100+50+20+10 = 180

(i) Total number of 50 p coin = 100

P (50 p coin) = 100/180 = 5/9 = 0.55

(ii) Total number of ₹5 coin = 10

P (₹5 coin) = 10/180 = 1/18 = 0.055

∴ P (not ₹5 coin) = 1-P (₹5 coin) = 1-0.055 = 0.945

11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see Fig. 15.4). What is the probability that the fish taken out is a male fish?

The total number of fish in the tank = 5+8 = 13

Total number of male fish = 5

P (male fish) = 5/13 = 0.38

12. A game of chance consists of spinning an arrow which comes to rest, pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5), and these are equally likely outcomes. What is the probability that it will point at

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

Total number of possible outcomes = 8

(i) Total number of favourable events (i.e. 8) = 1

∴ P (pointing at 8) = ⅛ = 0.125

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P (pointing at an odd number) = 4/8 = ½ = 0.5

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P (pointing at a number greater than 4) = 6/8 = ¾ = 0.75

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P (pointing at a number less than 9) = 8/8 = 1

13. A die is thrown once. Find the probability of getting

(i) a prime number

(ii) a number lying between 2 and 6

(iii) an odd number

Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

(i) Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = ½ = 0.5

(ii) Total numbers lying between 2 and 6 = 3 (3, 4 and 5)

P (getting a number between 2 and 6) = 3/6 = ½ = 0.5

(iii) Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = ½ = 0.5

14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Total number of possible outcomes = 52

(i) Total number of the king of red colour = 2

P (getting a king of red colour) = 2/52 = 1/26 = 0.038

(ii) Total number of face cards = 12

P (getting a face card) = 12/52 = 3/13 = 0.23

(iii) Total number of red face cards = 6

P (getting a king of red colour) = 6/52 = 3/26 = 0.11

(iv) Total number of jack of hearts = 1

P (getting a king of red colour) = 1/52 = 0.019

(v) Total numbers of the king of spade = 13

P (getting a king of red colour) = 13/52 = ¼ = 0.25

(vi) Total number of the queen of diamonds = 1

15. Five cards, the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Total number of cards = 5

(i) Number of queens = 1

P (picking a queen) = ⅕ = 0.2

(ii) If the queen is drawn and put aside, the total number of cards left is (5-4) = 4

(a) Total numbers of ace = 1

P (picking an ace) = ¼ = 0.25

(b) Total number of queens = 0

P (picking a queen) = 0/4 = 0

16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Numbers of pens = Numbers of defective pens + Numbers of good pens

∴ Total number of pens = 132+12 = 144 pens

P(picking a good pen) = 132/144 = 11/12 = 0.916

17. (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

(i) Number of defective bulbs = 4

The total number of bulbs = 20

∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = ⅕ = 0.2

(ii) Since 1 non-defective bulb is drawn, then the total number of bulbs left is 19

So, the total number of events (or outcomes) = 19

Number of non-defective bulbs = 19-4 = 15

So, the probability that the bulb is not defective = 15/19 = 0.789

18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5

The total number of discs = 90

(i) Total number of discs having two digit numbers = 81

(Since 1 to 9 are single-digit numbers and so, total 2-digit numbers are 90-9 = 81)

P (bearing a two-digit number) = 81/90 = 9/10 = 0.9

(ii) Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) = 9/90 = 1/10 = 0.1

(iii) Total numbers which are divisible by 5 = 18 (5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90)

P (getting a number divisible by 5) = 18/90 = ⅕ = 0.2

19. A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting

The total number of possible outcomes (or events) = 6

(i) The total number of faces having A on it = 2

P (getting A) = 2/6 = ⅓ = 0.33

(ii) The total number of faces having D on it = 1

P (getting D) = ⅙ = 0.166

20. Suppose you drop a die at random on the rectangular region shown in Fig. 15.6. What is the probability that it will land inside the circle with a diameter 1m?

First, calculate the area of the rectangle and the area of the circle. Here, the area of the rectangle is the possible outcome, and the area of the circle will be the favourable outcome.

So, the area of the rectangle = (3×2) m 2 = 6 m 2

The area of the circle = πr 2 = π(½) 2 m 2 = π/4 m 2 = 0.78

∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 or, 0.78/6 = 0.13

21. A lot consists of 144 ball pens, of which 20 are defective, and the others are good. Nuri will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

The total number of outcomes, i.e. pens = 144

Given, the number of defective pens = 20

∴ The numbers of non defective pens = 144-20 = 124

(i) Total number of events in which she will buy them = 124

So, P (buying) = 124/144 = 31/36 = 0.86

(ii) Total number of events in which she will not buy them = 20

So, P (not buying) = 20/144 = 5/36 = 0.138

22. Refer to Example 13. (i) Complete the following table:

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability of 1/11. Do you agree with this argument? Justify your Solution:

If 2 dice are thrown, the possible events are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

So, the total number of events: 6×6 = 36

(i) It is given that to get the sum as 2, the probability is 1/36 as the only possible outcomes = (1,1)

For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)

So, P(sum 3) = 2/36

E (sum 4) = (1,3), (3,1), and (2,2)

So, P (sum 4) = 3/36

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

So, P (sum 5) = 4/36

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

So, P (sum 6) = 5/36

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

So, P (sum 7) = 6/36

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

So, P (sum 8) = 5/36

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

So, P (sum 9) = 4/36

E (sum 10) = (4,6), (6,4), and (5,5)

So, P (sum 10) = 3/36

E (sum 11) = (5,6), and (6,5)

So, P (sum 11) = 2/36

E (sum 12) = (6,6)

So, P (sum 12) = 1/36

So, the table will be as:

(ii) The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.

23. A game consists of tossing a one-rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (losing the game) = 6/8 = ¾ = 0.75

24. A die is thrown twice. What is the probability that

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Outcomes are:

So, the total number of outcomes = 6×6 = 36

(i) Method 1:

Consider the following events.

A = 5 comes in the first throw,

B = 5 comes in the second throw

P(A) = 6/36,

P(B) = 6/36 and

P(not B) = 5/6

So, P(not A) = 1-(6/36) = 5/6

∴ The required probability = (5/6)×(5/6) = 25/36

Let E be the event in which 5 does not come up either time.

So, the favourable outcomes are [36–(5+6)] = 25

∴ P(E) = 25/36

(ii) Number of events when 5 comes at least once = 11(5+6)

∴ The required probability = 11/36

25. Which of the following arguments are correct and which are not correct? Give reasons for your solution:

(i) If two coins are tossed simultaneously, there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

(i) All the possible events are (H,H); (H,T); (T,H) and (T,T)

So, P (getting two heads) = ¼

and, P (getting one of each) = 2/4 = ½

∴ This statement is incorrect.

(ii) Since the two outcomes are equally likely, this statement is correct.

Exercise: 15.2 (Page No: 311)

1. Two customers, Shyam and Ekta, are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

(i) the same day?

(ii) consecutive days?

(iii) different days?

Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5×5 = 25

(i) The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)

So, P (both visiting on the same day) = 5/25 = ⅕

(ii) The number of favourable events = 8 (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Fri., Thu.), (Thu., Wed.), and (Wed., Tue.)

So, P(both visiting on the consecutive days) = 8/25

(iii) P (both visiting on different days) = 1-P (both visiting on the same day)

So, P (both visiting on different days) = 1-(⅕) = ⅘

2. A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times, and the total score in two throws is noted. Complete the following table, which gives a few values of the total score on the two throws:

What is the probability that the total score is

(iii) at least 6?

The table will be as follows:

(i) E (Even) = 18

P (Even) = 18/36 = ½

(ii) E (sum is 6) = 4

P (sum is 6) = 4/36 = 1/9

(iii) E (sum is atleast 6) = 15

P (sum is atleast 6) = 15/36 = 5/12

3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball

is double that of a red ball, determine the number of blue balls in the bag.

It is given that the total number of red balls = 5

Let the total number of blue balls = x

So, the total no. of balls = x+5

∴ P (drawing a blue ball) = [x/(x+5)] ——–(i)

P (drawing a red ball) = [5/(x+5)] ——–(i)

From equations (i) and (ii)

So, the total number of blue balls = 10

4. A box contains 12 balls, out of which x are black. If one ball is drawn at random from the

box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball is now

double of what it was before. Find x

Total number of black balls = x

Total number of balls = 12

P (getting black balls) = x/12 ——————-(i)

Now, when 6 more black balls are added,

Total balls become = 18

∴ Total number of black balls = x+6

Now, P (getting black balls) = (x+6)/18 ——————-(ii)

It’s given that the probability of drawing a black ball now is double of what it was before.

(ii) = 2 × (i)

(x+6)/18 = 2 × (x/12)

5. A jar contains 24 marbles, some are green, and others are blue. If a marble is drawn at

random from the jar, the probability that it is green is ⅔. Find the number of blue balls

in the jar.

Total marbles = 24

Let the total green marbles = x

So, the total blue marbles = 24-x

P(getting green marble) = x/24

From the question, x/24 = ⅔

So, the total green marbles = 16

And, the total blue marbles = 24-16 = 8

Chapter 15 of Class 10, Probability, belongs to the 6th Unit – Statistics and Probability, which adds up to 10 marks out of the total 80 marks in the examination. Unit 6 is one such unit in which marks can be easily scored once students get familiar with the method or the idea. Solving the NCERT exercise questions will enable the students to get thorough with the methods used in solving the problems of this chapter, in turn aiding the students to prepare well for the Class 10 Maths CBSE Board examination.

The main topics covered in NCERT Class 1o Maths Chapter 15 include:

15.1 Introduction

15.2 Probability – A Theoretical Approach

List of Exercises in Class 10 Maths Chapter 15

Exercise 15.1 Solutions 25 Questions (1 MCQ, 21 Short Answer Questions, 2  Long Answer Questions, 1 Main Question with 4 Sub-questions) Exercise 15.2 Solutions 5 Questions (5 Short Answer Questions)

Practising these  NCERT Solutions for Class 10 is very important from the CBSE Board examination point of view. Probability is one of the chapters in Class 10 Maths that contains the concepts that are most likely to be used in daily life. The chapter deals with different topics related to probability, which include:

• The difference between experimental probability and theoretical probability
• Why the probability of a sure event (or certain event) is 1.
• How the probability of an impossible event is 0?
• Elementary events.
• Complementary events.
• Finding the probability of different events.

Key Features of NCERT Solutions for Class 10 Maths Chapter 15 – Probability

• In-depth knowledge of the chapter/problems is provided in easy language.
• It is based on the latest CBSE Syllabus (2023-24).
• Best study material to prepare for the Class 10 CBSE Board exam.
• All the doubts that might arise while solving the exercise questions will get cleared by going through these solutions.

Along with the NCERT Solutions, students can also practise the other textbook solutions to understand the concepts in a much better way.

• RD Sharma Solutions for Class 10 Maths Chapter 13 Probability

Disclaimer – 

Dropped Topics – 

15.1 Introduction Exercise 15.2 (Optional)

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 15

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NCERT Solutions for Class 10 Maths Chapter 15 Probability

NCERT Solutions Class 10 Maths Chapter 15 Probability are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed the theoretical approach of probability in details. CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

CBSE Class 10 Maths Chapter 15 Probability Solutions

Below we have given the answers to all the questions present in Probability in our NCERT Solutions for Class 10 Maths chapter 15. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.1

NCERT Solutions for Class 10 Maths Chapter 15 Exercise 15.2

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CBSE Class 10 Mathematics paper was easy but lengthy with questions from NCERT

The cbse class 10 mathematics board exam, held on march 11, 2024, was considered reasonably easy though a little lengthy. the paper was of easy to moderate level in difficulty with ncert questions..

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The CBSE board exam 2024 began for both Classes 10 and 12 on February 15 and the Class 10 Mathematics paper was held on March 11. Students said that the Math paper was of easy to moderate level in difficulty.

The paper was a little lengthy but not so much that students couldn't complete it. Several students noted that the long answer questions or case studies were a little tough.

Questions were given from the NCERT and Exemplar questions. Those who had prepared well found the paper moderately easy.

Ansh from Cambridge School says most of the questions were from the NCERT: "Around 30%-40% of the questions were based on the sample papers and PYQs only."

"It was an easy paper and wasnt too lenthy. The case study section was a little tricky," he adds.

Koustav from Bal Bharti found the Math Basic paper quite easy but lengthy. "The long-answer section was little tricky," he says.

"The paper was very easy. I completed with 30 minutes left. The section with the case studies was the trickiest," he adds.

Yadhvi from Bal Bharti Public School, who appeared for the Mathematics Standard paper, says she had a great exam.

"The paper was lengthy. One 5-marker question from Triangles chapter was very difficult. But the rest of the paper was very easy," she says.

Amit from Lord Mahavira School, who appeared for the Standard paper, says the Math paper was lengthy but easy.

"There were many NCERT Exemplar-based questions in the paper. There was nothing unexpected. Section D was a little tough," he says.

Sayona from Bal Bharti Public School found the Math Standard paper quite touch. "The MCQs were very lengthy. The 2- and 3-mark questions were lengthy as well. I found the sample paper easier than the paper itself."

WHAT TEACHERS HAVE TO SAY

The Class 10 Mathematics paper received positive feedback from both students and teachers, aligning well with CBSE sample papers. The timely completion by students and the ease of the case study-based questions suggest a well-balanced and accessible paper,” says Harmeet Kaur, HOD Mathematics at Silverline Prestige School, Ghaziabad

“The provision of appropriate choices in the paper further contributed to student satisfaction, reflecting a successful examination,” Kaur adds.

“The CBSE Class 10 Mathematics paper maintained an easy to moderate difficulty level, effectively testing students' application and problem-solving skills,” says Jayant Kumar, TGT Mathematics, Seth Anandram Jaipuria School, Lucknow.

“Students who had a good grasp of core concepts found the paper manageable, although some areas were noted to be tricky,” he says.

“Overall, the exam was aligned with the sample paper format and marking scheme, offering a mix of easy and engaging questions,” he adds.

"The CBSE Class 10 Math paper (Standard) was moderate and encompassed all the necessary skills outlined in the syllabus. Questions were based on the NCERT textbooks, ensuring familiarity for students. Notably, there were no unexpected or unfamiliar questions that could unsettle examinees," says Anjani Rai, Coordinator, VidyaGyan School, Bulandshahr (UP).

"Most questions followed a clear format, allowing students who had diligently practiced from NCERT and similar resources to attempt and answer them confidently. Students encountered difficulty with a case study question from the Real Number chapter, describing it as tricky,' she says.

"Compared to the previous year's paper, this one was considered a little difficult, with focus on conceptual understanding rather than numeracy skills," she adds.

"In one case study, in a question from areas related to circles, some misprint in numbers were observed. The geometry questions were directly sourced from the NCERT book. It is expected that sincere students would have completed the paper well within the allotted time," Rai says.

"The paper maintained a balanced mix of question types, consistent with the sample question papers provided by CBSE," she adds.

Regarding the CBSE Class 10 Basic Mathematics paper, Rai says it was well-balanced with easy and straightforward questions.

"None of the questions were excessively time-consuming, maintaining a fair distribution of question types as per CBSE's sample question papers. Overall, the paper was well-structured and aligned with the expectations of students preparing for the Basic Mathematics examination," Rai adds.

“The CBSE Mathematics paper (Standard and Basic) can be considered moderate. The questions were as per the CBSE pattern. The syllabus was entirely covered in Standard as well as Basic Mathematics," says Mamta Chauhan, HOD Mathematics, Billabong High International School, Malad.

"In both Mathematics (Standard and Basic) some questions in Section B,C and D can be considered anticipated. Some MCQs in section A were tricky, though easy. There weren’t lengthy calculations. Some MCQs were too easy," Chauhan says.

"Overall Section B was easy to solve. Section C contained regularly practiced questions, Trigonometry proof was moderate. The difficulty level of questions in Section D was moderate," she adds.

"Overall Section E was easy to solve. While in Mathematics (Basic), Section A had MCQs can be considered as easy to average, Section B had Overall Section B was easy to solve. Sections C and D were moderately difficult while Section E was easy to solve," Chauhan says.

“The CBSE Class 10 Math question paper was easier than the CBSE Class 10 sample question paper. The paper was not lengthy at all, Most students felt Section-E was not difficult but time-consuming,” says Ajit Pratap Singh, PGT Maths, KIIT World School, Gurugram.

“The questions on the exam ranged in difficulty from easy and done within 3 hours. Case study-based questions required students to use some extra effort and time but they were easy,” he adds.

“The CBSE class 10 Basic Math paper struck a moderate balance. It featured logical questions that required a comprehensive understanding of the chapters. One question was from an exercise that had been removed from the NCERT,” says Ajay Pal Singh, Principal of Bhai Parmanand Vidya Mandir.

“Section A presented an easy level. Section B was comparatively simpler, whereas Sections C and D posed an above-average challenge, demanding careful reading of the questions. Despite being sourced from NCERT, the questions proved to be of a challenging difficulty level,” Singh says.

“Section E attained a moderate difficulty level. The exam's design, aimed at assessing students' competency, was challenging. Overall, it encompassed various types and topics of Mathematics,” he adds.

"The overall difficulty of the question paper was moderate, offering a fair evaluation for all students. This allowed above-average students to showcase their skills while remaining accessible to those with a solid grasp of the concepts," says Benasir PA, TGT Mathematics, JAIN International Residential School, Bengaluru.

"It's commendable that the paper maintained consistency within question sets, ensuring fairness and preventing discrepancies in difficulty levels. However, Set 3 questions were slightly more challenging compared to Sets 1 and 2. Care should be taken to ensure that any increase in difficulty is reasonable and doesn't unfairly disadvantage students," the teacher says.

"The distribution of weightage between textbook-based questions and multiple-choice questions (MCQs) in the basic question paper seemed uneven. While it's important to include textbook-aligned questions to assess understanding, MCQs should be designed with basic-level students in mind, ensuring clarity and accessibility," the teacher adds.

"The mathematics paper for the Class 10 board examination was moderate, with most of the questions found to be from the syllabus. However, the paper had an inclusion of both theoretical and application-based problems," states BYJU'S instructors.

"Overall, the mathematics paper was found to be in line with expectations," they add.

HBSE Class 12 Maths Sample Paper 2024 with Answer Key PDF

The HBSE Class 12 Maths Sample Paper 2024 PDF is given here with answers. Download & solve Haryana Board 12th Maths sample paper to get good marks in board exam. Get HBSE 12th Maths Paper Pattern 2024

Table of Contents

The Haryana Board Class 12 Mathematics exam is being organized for both the commerce and science stream students on 27 March 2024. The Board of School Education , Haryana (BSEH) is conducting the Haryana 12th class Mathematics exam paper in offline mode (Pen and Paper Based exam). The examination is being conducted in single shift from 12:30 PM to 3:30 PM. For better preparation, the HBSE Class 12 Maths sample paper 2024 is provided on this page.

HBSE Class 12 Maths Sample Paper 2024

The HBSE Class 12 Maths sample paper 2024 has been developed in four different codes, just like the actual exam paper. The Haryana Board 12th mathematics board sample paper contains different types of questions ranging from objective multiple choice questions to subjective essay type questions. The HBSE class 12th sample paper, also known as model papers or practice papers, are formulated based on the standard syllabus prescribed by the Haryana Board.

The HBSE class 12th Maths sample paper have been designed by the expert faculty for mathematics who are also responsible for formulating the actual board paper. That is why, the questions present in the board exam paper and the sample paper seems identical. Therefore, every student who have taken mathematics as a subject in their 12th grade must solve the official sample paper to score high marks in the board exam.

HBSE 12th Maths Sample Paper 2024

The HBSE 12th Maths sample paper for the year 2024 contains exactly the same number of questions and has the same pattern as the board exam paper. By solving the sample papers, students can also get sense of the difficulty level of the final exam. Students can get a real sense of the nature of questions that the HBSE asks in the board exam by solving through the model question papers of Mathematics. This help students increase their board exam marks considerably.

Check: HBSE Class 12 Date Sheet 2024

Haryana Board 12th Class Maths Paper 2024 Overview

The Haryana Board 12th class Maths paper 2024 has the weightage of 100 marks in total. The theory exam paper is held for 80 marks while the rest of the 20 marks is awarded on the basis of internal assessment. There is no negative marking for the Mathematics exam paper. The official answer key for the Haryana  Board 12th Class Mathematics question paper 2024 is expected to be out in April 2024. The overview of the HBSE 12th Maths paper is given below.

Class 12 Maths Important Questions 2024 Haryana Board

The important questions for the Haryana Board 12th Mathematics final exam is given below. Students should solve these questions to score high grades in the exam.

Q) If the vertices of a triangle are (3, 8), (-4, 2) and (5, 1), then by using determinants its area is (A) 60/2 (B) 61/2 (C) 62/2 (D) none of the above

Q) The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0

Q) Assertion (A): Let L be the collection of all lines in a plane and R₁ be the relation on L as R₁ = {(L1, L2): L₁ ⊥ L₂) is a symmetric relation. Reason (R) : A relation R is said to be symmetric if (a, b) ∈ R => (b,a) ∈ R.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A) (C) Assertion (A) is true and Reason (R) is false. (D) Assertion (A) is false and Reason (R) is true

Q) Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that first ball is black and second is red. 

Q) Check the injectivity and surjectivity of the function f: Z→Z given by f(x) = |x|

Q) A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Q) Find the area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0, x = 2.

Q) How many number of arbitrary constants are there in the general solution of a differential equation of fourth order? 

Q) If A and B are any two events such that P(A) + P(B) – P(A∩B) = P(A), then P(A|B) ____.

Q) The projection vector of 𝑎⃗ = 𝑖̂+ 3𝑗̂+7𝑘̂ on 𝑏⃗⃗ = 7𝑖̂- 𝑗̂+ 8𝑘̂ is ________ .

HBSE Class 12th Maths Paper 2024 Pattern

The Haryana Board Class 12 Mathematics question paper 2024 comprises 38 questions in total. The marks for each questions varies from 1 to 5. The question paper also consists of case study based questions. The question paper contains questions in both Hindi and English medium. The entire question paper is divided into five sections. The HBSE class 12th maths paper 2024 pattern is given herein.

• The mathematics paper has 38 questions overall
• The Haryana 12th mathematics question paper has 5 sections from Section A to Section E.
• The first 20 questions of section A from Question Number 1 to Question Number 20 consists of MCQs and Reason-Assertions. Each question in section is worth 1 mark
• The section B of the question paper has five questions. Question Number 21 to 25 of this section are short answer type questions for 2 marks each
• The section C of the question contains six short answer type questions. Question numbers 26 to 31 of this section has the weightage of 3 marks each
• The section D of mathematics paper contain four long answer type questions. Question Number 32 to 35 of this section has the weightage of 5 marks each
• The section E of the HBSE 12th Maths question paper consists of questions based on Case Study. There are three questions in this section from question number 36 to 38. Every question of this section is worth 4 marks.

HBSE Sample Paper 2024 Class 12 PDF Download Mathematics

The Mathematics Class 12 Model paper 2024 pdf is given below for the Haryana Board final exam in March. The HBSE 12th mathematics model paper 2024 is released in four different codes. The pdfs for each of the codes is given below along with their answer key (in the subsequent section). Students must solve these questions to boost their confidence for the board exam.

Haryana Board Class 12 Maths Sample Paper 2024 Set A

Haryana Board Class 12 Maths Sample Paper 2024 Set B

Haryana Board Class 12 Maths Sample Paper 2024 Set C

Haryana Board Class 12 Maths Sample Paper 2024 Set D

HBSE Maths Sample Paper Answer Key 2024 Class 12 PDF Download

The Haryana Board Class 12 answer key for the Math sample paper 2024 given above is released in the pdf form. The answer key contains solutions for the questions present in the sample paper. Just like the sample papers, the answer key is also released for four different sets. Students must compare their answers to the answers provided in the answer key so that they can know the standard answer writing process. The answer key for all four sets of the sample paper given above is available for download below in the PDF format.

HBSE Class 12 Maths Sample Paper Answer Key 2024 Set A

HBSE Class 12 Maths Sample Paper Answer Key 2024 Set B

HBSE Class 12 Maths Sample Paper Answer Key 2024 Set C

HBSE Class 12 Maths Sample Paper Answer Key 2024 Set D

Haryana Board 12th Maths Question Paper 2024

The questions asked in the Haryana Board 12th Mathematics question paper on 27 March 2024 will be provided below for students in the PDF format. The question paper PDF provided here can be used as a reference for checking their answers with the help of the answer key. The question paper can be downloaded after the conclusion of the examination on the exam day itself. Students should keep checking this page to get the Haryana Board 12th maths question paper 2024.

HBSE 12th Maths Answer Key 2024

The HBSE 12th Maths answer key 2024 for the questions present in the March final question paper is provided here for students. The answers mentioned here is formulated by expert teachers, so that students can get the correct answer for all the questions. By going through the answer key, students will be able to assess their performance and calculate marks for the mathematics exam paper. The answer key will be made available after the conclusion of the exam.

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How can students get solutions for the HBSE 12th Maths sample paper 2024?

Students can download and view the HBSE 12th Maths sample paper 2024 in the above article in the PDF form.

The Haryana Board 12th Maths question paper contain how many sets?

The Haryana Board 12th Maths question paper contains four sets or codes from A to D.

Does the mathematics exam paper of the Haryana Board Class 12 contains multiple choice questions?

Yes, the Haryana Board Class 12 mathematics exam paper 2024 contains multiple choice questions in the first section of the question paper.

How many questions does the HBSE 12th Mathematics question paper contain?

The HBSE 12th mathematics question paper 2024 contains a total of 38 questions.

What is the maximum marks allocated for the Haryana Board 12th math exam paper?

The Haryana Board 12th Math board exam is held for 80 marks.

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