case study of probability class 10

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

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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:

(a) 18	(b) 20
(c) 22	(d) 30

Answer: (a) 18

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

(a) 18	(b) 25
(c) 24	(d) 36

Answer: (c) 24

(iii) The probability of distributing white chocolates is

(a) 11/27	(b)8/21
(c) 1/9	(d) 2/9

Answer: (d) 2/9

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

(a) 3/17	(b) 5/9
(c) 1/3	(d) 1/27

Answer: (b) 5/9

(v) The probability of distributing all the chocolates is

(a) 0	(b) 1
(c) 1/2	(d) 3/4

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 Expert

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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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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Case Study Class 10...

Case Study Class 10 Maths Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

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.


I	NUMBER SYSTEMS	06
II	ALGEBRA	20
III	COORDINATE GEOMETRY	06
IV	GEOMETRY	15
V	TRIGONOMETRY	12
V	MENSURATION	10
VI	STATISTICS & PROBABILITY	11

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

$case study of probability class 10$

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

$case study of probability class 10$

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

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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Case Study and Passage Based Questions for Class 10 Maths Chapter 15 Probability

Last modified on: 1 year ago
Reading Time: 3 Minutes

Case Study Questions:

Question 1:

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

(i) Find the probability of getting a king of red colour. (a) 1/26 (b) 1/13 (c) 1/52 (d) 1/4

(ii) Find the probability of getting a face card. (a) 1/26 (b) 1/13 (c) 2/13 (d) 3/13

(iii) Find the probability of getting a jack of hearts. (a) 1/26 (b) 1/52 (c) 3/52 (d) 3/26

(iv) Find the probability of getting a red face card. (a) 3/26 (b) 1/13 (c) 1/52 (d) 1/4

(v) Find the probability of getting a spade. (a) 1/26 (b) 1/13 (c) 1/52 (d) 1/4

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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.


Real Number
Polynomials ( )
Pair of Linear Equations in Two Variables (MCQ, Case-Based, Assertion & Reasoning)
Coordinate Geometry (MCQ, Case-Based, Assertion & Reasoning)
Triangles
Introduction to Trigonometry (MCQ, Case-Based, Assertion & Reasoning)
Areas Related to Circles (MCQ, Case-Based, Assertion & Reasoning)
Probability (MCQ, Case-Based, Assertion & Reasoning)
Quadratic Equation (MCQ)
Arithmetic Progression (MCQ)
Some Application of Trigonometry (MCQ)
Circles (MCQ)
Constructions (MCQ)
Surface Areas and Volumes (MCQ)
Statistics (MCQ)

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.

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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

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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Class 10 Maths

Important Questions Class 10 Maths Chapter 15 Probability

Important questions for Class 10 Maths Chapter 15 Probability are given here based on the weightage prescribed by CBSE. The questions are framed as per the revised CBSE 2022-2023 Syllabus and latest exam pattern. Students preparing for the CBSE class 10 board exams are advised to go through these Probability questions to get the full marks for the questions from this chapter.

Students can also refer to the solutions prepared by BYJU’S experts for all the chapters of Maths. Important questions for class 10 maths all chapters are also available to help the students in their examination preparation. The more students will practice, the more they can score marks in the exam. Students will also find Important Questions of class 10 Maths Chapter 15 Probability along with detailed solutions. So, if students could not solve any question, they can refer to the solution and understand it easily.

Also, check: Class 10 Maths Chapter 15 Probability MCQs

Important Questions & Answers For Class 10 Maths Chapter 15 Probability

Q. 1: Two dice are thrown at the same time. Find the probability of getting

(i) the same number on both dice.

(ii) different numbers on both dice.

Given that, Two dice are thrown at the same time.

So, the total number of possible outcomes n(S) = 6 2 = 36

(i) Getting the same number on both dice:

Let A be the event of getting the same number on both dice.

Possible outcomes are (1,1), (2,2), (3, 3), (4, 4), (5, 5) and (6, 6).

Number of possible outcomes = n(A) = 6

Hence, the required probability =P(A) = n(A)/n(S)

(ii) Getting a different number on both dice.

Let B be the event of getting a different number on both dice.

Number of possible outcomes n(B) = 36 – Number of possible outcomes for the same number on both dice

= 36 – 6 = 30

Hence, the required probability = P(B) = n(B)/n(S)

Q. 2: A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the

(i) yellow ball?

(ii) red ball?

(iii) blue ball?

Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them from the bag.

Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken out is blue’, and R be the event ‘the ball taken out is red’.

The number of possible outcomes = Number of balls in the bag = n(S) = 3.

(i) The number of outcomes favourable to the event Y = n(Y) = 1.

So, P(Y) = n(Y)/n(S) =1/3

Similarly, (ii) P(R) = 1/3

and (iii) P(B) = ⅓

Q.3: One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will

(i) be an ace,

(ii) not be an ace.

Well-shuffling ensures equally likely outcomes.

(i) Card drawn is an ace

There are 4 aces in a deck.

Let E be the event ‘the card is an ace’.

The number of outcomes favourable to E = n(E) = 4

The number of possible outcomes = Total number of cards = n(S) = 52

Therefore, P(E) = n(E)/n(S) = 4/52 = 1/13

(ii) Card drawn is not an ace

Let F be the event ‘card drawn is not an ace’.

The number of outcomes favourable to the event F = n(F) = 52 – 4 = 48

Therefore, P(F) = n(F)/n(S) = 48/52 = 12/13

Q.4: Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown, and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Number of total outcome = n(S) = 36

(i) Let E 1 be the event ‘getting sum 2’

Favourable outcomes for the event E 1 = {(1,1),(1,1)}

n(E 1 ) = 2

P(E1) = n(E1)/n(S) = 2/36 = 1/18

(ii) Let E 2 be the event ‘getting sum 3’

Favourable outcomes for the event E 2 = {(1,2),(1,2),(2,1),(2,1)}

n(E 2 ) = 4

P(E 2 ) = n(E 2 )/n(S) = 4/36 = 1/9

(iii) Let E 3 be the event ‘getting sum 4’

Favourable outcomes for the event E 3 = {(2,2)(2,2),(3,1),(3,1),(1,3),(1,3)}

n(E 3 ) = 6

P(E 3 ) = n(E 3 )/n(S) = 6/36 = 1/6

(iv) Let E 4 be the event ‘getting sum 5’

Favourable outcomes for the event E 4 = {(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}

n(E 4 ) = 6

P(E 4 ) = n(E 4 )/n(S) = 6/36 = 1/6

(v) Let E 5 be the event ‘getting sum 6’

Favourable outcomes for the event E 5 = {(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)}

n(E 5 ) = 6

P(E 5 ) = n(E 5 )/n(S) = 6/36 = 1/6

(vi) Let E 6 be the event ‘getting sum 7’

Favourable outcomes for the event E 6 = {(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}

n(E 6 ) = 6

P(E 6 ) = n(E 6 )/n(S) = 6/36 = 1/6

(vii) Let E 7 be the event ‘getting sum 8’

Favourable outcomes for the event E 7 = {(5,3),(5,3),(6,2),(6,2)}

n(E 7 ) = 4

P(E 7 ) = n(E 7 )/n(S) = 4/36 = 1/9

(viii) Let E 8 be the event ‘getting sum 9’

Favourable outcomes for the event E 8 = {(6,3),(6,3)}

n(E 8 ) = 2

P(E 8 ) = n(E 8 )/n(S) = 2/36 = 1/18

Q.5: A coin is tossed two times. Find the probability of getting at most one head.

When two coins are tossed, the total no of outcomes = 2 2 = 4

i.e. (H, H) (H, T), (T, H), (T, T)

H represents head

T represents the tail

We need at most one head, which means we need one head only otherwise no head.

Possible outcomes = (H, T), (T, H), (T, T)

Number of possible outcomes = 3

Hence, the required probability = ¾

Q.6: An integer is chosen between 0 and 100. What is the probability that it is

(i) divisible by 7?

(ii) not divisible by 7?

Number of integers between 0 and 100 = n(S) = 99

(i) Let E be the event ‘integer divisible by 7’

Favourable outcomes to the event E = 7, 14, 21,…., 98

Number of favourable outcomes = n(E) = 14

Probability = P(E) = n(E)/n(S) = 14/99

(ii) Let F be the event ‘integer not divisible by 7’

Number of favourable outcomes to the event F = 99 – Number of integers divisible by 7

= 99-14 = 85

Hence, the required probability = P(F) = n(F)/n(S) = 85/99

Q. 7: 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)

P(not E) = 1 – 0.05

∴ P(not E) = 0.95

Q. 8: 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 is 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(E) = (Number of favourable outcomes) / (Total number of outcomes)

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

Q. 9: 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?

Outcomes 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 outcomes = 6 × 6 = 36

(i) Method 1:

Consider the following events.

A = 5 comes in the first throw,

B = 5 comes in second throw

P(A) = 6/36,

P(B) = 6/36 and

P(not B) = 5/6

So, P(notA) = 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

Q.10: A die is thrown once. What is the probability of getting a number less than 3?

Given that a die is thrown once.

Total number of outcomes = n(S) = 6

i.e. S = {1, 2, 3, 4, 5, 6}

Let E be the event of getting a number less than 3.

n(E) = Number of outcomes favourable to the event E = 2

Since E = {1, 2}

Hence, the required probability = P(E) = n(E)/n(S)

Q.11: If the probability of winning a game is 0.07, what is the probability of losing it?

Given that the probability of winning a game = 0.07

We know that the events of winning a game and losing the game are complementary events.

Thus, P(winning a game) + P(losing the game) = 1

So, P(losing the game) = 1 – 0.07 = 0.93

Q.12: The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 1/5. The probability of selecting a black marble at random from the same jar is 1/4. If the jar contains 11 green marbles, find the total number of marbles in the jar.

Given that,

P(selecting a blue marble) = 1/5

P(selecting a black marble) = 1/4

We know that the sum of all probabilities of events associated with a random experiment is equal to 1.

So, P(selecting a blue marble) + P(selecting a black marble) + P(selecting a green marble) = 1

(1/5) + (1/4) + P(selecting a green marble) = 1

P(selecting a green marble) = 1 – (1/4) – (1/5)

= (20 – 5 – 4)/20

P(selecting a green marble) = Number of green marbles/Total number of marbles

11/20 = 11/Total number of marbles {since the number of green marbles in the jar = 11}

Therefore, the total number of marbles = 20

Q.13: The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?

Total number of apples in the heap = n(S) = 900

Let E be the event of selecting a rotten apple from the heap.

Number of outcomes favourable to E = n(E)

P(E) = n(E)/n(S)

0.18 = n(E)/900

⇒ n(E) = 900 × 0.18

⇒ n(E) = 162

Therefore, the number of rotten apples in the heap = 162

Q.14: A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

Number of white balls = 15

Let x be the number of black balls.

Total number of balls in the bag = 15 + x

Also, the probability of drawing a black ball from the bag is thrice that of drawing a white ball.

⇒ x/(15 + x) = 3[15/(15 + x)]

⇒ x = 3 × 15 = 45

Hence, the number of black balls in the bag = 45.

Practice Questions for Class 10 Maths Chapter 15 Probability

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, find the number of blue balls in the bag.
A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What are the odds against his winning this bet?
A bag contains 12 balls out of which x are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball? (ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double that in case (i). Find x.
Five male and three female candidates are available for selection as a manager in a company. Find the probability that a male candidate is selected.
A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number that is a perfect square.
In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11?

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28. Case study: Probability

Exercise condition:.

CBSE Class 10 QnA

CBSE Class 10 Maths Chapter 15 Important Questions and Answers: Probability

Cbse class 10 maths chapter 15 important questions and answers: in this article we will go through important questions from unit 7 chapter 15 probability. the questions range from mcqs to 4 mark questions. the answers are also included and can be accessed from thre pdf attached towards the end..

CBSE Class 10 Maths Chapter 15 Important Questions and Answers: Probability is the second chapter in the 7th unit of CBSE Class 10 Mathematics syllabus. This unit carries a weightage of 11 marks. In this article, we are providing Multiple Choice Questions, 1 mark objective type questions, short answer questions of 2 and 3 marks, long answer questions and some case study questions from Chapter 15 Probability.

Also check carefully the latest resources provided by CBSE for Class 10 Mathematics 2023

CBSE Class 10 Maths Chapter 15 Important Questions

Multiple choice questions.

1 The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs in the lot is

2 A letter of the English alphabet is chosen at random. The probability that the chosen letter is a consonant is

(D) None of these

3 If P(E ) = 0.05 , then the probability of P ( not E) is

4 The probability that a non – leap year has 53 Sundays is

5 In a single throw of a die , the probability of getting a multiple of 3 is

OBJECTIVE TYPE QUESTIONS

1 A box contains cards numbered 6 to 50. A card is drawn at random from the box. Find the probability that the card drawn has a number which is a perfect square.

2 A card is drawn from a pack of 52 cards. Find the probability of getting a king of red colour.

3 A die is thrown twice. Find the probability of getting a sum less than 8

4 A card is drawn from a pack of 52 cards. Find the probability that the card drawn is not a face card

5 A number is selected from the first 50 natural numbers. What is the probability that it is a multiple of 3 or 5?

SHORT ANSWER QUESTIONS (2 marks)

1 A letter is chosen at random from the letters of the word “ASSASSINATION”, then the probability that the letter chosen is a vowel is in the form of 6/ (2x+1), if so find the value of x

2 A coin & a die are tossed simultaneously. Find the probability that a tail & a prime number turns up.

3 A letter of the English alphabet is chosen at random. Determine the probability that the chosen letter is a vowel

4 A box contains 20 cards numbered from 1 to 20 . A card drawn at random from the box. Find the probability that the card drawn at random is divisible by 2 or 3.

5 All cards of ace , jack and queen are removed from a deck of playing cards. One card is drawn at random from the remaining cards. Find the probability that the card drawn is

(i) a face card

SHORT ANSWER QUESTIONS (3 marks)

1 The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 1/5. The probability of selecting a black marble at random from the same jar is 1/4. If the jar contains 11 green marbles, find the total number of marbles in the jar.

2 A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is:

(i) acceptable to. Varnika?

(ii) acceptable to the trader?

3 A bag contains 24 balls of which x are red 2x are white and 3x are blue. A ball is selected at random. What is the probability that

(i) it is red

(ii) it is blue

(iii) neither red nor blue

4 A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

5 All the three face cards of spades are removed from a well- shuffled pack of 52 cards. A card is drawn at random from the remaining pack. Find the probability of getting

(i) a black face cards

(ii) a queen

LONG ANSWER QUESTION (4 marks)

1 Cards marked with the number 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from the box. Find the probability that the number on the card is:

(i) An even number

(ii) A number less than 14

(iii) A number is perfect square

(iv) A prime number less than 20

2 Cards bearing numbers 3, 5... 35 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing (a) a prime number less than 15 (b) a number divisible by 3 and 5

3 Three coins are tossed simultaneously. Find the probability of getting

(i) Exactly 2 heads (ii) at least 1 head (iii) at most 2 tails (iv) exactly 3 heads

4 All the black Ace cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting

i) a Ace card

ii) a red card

iii) a black card

5 The Ace, number 10 and jack of clubs are removed from a deck of 52 playing cards and remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of

CASE STUDY QUESTIONS

Find the probability of getting no heads
Find the probability of getting one tail

2 Five cards – ten, jack, queen, king, and an ace of diamonds are shuffled face downwards. One card is picked at random.

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

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CBSE Class 10 Maths Preparation Tips to Score 95+ Marks in CBSE Class 10 Maths Board Exam 2023

As the students preparing to appear for CBSE class 10 board examinations must be aware of the updated and rationalised syllabus, it is very important for the students to practise from the resources that have been prepared according to the changes in the latest syllabus.

These important questions have been prepared according to the changes based on the revised syllabus .

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Case Based Questions Test: Probability - Class 10 MCQ

10 questions mcq test - case based questions test: probability, read the following text and answer the following questions on the basis of the same: on a weekend rani was playing cards with her family. the deck has 52 cards.if her brother drew one card. q. find the probability of getting a king of red colour .

No. of cards of a king of red colour = 2

Total no. of cards = 52

Probability of getting a king of red colour

= No. of king of red colour/Total number of cards

= 2/52 = 1/26

Read the following text and answer the following questions on the basis of the same: On a weekend Rani was playing cards with her family. The deck has 52 cards.If her brother drew one card. Find the probability of getting a face card.

Face cards are King, Queen and Jack

Each type has 4 cards

Total number of face cards = 3 x 4 = 12

P (getting a face card) = Number of face cards/Total number of card

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Read the following text and answer the following questions on the basis of the same: On a weekend Rani was playing cards with her family. The deck has 52 cards.If her brother drew one card. Find the probability of getting a jack of hearts.

We have 4 Jack cards

Total number of Jack of hearts = 1

P (getting a face card) = Number of jack of heart cards/Total number of cards

Read the following text and answer the following questions on the basis of the same:

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

Find the probability of getting a red face card.

No. of red face card = 6

Total no of cards = 52

Probability of getting a face card

= No. of red face cards/Total no. of cards

= 6/52 = 3/26

Find the probability of getting a spade.

No. of spade card = 13

Probability of getting a spade card

= No. of spade cards/Total no. of cards

= 13/52 = 1/4

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

Ravi got the 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?

= (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)

Total outcomes = 36

No. of outcomes when the sum is 8 = 5

Probability = 5/36

Rahul got the 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?

Probability = 0/35 = 0

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 ?

probability = 36/36 = 1

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 ?

Number of outcomes where sum is 7 = 6

Probability that sum of two numbers is 7

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 ?

Number of outcomes where sum is greater than 8

= 4 + 3 + 2 + 1

Probability that sum of two numbers is greater than 8

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Important Questions for Case Based Questions Test: Probability

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CBSE 10th Standard Maths Subject Probability Case Study Questions With Solution 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

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10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(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) a monkey

(iii) a teddy bear

(iv) not a monkey

(v) not a pokemon

(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) The probability of getting exactly 1 head is

(iii) The probability of getting exactly 3 tails is

(iv) The probability of getting atmost 3 heads is

(v) The probability of getting atleast two heads is

(ii) not of yellow colour

(iii) of green colour

(iv) of yellow colour

(v) not of blue colour

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

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Probability Class 10 Extra Questions Maths Chapter 15 Solutions

January 7, 2023 by Bhagya

Extra Questions for Class 10 Maths Probability with Answers

Extra Questions for Class 10 Maths Chapter 15 Probability. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download Class 10 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Probability Class 10 Extra Questions Very Short Answer Type

Question 1. A number is chosen at random from the numbers -3, -2, -1,0,1,2,3. What will be the probability that square of this number is less than or equal to 1 ? [CBSE Delhi 2017] Answer: Here total outcomes = 7, favourable outcomes are -1, 0, 1 Reqd. probability = $\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}=\frac{3}{7}$

Question 2. The probability that it will rain today is 0.75. /£) What is the probability that it will not rain today? Answer: P(rain today) = 0.75 P(not rain today) = 1 – P(rain today) = 1 – 0.75 = 0.25

Question 3. The probability of getting a bad egg in a lot of 500 eggs is 0.028. Find the number of good eggs in the lot. Answer: Total eggs = 500 Let no. of bad eggs = x P(bad eggs) = $\frac{x}{500}$ 0.028 = $\frac{x}{500}$ ⇒ x = 14 Number of good eggs = 500 – 14 = 486 Hence, number of good eggs is 486.

Question 4. A card is selected at random from a deck of 52 cards. Find the probability that the selected card is red face card. Answer: Total cards = 52 Red face cards = 6 P(a red face card) = $\frac{6}{52}=\frac{3}{26}$

Question 5. A card is drawn at random from a well shuffled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen. [CBSE Outside Delhi 2016] Answer: Let E be the event of getting neither red card nor queen No. of red cards = 26 (including 2 queens) Remaining (black) queens = 2 Neither red nor queen = 52 – (26 + 2) = 52 – 28 = 24 24 6 P(E) = $\frac{24}{52}=\frac{6}{13}$

Question 6. Cards marked with number 3,4,5……………. ,50 are placed in a box and mixed thoroughly. A card is drawn at random from the box. Find the probability that the selected card bears a perfect square number. [CBSE 2016] Answer: It is given that the box contains cards marked with numbers 3,4,5, ………… , 50. ∴ Total number of outcomes 48 There are six perfect squares, i.e., 4, 9, 16, 25, 36 and 49. ∴ Number of favourable outcomes = 6 Probability that a card drawn at random bears a perfect square = $\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{6}{48}=\frac{1}{8}$

Question 7. What is the probability that a doublet occurs on throwing two dice? Answer: Total outcomes = 36 Doublets = 6 {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)} P(a doublets) = $\frac{6}{36}=\frac{1}{6}$

Question 8. In a throw of a coin, find the probability of getting a head. Answer: Here, S = {H, T} and E = {H} ∴ P(E) = $\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{1}{2}$

Question 9. Two unbiased coins are tossed. What is the probability of getting at most one head? Answer: Here, S = {HH, HT, TH, TT). Let E = event of getting at most one head. ∴ E = {TT, HT, TH}. ∴ P(E) = $\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{3}{4}$

Question 10. An unbiased die is tossed. Find the probability of getting a multiple of 3? Answer: Here, S = {1, 2, 3, 4, 5, 6} Then, E = {3, 6} ∴ P(E) = $\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{2}{6}=\frac{1}{3}$

Question 11. In a well shuffled pack of cards, a card is © drawn at random. Find the probability of getting a black queen. Answer: Total cards = 52 Total no. of black queens = 2 ∴ P(a black queen) = $\frac{2}{52}=\frac{1}{26}$

Question 12. Two coins are tossed simultaneously. What is the probability of getting exactly one ‘. head? Answer: Here, sample space S = {HH, HT, TH, TT} ∴ n(S) = 4 and E = {HT, TH} ∴ n(E) = 2 ∴ Reqd. Prob. = P(E) = $\frac{2}{4}=\frac{1}{2}$

Question 13. In a lottery there are 13 prizes and 117 blanks. What is the probability of not winning a prize? Answer: No. of ways of not winning a prize = No. of blanks = 117 Total cases = 13 + 117 = 130 P(not winning) = $\frac{117}{130}=\frac{9}{10}$

Question 14. What is the probability of getting “a black¬face card’ from a well shuffled deck of 52 playing cards? Answer: There are 6 black face cards. (2 red kings, 2 red queens and 2 jacks) Total cards = 52 ∴ Required Probability = $\frac{6}{52}=\frac{3}{26}$

Question 15. Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken out at random from the box. Find the probability that the bulb drawn is non-defective. Answer: No. of non-defective bulbs = 400 – 15 = 385 Total bulbs = 400 ∴ P(a non-defective bulb) = $\frac{385}{400}=\frac{77}{80}$

Question 16. If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Give reasons. Answer: No, the outcomes ‘head’ and ‘tail’ are equally likely every time regardless of what you get in a few tosses.

Probability Class 10 Extra Questions Short Answer Type-1

Question 1. Jayanti throws a pair of dice and records the product of the numbers appearing on the dice. Pihu throws 1 dice and records the square of the number that appears on it. Who has the better chance of getting 36? Justify Or An integer is chosen between 70 and 100. Find the probability that it is (i) a prime number (ii) divisible by 7. [CBSE SQP 2019-20 (Standard)] Answer: For Jayanti: n (S) = 36 Favourable outcome for getting product 36 is 6 on both dice which is (6,6) just one ∴ P (Jayanti getting 36) = $\frac{1}{6}$

For Pihu: Total number of outcomes = 6 Favourable outcome is 6 i.e., 1. Probability (getting the number 36) = $\frac{1}{6}$ ∵ $\frac{1}{6}>\frac{1}{36}$ Pihu has the better chance. Or Total number of possible outcomes = Number of integers between 70 and 100 = 29 (i) Prime numbers between 70 and 100 are 71, 73, 79, 83, 89, 97 i.e., 6 in all. ∴ P (Prime number) = $\frac{6}{29}$

(ii) Numbers lying between 70 and 100 and divisible by 7 are 77,84, 91,98 i.e., 4 in all. ∴ P(Divisible by 7) = $\frac{4}{29}$

Question 2. A game consists of tossing a coin 3 times and noting the outcome each time. If getting the same result in all the tosses is a success, find the probability of lossing the game. [CBSE 2019] Answer: Here, sample space S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Total number of outcomes n(S) = 8 If E be the given event of success then E = {HHH, TTT} Total number of favourable outcomes, n(E) = 2 P (Losing the game) = 1 – P (Success) = 1 – $\frac{n(\mathrm{E})}{n(\mathrm{~S})}$ = 1 – $\frac{2}{8}$ = 1 – $\frac{1}{4}=\frac{3}{4}$

Question 3. A die is thrown once. Find the probability of getting a number which (i) is a prime number (ii) lies between 2 and 6. [CBSE 2019] Answer: (i) Here, total number of outcomes, n (S) = 6 Let E be the event of getting prime number, then E = {2,3,5} Total number of favourable outcomes, n(E) = 3

(ii) Let F be the event of getting a number between 2 and 6. Then, F = {3, 4, 5} ⇒ Total number of favourable outcomes n(F) = 3 ∴ P(F) = $\frac{n(\mathrm{~F})}{n(\mathrm{~S})}=\frac{3}{6}=\frac{1}{2}$

Question 4. A pair of dice is thrown once. Find the probability of getting (i) even number on each dice (») a total of 9. [CBSE 2019 (c)] Answer: Here, n(S) = 36 Let E 1 be event of getting even number on each dice. ∴ E 1 ={(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)} ∴ n(E 1 ) = 9 ∴ P(even number on each die) P(E 1 ) = $\frac{n\left(\mathrm{E}_1\right)}{n(\mathrm{~S})}$ = $\frac{9}{36}=\frac{1}{4}$

(ii) Let E 2 be the event of getting a total of 9. ∴ E 2 = {(3,6), (4,5), (5,4), (6,3)}, ∴ P(E 2 ) = 4 ∴ P(a total of 9) = P(E 2 ) = $\frac{n\left(\mathrm{E}_2\right)}{n(\mathrm{~S})}=\frac{4}{36}=\frac{1}{9}$

Question 5. A bag contains some balls of which x are white, 2x are black and 3x are red. A ball is selected at random. What is the probability that it is (i) not red (it) white? [CBSE 2019 (C)] Answer: Total no. of balls = x + 2x + 3x = 6x (i) P(not red) = 1 – P(red) = 1 – $\frac{\text { No. of red balls }}{\text { Total no. of balls }}$ = 1 – $\frac{3 x}{6 x}$ = 1 – $\frac{1}{2}=\frac{1}{2}$

(ii) P(white) = $\frac{\text { No. of white balls }}{\text { Total no. of balls }}$ = $\frac{x}{6 x}$ = $\frac{1}{6}$

Question 6. In a simultaneous throw of a pair of dice, find the probability of getting a total more than 7. Answer: Here, n (S) = (6 x 6) = 36. Let E = Event of getting a total more than 7 = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} ∴ P(E) = $\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{15}{36}=\frac{5}{12}$

Question 7. A bag contains 6 white and 4 black balls. Two balls are drawn at random. Find the probability that they are of the same colour. Answer: Let S be the sample space. Then n (S) = Number of ways of drawing 2 balls out of (6 + 4) = 10 C 2 = $\frac{10 !}{2 !(10-2) !}=\frac{10 \times 9 \times 8 !}{2 \times 8 !}$ = 45

Let E = Event of getting both balls of the same colour. Then n(E) = Number of ways of drawing 2 balls out of 4 = ( 6 C 2 + 4 C 2 ) = $\frac{(6 \times 5)}{(2 \times 1)}+\frac{(4 \times 3)}{(2 \times 1)}$ = (15 + 6) = 21 ∴ P(E) = $\frac{n(\mathrm{E})}{n(\mathrm{~S})}=\frac{21}{45}=\frac{7}{15}$

Question 8. Two friends were born in the year 2000. What is the probability that they have the same birthday? Answer: There are 366 days in 2000 (leap year) Total no. of ways in which both friends can have their birthday = 366 × 366 No. of ways in which they have different birthdays = 366 × 365 ∴ P(different birthday) = $\frac{366 \times 365}{366 \times 366}=\frac{365}{366}$ ⇒ P(same birthday) = 1 – P(different birthdays) = 1 – $\frac{365}{366}=\frac{1}{366}$

Question 9. 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. Answer: Total possible outcomes in throwing a single die are 1, 2, 3,4, 5, 6; which are six in number, (i) Outcomes favouring a prime number among these six outcomes are 2,3,5 So, number of outcomes favouring a prime number = 3 ∴ p(a prime number) = $\frac{\text { Number of outcomes favouring a prime number }}{\text { Total number of possible outcomes }}$ = $\frac{3}{6}=\frac{1}{2}$

(ii) Since numbers lying between 2 and 6 are 3,4,5 So, number of outcomes favouring an outcome between 2 and 6 = 3 ∴ P(a number lying between 2 and 6) = $\frac{3}{6}=\frac{1}{2}$

(iii) Since odd numbers on a die are 1,3,5 So, number of outcomes favouring an odd number = 3 ∴ P(an odd number) = $\frac{3}{6}=\frac{1}{2}$

Question 10. A box contains cards numbered 11 to 123. A card is drawn at random from the box. Find the probability that the number on the drawn card is: (i) a square number (ii) a multiple of 7 [CBSE Sample Paper-2017] Answer: Here n(s) = (123 – 10) = 113 Square numbers are E 1 = {16, 25, 36, 49, 64, 81, 100, 121} n(E 1 ) = 8 ∴ (i) p( a square number) = $\frac{n\left(\mathrm{E}_{1)}\right.}{n(\mathrm{~s})}=\frac{18}{113}$ Multiple of 7 are E2 = {14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119} n(E 2 ) = 16 (ii) P(a multiple of 7) = $\frac{n\left(\mathrm{E}_2\right)}{n(s)}=\frac{16}{113}$

Question 11. Cards, marked with numbers 5 to 50, are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is (i) a prime number less than 10. (ii) a number which is a perfect square. Answer: S = {5, 6, 7,…, 50} ⇒ n(S) = 46 (i) Prime number less than 10 are 5,7 i.e. two ∴ P(prime number less than 10) = $\frac{2}{46}=\frac{1}{23}$

(ii) Perfect square numbers are 9, 16, 25, 36, 49, i.e. five ∴ P(a perfect square) = $\frac{5}{46}$

Question 12. A bag contains 14 balls of which x are white. If 6 more white balls are added to the bag, the probability of drawing a white ball is $\frac{1}{2}$. Find the value of x. Answer: Total no. of balls = 14, out of which x are white balls, 6 more white balls are added then white balls = x + 6 P(a white ball) = $\frac{x+6}{20}=\frac{1}{2}$ ∴ x = 4

Question 14. Two players, Sangeeta and Reshma play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning? Answer: We denote the event ‘Reshma wins the tennis match’ by R. We are given that P(S) = 0.62. Now, the event S and R are complementary because at a time only one can happen. i.e. ‘not S’ means the event R and ‘not R’ means the event S. ⇒ P(S) + P(R) = 1 ⇒ 0.62 + P(R) = 1 ⇒ P(R) = 1 – 0.62 = 0.38

Question 15. Two different dice are tossed together. Find the probability: (i) of getting a doublet (ii) of getting a sum 10, of the numbers on the two dice. [CBSE 2018] Answer: Total number of possible outcomes = 36 (i) Doublets are (1,1) (2, 2) (3, 3) (4,4) (5,5) (6,6) Total number of doublets = 6 ∴ Prob (getting a doublet) = $\frac{6}{36}=\frac{1}{6}$

(ii) Favourable outcomes for getting sum 10 are (4, 6) (5, 5) (6, 4) i.e., 3 ∴ Prob (getting a sum) 10 = $\frac{3}{36}=\frac{1}{12}$

Question 16. An integer is chosen at random between 1 and 100. Find the probability that it is : (i) divisible by 8 (ii) not divisible by 8. [CBSE 2018] Answer: Total number of outcomes = 98 (i) Favourable outcomes are 8,16,24,…, 96 i.e., 12 ∴ Prob (integer is divisible by 8) = $\frac{12}{98}=\frac{6}{49}$

(ii) Prob (integer is not divisible by 8) = 1 – $\frac{6}{49}$ = $\frac{43}{49}$

Probability Class 10 Extra Questions Short Answer Type-2

Question 1. Two different dice are thrown together. Find the probability that the number obtained. (i) have a sum less than 7 (ii) have a product less than 16 (iii) is a doublet of odd numbers. [CBSE 2017] Answer: Total number of outcomes = 36 (i) Favourable outcomes are (1,1) (1,2) (1,3) (1,4) (1,5) (2,1) (2,2) (2,3) (2,4) (3,1) (3,2) (3,3) (4,1) (4,2) (5,1) no. of favourable outcomes = 15 ∴ P(sum lesfe than 7) = $\frac{15}{36}=\frac{5}{12}$

(ii) Favourable outcomes 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) (4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (6,1) (6.2) no. of favourable outcomes = 25 ∴ P(product less than 16) = $\frac{25}{36}$

(iii) Favourable outcomes are ∴ P(doublet of odd number) = $\frac{3}{36}=\frac{1}{12}$

Question 2. In a single throw of a pair of different dice, what is the probability of getting (?) a prime number on each dice? (it) a total of 9 or 11? [CBSE 2016] Answer: Total number of outcomes on throwing a pair of dice = 6 x 6 = 36 (i) Let E be the event of getting a prime number on each dice. ∴ Favourable outcomes = {(2, 2), (2, 3), (2, 5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5)} Number of favourable outcomes = 9 Now, P(E) = $\frac{9}{36}=\frac{1}{4}$ Thus, the probability of getting a prime number 1 on each dice is $\frac{1}{4}$

(ii) Let F be the event of getting a total of 9 or 11. ∴ Favourable outcomes = {(3, 6), (4, 5), (5, 4), (6,3), (5,6), (6,5)}. Number of favourable outcomes = 6 Now, P(F) = $\frac{6}{36}=\frac{1}{6}$ Thus, the probability of getting a total of 9 or 11 is $\frac{1}{6}$

Question 3. Three different coins are tossed together. Find the probability of getting (i) exactly two heads (it) at least two heads (iii) at least two tails. [CBSE Outside Delhi 2016] Answer: Here sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} ⇒ n(s) = 8

(i) Now, P (exactly two heads) = P ({TTH, THT, HTT, TTT}) = $\frac{3}{8}$

(ii) P(at least two heads) = P({HHT, HTH, THH, HHH}) = $\frac{4}{8}=\frac{1}{2}$

(iii) P(at least two tails) = P({TTH, THT, HTT, TTT}) = $\frac{4}{8}$

Probability Class 10 Extra Questions Maths Chapter 15 5

(ii) P(one boy and one girl) = P(B 1 G 1 , B 1 G2, BA, B2G2, B2G3) = $\frac{6}{10}=\frac{3}{5}$

(iii) P (at least one boy is selected) = P(one boy or both boys) = P(B 1 G 1 , B 1 G 2 , B 2 G 3 , B 2 G 1 , B 2 G 2 , B 2 G 3 , B 3 B 2 ) = $\frac{7}{10}$

Question 5. A card is drawn at random from a well shuffled deck of playing cards. Find the 1 probability of drawing a (>) face card (ii) card which is neiher a king nor a red card. Answer: (i) There are 12 face cards ∴ P(face card) = $\frac{12}{52}=\frac{3}{13}$

(ii) P(neither a king nor a red card) = 1 – P (red and king) = 1 – $\frac{28}{52}$ [as there are 26 red cards including 2 red kings and 2 black kings] = 1 – $\frac{7}{13}=\frac{6}{13}$

Question 6. A number ‘a’ is selected from the number [2,3] and then a second number V is selected from [1, 3, 7]. What is the probability that the product ab of two numbers is greater than 9? Answer: Number ‘a’ can be selected in two ways and corresponding to each such way there are three ways of selecting number ‘b’. Threrefore, two numbers can be selected in 2×3 = 6 ways as listed below: (2,1), (2, 3), (2, 7), (3,1), (3, 3), (3, 7) ∴ n(S) = 6 The product ab is greater than 9 if a and b are chosen in any one of following ways (2, 7), (3, 7) ∴ Favourable cases = 2 ∴ Required probability = $\frac{2}{6}=\frac{1}{3}$

Question 7. The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is; (i) a heart (ii) a king. Answer: After removal of three cards (king of clubs, queen of clubs and jack of clubs) we are left with 52 – 3 = 49 cards. Well shuffling of these left over 49 cards ensure equally likely outcomes i.e., each of the 49 cards are equally likely to be drawn. ∴ Total number of possible outcomes = 49

(i) Let ‘E 1 ‘ be the event of getting a heart. Since all the 13 cards of heart are intact among 49 cards. ∴ Number of outcomes favourable to E 1 = 13 ⇒ P(a heart) = P(E 1 )

(ii) Let ‘E 2 ‘ be the event of getting a king. Since there are 4 kings out of which king of clubs is removed. ∴ Number of king among left over 49 cards = 3 ⇒ Number of outcomes favourable to E2 = 3 ∴ P(a king) = P(E 3 ) = $=\frac{\text { Number of outcomes favourable to the event } \mathrm{E}_2}{\text { Total number of possible outcomes }}$ = $\frac{3}{49}$

Probability Class 10 Extra Questions Long Answer Type 1

Question 1. Peter throws two different dice together and find the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25. [CBSE Delhi 2017] Answer: For Peter, Total number of outcomes = 36 Favourable outcome is (5,5) i.e. 1 out of 36 ∴ P(Peter getting the number 25) = $\frac{1}{6}$ For Rina, total number of outcomes = 6 Favourable outcome is 5. i.e. 1 out of 6 ∴ P(Rina getting the number 25) = $\frac{1}{6}$ Rina has the better chance.

Probability Class 10 Extra Questions Maths Chapter 15 6

(ii) There are five numbers greater than 3, that is 4, 5, 6, 7 and 8. Probability that the arrow will point at a number greater than 3 is given by P (Arrow point at a number greater than 3) = $\frac{5}{8}$

(iii) All the numbers are less than 9 Probability that the arrow will point at a number less than 9 is given by P (Arrow point at a number less than 9) = $\frac{8}{8}$ = 1

Question 3. A number x is selected at random from the numbers 1, 2, 3 and 4. Another number y is selected at random from the numbers 1, 4, 9 and 16. Find the probability that product of x and y is less than 16. [CBSE Outside Delhi 2016] Answer: S = {(1,1) (1, 4) (1, 9) (1,16) (2,1) (2, 4) (2, 9), (2,16), (3,1) (3,4), (3,9), (3,16), (4,1), (4,4), (4,9),(4,16)} ⇒ n(S) = 16 Let E 1 be event that xy < 16 ∴ E 1 = {(1,1); (1,4); (1,9); (2,1); (2,4), (3,1), (3,4), (4,1)} ⇒ n(E 1 ) = 8

Probability Class 10 Extra Questions Maths Chapter 15 7

(ii) Possible outcomes are 1, 2, 3, ………….. ,80 Total number of favourable outcomes = 80 Now favourable outcomes are “perfect square”. Such outcomes are 1, 4, 9, 16, 25, 36, 49, 64. Number of favourable outcomes = 8 Hence, P(a perfect square number) = $\frac{8}{80}=\frac{1}{10}$

Question 5. A box contains 20 balls bearing numbers 1, 2, 3, 4, 5, ………, 18, 19, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is: (i) an odd number? (ii) divisible by 2 or 3? (iii) prime number? (iv) not divisible by 10? Answer: Since balls are marked as 1, 2, 3, 4, 5, 6, ………. , 17, 18,19, 20. ∴ Total number of balls = 20, random drawing of balls ensures equally likely outcomes. ⇒ Total number of possible outcomes = 20

(i) Let E denotes the event of drawing an odd number. The number of outcomes favourable to the event E = 10 [∵ 1,3, 5,7,9,11,13,15,17,19 are odd numbers from 1 to 20] P(an odd number) = P(E) = $\frac{\text { Number of outcomes favourable to the event } \mathrm{E}}{\text { Total number of possible outcomes }}$ = $\frac{10}{20}=\frac{1}{2}$

(ii) Let F denotes the event of drawing a ball divisible by 2 or 3. Since, there are 13 balls: 2,3,4,6,8,9,10,12,14, 15,16,18,20 which are divisible by 2 or 3. Number of outcomes favourable to the event F = 13. P(a number divisible by 2 or 3) = P(F) = $\frac{\text { Number of outcomes favourable to the event } F}{\text { Total number of possible outcomes }}$ = $\frac{13}{20}$

(iii) Let ‘G’ denotes the event of drawing a ball which is a prime number. There are 8 balls: 2, 3, 5, 7,11,13,17,19 which are prime. Number of outcomes favourable to the event G = 8 P(a prime number) = P(G) = $\frac{\text { Number of outcomes favourable to the event } G}{\text { Total number of possible outcomes }}$ = $\frac{8}{20}=\frac{2}{5}$

(iv) Let ‘H’ denote the event of drawing a ball divisible by 10 There are 2 balls: 10,20, which are divisible by 10. Number of outcomes favourable to the event H = 2. P(a number divisible by 10) = P(H) = $\frac{\text { Number of outcomes favourable to the event } \mathrm{H}}{\text { Total number of possible outcomes }}$ = $\frac{2}{20}=\frac{1}{10}$

P(number not divisible by 10) = 1 – P(a number divisible by 10) [VP(E) = 1-P(E)1 = 1 – P(H) = 1 – $\frac{1}{10}=\frac{9}{10}$

Probability Class 10 Extra Questions HOTS

Question 1. If 65% of the population of a town have black eyes, 25% have brown eyes and the remaining have blue eyes. What is the probability that a person selected at random has (i) Blue eyes (ii) Brown or black eyes (Hi) Blue or black eyes (iv) Neither blue nor brown eyes. Answer: Assuming there are 100 pairs of eyes we have: Number of pairs of black eyes = 65 Number of pairs of brown eyes = 25 Number of pairs of blue eyes = 10 (i) P(blue eyes) = $\frac{\text { Number of pairs of blue eyes }}{\text { Total pairs of eyes }}=\frac{10}{100}=\frac{1}{10}$

(ii) P(brown or black eyes) = $\frac{\text { Number of pairs of brown or black eyes }}{\text { Total pairs of eyes }}$ = $\frac{25+65}{100}=\frac{90}{100}=\frac{9}{10}$

(iii) P(blue or black eyes) = $\frac{\text { Number of pairs of blue or black eyes }}{\text { Total pairs of eyes }}$ = $\frac{65+10}{100}=\frac{75}{100}=\frac{3}{4}$

(iv) P(neither blue nor brown eyes) . = 1 – P (either blue or brown eyes) = 1 – $\frac{\text { Number of pairs of blue or brown eyes }}{\text { Total pairs of eyes }}$ = 1 – $\left(\frac{25+10}{100}\right)$ = 1 – $\frac{35}{100}=\frac{65}{100}=\frac{13}{20}$

Probability Class 10 Extra Questions Maths Chapter 15 8

Question 3. Find the probability that all three children in a family have different birthdays. (Take 1 year = 365 days). Answer: The first child may be bom on any of the 365 days of the year. Similarly, the second as well as the third one may also be bom on any of the 365 days of respective year of birth. Thus, the total number of possible ways in which the three children have birthdays is 365 * 365 * 365.

These cases are mutually exclusive, exhaustive and equally likely. Thus, total number of cases = 365 × 365 × 365.

For the number of favourable cases, we note that the first child may have any one of the 365 days of the year as its birthday. In order that the second child has a birthday different from that of the first, it should have been bom on any one of the 364 remaining days of the year. Similarly, the third one should have been bom on any one of the remaining 363 days of the year. So the number of cases favourable to the event ‘different birthdays’ is clearly [365 × 364 × 363.] ∴ Required probability = $\frac{365 \times 364 \times 363}{365 \times 365 \times 365}=\frac{132132}{133225}$

Multiple Choice Questions

Choose the correct option out of four given in each of the following: Question 1. An event is very unlikely to happen. Which of the following number is closest to its probability? (a) 0.0001 (b) 0.1 (c) 1.0 (d) 0.001 Answer: (a) 0.0001

Question 2. Which of the following can’t be a probability of an event? (a) $\frac{3}{7}$ (b) 0.71 (c) 41% (d) $\frac{17}{16}$ Answer: (d) $\frac{17}{16}$

Question 3. Minimum value of probability of an event is (a) -1 (b) 1 (c) $\frac{1}{2}$ (d) 0 Answer: (d) 0

Question 4. Maximum value of probability of an event is (a) +1 (b) 100% (c) $\frac{1}{2}$ (d) (a) and (b) Answer: (d) (a) and (b)

Question 5. Which of the following number best expresses the probability of a man dying before completing 150 years of age? (a) 0.5 (b) 0.6 (c) 0.2 (d) 1 Answer: (d) 1

Question 6. Which of the following number best expresses the probability of the sun rising from west? (a) 1 (b) 0 (c) $\frac{1}{2}$ (d) can’t say Answer: (b) 0

Question 7. Two coins are tossed. The probability of getting atmost one head is (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (c) $\frac{3}{4}$ (d) $\frac{1}{3}$ Answer: (c) $\frac{3}{4}$

Question 8. Two players X and Y play a game of chess. The probability of X winning the game is 0.67. The probability of Y losing the game i (a) 0.37 (b) 0.67 (c) 0.33 (d) none of these Answer: (b) 0.67

Question 9. A lot of 25 bulbs contains 5 defective bulbs. A bulb is drawn at random from the lot. The probability of having a good bulb is (a) $\frac{1}{5}$ (b) $\frac{4}{5}$ (c) $\frac{2}{5}$ (d) $\frac{3}{5}$ Answer: (b) $\frac{4}{5}$

Question 10. A jar contains 6 red, 5 black and 3 green marbles of equal size. The probability that a randomly drawn marble would be green in colour (a) $\frac{5}{14}$ (b) $\frac{11}{14}$ (c) $\frac{3}{14}$ (d) $\frac{6}{14}$ Answer: (c) $\frac{3}{14}$

Probability Class 10 Extra Questions Maths Chapter 15 1

Question 12. In a bag, there are 100 bulbs out of which 30 are bad ones. A bulb is taken out of the bag at random. The probability of the selected bulb to be good is (a) 0.50 (b) 0.70 (c) 0.30 (d) none of these Answer: (b) 0.70

Probability Class 10 Extra Questions Maths Chapter 15 2

Question 15. In a class of 30 students, there are 16 girls and 14 boys. Five are A grade students, and three of these students are girls. If a student is chosen at random, the probability of it being a girl student or A grade student is (a) $\frac{3}{5}$ (b) $\frac{4}{5}$ (c) $\frac{8}{15}$ (d) $\frac{7}{15}$ Answer: (a) $\frac{3}{5}$

Fill in the Blanks

Question 1. An experiment whose outcome can’t be predicted yet it is one of the several possible outcomes is called ___________ experiment. Answer: random

Question 2. Probability of ___________ event is 100%. Answer: sure

Question 3. Probability has two approaches namely ___________ probability and ___________ probability. Answer: empirical, theoretical

Question 4. ___________ probability does not require any experiment to be performed. Answer: Theoretical

Question 5. A number is choosen at random from 1 to 9, the probability that it being prime is ___________ Answer: $\frac{4}{9}$

Question 6. A bag contains 7 red and 7 black balls then the outcomes of getting a red and a black ball are ___________ outcomes. Answer: equally likely

Question 7. Sample space, when two coins are tossed is ___________. Answer: {HH, HT, TH, TT}

Question 8. Event of getting odd number and event of getting even number on die are an example of ___________ events. Answer: mutually exclusive

Question 9. Probability of getting a red card in a single draw from a well shuffled deck of 52 cards ___________. Answer: $\frac{1}{2}$

Question 10. Probability of getting a number less than 100 on die is ___________. Answer: 1

Extra Questions for Class 10 Maths

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10th Class Mathematics Probability Question Bank

Done case based (mcqs) - probability total questions - 40.

Study Case : Q. 1 to 5

In a club, men are playing the card game. A man named Sanjeev draw a card from a well shuffled deck of cards.

Based on the above information, give the answer of the following questions:

A) \[\frac{1}{26}\] done clear

B) \[\frac{1}{12}\] done clear

C) \[\frac{1}{52}\] done clear

D) \[\frac{1}{4}\] done clear

question_answer 2) Find the probability of getting a face card:

B) \[\frac{1}{13}\] done clear

C) \[\frac{2}{13}\] done clear

D) \[\frac{3}{13}\] done clear

question_answer 3) Find the probability of getting a jack of clubs.

B) \[\frac{1}{52}\] done clear

C) \[\frac{3}{52}\] done clear

D) \[\frac{3}{26}\] done clear

question_answer 4) Find the probability of getting a black face card:

A) \[\frac{3}{13}\] done clear

question_answer 5) Find the probability of getting a diamond:

Study Case : Q. 6 to 10

An unbiased game to chance as shown below consists of spinning the wheel on which different areas have been marked with different fruits such as apple, grapes, mange and orange denoted by the alphabets A, G, M and O on the wheel respectively. Numbers have been marked or different parts and each of theese is equally likely, the prize depends on the number at which the arrow points once the wheel comes to a rest.

Based on the given information, give the answer of the following questions:

A) 0 done clear

B) \[\frac{3}{10}\] done clear

C) \[\frac{1}{2}\] done clear

D) \[\frac{3}{5}\] done clear

question_answer 7) The probability that the arrow will point a perfect square number is:

A) \[\frac{2}{5}\] done clear

B) \[\frac{3}{5}\] done clear

D) \[\frac{7}{10}\] done clear

question_answer 8) The probability of arrow pointing towards grapes 'G' is:

A) \[\frac{1}{2}\] done clear

C) \[\frac{2}{5}\] done clear

D) \[\frac{1}{10}\] done clear

question_answer 9) The probability of arrow pointing towards a number divisible by 4 is:

A) \[0\] done clear

C) \[\frac{3}{4}\] done clear

D) \[\frac{2}{5}\] done clear

question_answer 10) The probability of arrow pointing towards Banana B is:

B) 1 done clear

Study Case : Q. 11 to 15

Shivam sees a game in a fair. He is keenly interested to play it. He asks the rules of game from the owner. Owner says that you will move this spinner first, if it stops on a prime number, then you are allowed to pick a marble from a bag and bag has only white and black marbles. Prizes are given when a white marble is picked randomly.

Based on the above information, answer the following questions:

A) \[\frac{3}{8}\] done clear

B) \[\frac{5}{8}\] done clear

D) 1 done clear

question_answer 12) What is the probability that Shivam will be allowed to pick a marble from a bag?

B) \[\frac{1}{2}\] done clear

C) \[\frac{5}{8}\] done clear

D) \[\frac{3}{7}\] done clear

question_answer 13) The probability that Shivam will pick a black marble from the bag is:

A) \[\frac{7}{12}\] done clear

B) \[\frac{5}{12}\] done clear

D) \[\frac{2}{3}\] done clear

question_answer 14) The probability that Shivam will get a prize is:

A) \[\frac{3}{5}\] done clear

B) \[\frac{3}{4}\] done clear

C) \[\frac{5}{12}\] done clear

D) \[\frac{7}{12}\] done clear

question_answer 15) The probability that Shivam will pick a red marble from the bag is:

A) 1 done clear

B) 0 done clear

Study Case : Q. 16 to 20

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:

Days	Sunny	Cloudy	Partially cloudy	Rainy
Probability	\[\frac{1}{2}\]	X	\[\frac{1}{5}\]	y

A) 5 done clear

B) 10 done clear

C) 15 done clear

D) 20 done clear

question_answer 17) If the number of cloudy days in June is 5, then x =

A) \[\frac{1}{4}\] done clear

B) \[\frac{1}{6}\] done clear

C) \[\frac{1}{8}\] done clear

question_answer 18) The probability that the day is not rainy is:

A) \[\frac{13}{15}\] done clear

B) \[\frac{11}{15}\] done clear

C) \[\frac{1}{15}\] done clear

D) None of these done clear

question_answer 19) If the sum of x and y is \[\frac{3}{10},\] then the number of rainy days in June is:

B) 2 done clear

C) 3 done clear

D) 4 done clear

question_answer 20) Find the number of partially cloudy days:

A) 2 done clear

B) 4 done clear

C) 6 done clear

D) 8 done clear

Study Case : Q. 21 to 25

A man is sitting on a helicopter which is positioned in such a way that the entire garden is visible to him. The garden is rectangular in shape with a pond of diameter 2 m as shown in the figure. There is a semi-circular concrete patio at one end and two flower beds at the other two comers as shown. The remaining area has beautiful lush green grass. The man drops a ball from the helicopter.

Based on the above information, answer the follows questions:

A) \[\frac{11}{140}\] done clear

B) \[\frac{11}{280}\] done clear

C) 0 done clear

question_answer 22) The probability that the ball dropped will fall in the patio is:

B) 1 done clear

C) \[\frac{11}{224}\] done clear

D) \[\frac{55}{224}\] done clear

question_answer 23) The probability that the ball dropped will not fall in pond, is:

A) \[\frac{229}{240}\] done clear

C) \[\frac{11}{240}\] done clear

D) \[\frac{11}{120}\] done clear

question_answer 24) The probability that the ball dropped will fall either in pond or in patio, is:

A) \[\frac{55}{224}\] done clear

C) \[\frac{363}{1120}\] done clear

question_answer 25) The probability that the ball dropped will fall on the grass in the garden area.

A) \[\frac{363}{1120}\] done clear

B) \[\frac{723}{1120}\] done clear

C) \[\frac{11}{280}\] done clear

D) \[\frac{229}{240}\] done clear

Study Case : Q. 26 to 30

Sonny goes to a store to purchase juice cartons for his Shop. The store has 80 cartons of Litchi juice, 90 cartons of pineapple juice. 38 cartons of mango juice and 42 cartons of banana juice. If Sunny chooses a carton at random, then answer the following questions:

A) \[\frac{1}{25}\] done clear

B) \[\frac{8}{25}\] done clear

C) \[\frac{13}{25}\] done clear

D) \[\frac{9}{25}\] done clear

question_answer 27) The probability that the selected carton is not of Litchi juice is:

A) \[\frac{14}{25}\] done clear

B) \[\frac{11}{25}\] done clear

C) \[\frac{17}{25}\] done clear

D) \[\frac{4}{25}\] done clear

question_answer 28) The probability of selecting a carton of banana juice is:

A) \[\frac{51}{125}\] done clear

B) \[\frac{16}{125}\] done clear

D) \[\frac{21}{125}\] done clear

question_answer 29) Sunny buys 4 cartons of pineapple juice, 3 cartons of Litchi juice and 3 cartons of banana juice. A customer comes to Sunny's shop and picks a tetrapack of juice at random. The probability that the customer picks a banana juice, if each carton has 10 tetrapacks of juice, is:

A) \[\frac{1}{10}\] done clear

B) \[\frac{2}{10}\] done clear

C) \[\frac{3}{10}\] done clear

question_answer 30) If the storekeeper bought 14 more cartons of pineapple juice, then the probability of selecting a tetrapack of pineapple juice from the store is:

A) \[\frac{25}{127}\] done clear

B) \[\frac{50}{127}\] done clear

C) \[\frac{75}{127}\] done clear

D) \[\frac{100}{127}\] done clear

Study Case : Q. 31 to 35

Three persons toss 3 coins simultaneously and note the outcomes. Then, they ask few questions to one another.

Help them in finding the answer of the following questions:

question_answer 32) The probability of getting exactly 1 head is:

B) \[\frac{1}{4}\] done clear

D) \[\frac{3}{8}\] done clear

question_answer 33) The probability of getting exactly 3 tails is:

C) \[\frac{1}{4}\] done clear

D) \[\frac{1}{8}\] done clear

question_answer 34) The probability of getting at most 3 heads is:

question_answer 35) The probability of getting at least two heads is:

Study Case : Q. 36 to 40

Sanjeev is very found of collecting balls of different colours. He has a total of 25 balls in his basket out of which five balls are red in colour and eight are white. Out of the remaining balls, some are green in colour and the rest are pink:

A) 4 done clear

B) 8 done clear

C) 10 done clear

D) 12 done clear

question_answer 37) The probability of drawing a ball of colour other than green colour is:

B) \[\frac{4}{25}\] done clear

C) \[\frac{21}{25}\] done clear

D) \[\frac{17}{25}\] done clear

question_answer 38) The probability of drawing either a green or white ball is:

B) \[\frac{12}{25}\] done clear

question_answer 39) The probability of drawing neither a pink nor a white ball is:

A) \[\frac{9}{25}\] done clear

C) \[\frac{16}{25}\] done clear

question_answer 40) The probability of not drawing a red ball is:

A) \[\frac{1}{5}\] done clear

B) \[\frac{2}{5}\] done clear

C) \[\frac{3}{5}\] done clear

D) \[\frac{4}{5}\] done clear

Study Package

Case Based (MCQs) - Probability

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Delineating bacterial genera based on gene content analysis: a case study of the Mycoplasmatales-Entomoplasmatales clade within the class Mollicutes

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Genome-based analysis allows for large-scale classification of diverse bacteria and has been widely adopted for delineating species. Unfortunately, for higher taxonomic ranks such as genus, establishing a generally accepted approach based on genome analysis is challenging. While core-genome phylogenies depict the evolutionary relationships among species, determining the correspondence between clades and genera may not be straightforward. For genotypic divergence, percentage of conserved proteins (POCP) and genome-wide average amino acid identity (AAI) are commonly used, but often do not provide a clear threshold for classification. In this work, we investigated the utility of global comparisons and data visualization in identifying clusters of species based on their overall gene content, and rationalized that such patterns can be integrated with phylogeny and other information such as phenotypes for improving taxonomy. As a proof of concept, we selected 177 representative genome sequences from the Mycoplasmatales-Entomoplasmatales clade within the class Mollicutes for a case study. We found that the clustering patterns corresponded to the current understanding of these organisms, namely the split into three above-genus groups: Hominis, Pneumoniae, and Spiroplasma-Entomoplasmataceae-Mycoides (SEM). However, at the genus level, several important issues were found. For example, recent taxonomic revisions that split the Hominis group into three genera and Entomoplasmataceae into five genera are problematic, as those newly described or emended genera lack clear differentiations in gene content from one another. Moreover, several cases of mis-classification were identified. These findings demonstrated the utility of this approach and the potential application for other bacteria.

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Study Case : Q. 6 to 10. An unbiased game to chance as shown below consists of spinning the wheel on which different areas have been marked with different fruits such as apple, grapes, mange and orange denoted by the alphabets A, G, M and O on the wheel respectively. Numbers have been marked or different parts and each of theese is equally ...
Delineating bacterial genera based on gene content analysis: a case
As a proof of concept, we selected 177 representative genome sequences from the Mycoplasmatales-Entomoplasmatales clade within the class Mollicutes for a case study. We found that the clustering patterns corresponded to the current understanding of these organisms, namely the split into three above-genus groups: Hominis, Pneumoniae, and ...