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

Class 10 Case Study Based Questions Chapter 15 Probability CBSE Board Term 1 with Answer Key

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

Last modified on: 8 months 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 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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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

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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:

case study questions in probability class 10

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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Class 10 Maths Chapter 14 Case Based Questions - Probability

Study case - 1.

Q2: The probability of drawing a ball of colour other than green colour is: (a) 0 (b) 4/25 (c) 21/25 (d) 17/25 Ans: (c) Explanation: From part (A), number of green balls = 4. ∴ Number of balls of colour other than = 25 − 4 = 21. ∴ Probability of drawing a ball of colour other than green colour = 21/25.

Q3: The probability of drawing either a green or white ball is: (a) 0 (b) 12/25 (c) 13/25 (d) 17/25 Ans: (b) Explanation: The number of green balls = 4 and number of white balls =8. Therefore, total number of green balls + white balls = 4 + 8 = 12.. ∴ Probability of drawing either a or a white ball = 12/25.

Study Case - 2

Q1: The probability of getting almost one tail is: (a) 0 (b) 1 (c) 1/2 (d) 1/4 Ans: (c) Explanation: Sample space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} ⇒ n(S) = 8 Let A be the event of getting at most one tail. ∴ A = {HHH, HHT, HTH, THH} ⇒ n(A) = 4 ∴ Required probability = 4/8 = 1/2

Q2: The probability of getting exactly 1 head is: (a) 1/2 (b) 1/4 (c) 1/8 (d) 3/8 Ans: (d) Explanation: Let B be the event of getting exactly 1 head. ∴B = {HTT, THT, TTH} ⇒ n(B) = 3 ∴ Required probability = 3/8 Q3: The probability of getting exactly 3 tails is: (a) 0 (b) 1 (c) 1/4 (d) 1/8 Ans: (d) Explanation: Let C be the event of getting exactly 3 tails. ∴ C = {TTT} ⇒ n(C) = 1 ∴ Required probability = 1/8 Q4: The probability of getting at most 3 heads is: (a) 0 (b) 1 (c) 1/2 (d) 1/8 Ans: (b) Explanation: Let D be the event of getting atmost 3 heads. ∴ D = {HHH, HHT, HTH, HTT, THH, THT, TTT} ⇒ n(D) = 8 ∴ Required probability = 8/8 = 1.

Q5: The probability of getting at least two heads is: (a) 0 (b) 1 (c) 1/2 (d) 1/4 Ans: (c) Explanation: Let E be the event of getting at least two heads. ∴E = {HHT, HTH, THH, HHH} ⇒ n(E) = 4 ∴ Required probability = n(E)/n(S) = 4/8 = 1/2.

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Class 10th Maths - Probability Case Study Questions and Answers 2022 - 2023

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Probability, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Probability case study questions with answer key.

10th Standard CBSE

Final Semester - June 2015

(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

In a party, some children decided to play musical chair game. In the game the person playing the music has been advised to stop the music at any time in the interval of 3 mins after he start the music in each turn. On the basis of the given information, answer the following questions. (i) What is the probability that the music will stop within first 30 sees after starting?

(ii) The probability that the music will stop within 45 sees after starting is

(iii) The probability that the music will stop after 2 mins after starting is

(iv) The probability that the music will not stop within first 60 sees after starting is

(v) The probability that the music will stop within first 82 sees after starting 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

Rahul goes to a fete in Mussoorie. There he saw a game having prizes - wall clocks, power banks, puppets and water bottles. The game consists of a box having cards inside it, bearing the numbers 1 to 200, one on each card. A person has to select a card at random. Now, the winning of prizes has the following conditions:

Wall clock - If the number on the selected card is a perfect square.
Power bank - If the number on the selected card is multiple of 3.
Puppet - If the number on selected card is divisible by 10.
Water bottle - If the number on the selected card is a prime number more than 100 but less than 150.
Better luck next time - If the number on the selected card is a perfect cube.

(ii) The probability of winning a water bottle is

(iii) The probability of winning a power bank is

(iv) The probability of winning a wall clock is

(v) The probability of getting 'Better Luck next time' is

(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

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

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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

$case study questions in probability class 10$

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

$case study questions in probability class 10$

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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Chapter 14 Class 10 Probability

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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
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Probability of Impossible and sure events
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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.

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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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Important Questions for CBSE Class 10 Maths Chapter 15 - Probability

CBSE Class 10 Maths Chapter 15: Probability Important Questions: Free PDF Download

The probability class 10 important questions are available for students to help them get a thorough knowledge of the concept. It is advised that students should practice from the given important questions to score better marks in the examination. These essential questions from the probability chapter can also be downloaded in PDF version for the convenience of the students. The more they refer to these questions, the better grasp they will have on this topic. All these questions are framed by subject experts following the latest guidelines and patterns of the CBSE Boards . Even, there are solutions provided with the questions so that students can review their answers. Besides, they can also make notes using these questions.

Vedantu is a platform that provides free NCERT Book Solutions and other study materials for students. You can Download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.

Download CBSE Class 10 Maths Important Questions 2023-24 PDF

Also, check CBSE Class 10 Maths Important Questions for other chapters:

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Study Important Questions for Class 10 Maths Chapter 15 – Probability

1. An integer is chosen at random from the first two hundreds digit. What is the probability that the integer chosen is divisible by ${6}$ or ${8}$.

Ans: First $200$ integers that are multiples of $6$ are listed as

\[\text{6, 12, 18, }\underline{\text{24}}\text{, 30, 36, 42, }\underline{\text{48}}\text{, 54, 60, 66, }\underline{\text{72}}\text{, 78, 84, 90, }\underline{\text{96}}\text{, 102, 108, 114,}\]\[\underline{\text{120}}\text{, 126, 132, 138, }\underline{\text{144}}\text{, 150, 156, 162, }\underline{\text{168}}\text{, 174, 180, 186, }\underline{\text{192}}\text{, 198}\].

First $200$ integers that are multiples of $8$ are listed as

\[\text{8, 16, }\underline{\text{24}}\text{, 32, 40, }\underline{\text{48}}\text{, 56, 64, }\underline{\text{72}}\text{, 80, 88, }\underline{\text{96}}\text{, 104, 112, }\underline{\text{120}}\text{, 128, 136, }\underline{\text{144}}\text{,}\]\[\text{152, 160, }\underline{\text{168}}\text{, 176, 184, }\underline{\text{192}}\text{, 200}\].

Thus, there are total $50$ numbers which are multiples of $6$ or $8$.

So, the probability that the integer chosen is divisible by $6$ or $8$ is $=\frac{50}{200}=\frac{1}{4}$.

2. 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 out in the box. the probability of drawing a black ball is now double of what it was before. Find ${x}$.

Ans: When some balls are drawn randomly, then it ensures equal likely outcomes.

There are a total $12$ balls.

So, $12$ possible outcomes can occur.

Now, let there are $x$ balls, which are black.

Thus, the probability that a ball drawn is black $\begin{align} & =\frac{\text{Number of favourable outcomes}}{Total\text{ n}umber\text{ of possible outcomes}} \\ & =\frac{x}{12} \\ \end{align}$

Since, $6$ more black balls are out in the box, so there will be a total $x+6$ black balls out of the total balls $12+6=18$.

By the given condition, the probability obtained for drawing black ball in the second case is $=2\times $ the probability obtained for drawing of black ball in the first case.

$\frac{x+6}{18}=2\times \frac{x}{12}$

$\Rightarrow \frac{x+6}{18}=\frac{x}{6}$

$\Rightarrow 6x+36=18x$

$\Rightarrow x=3$

Thus, there are $3$ black balls in the box.

3. A bag contains ${8}$ red balls and ${x}$ blue balls, the odd against drawing a blue ball are \[{2}:{5}\]. What is the value of ${x}$?

Ans: Since, there are $x$ blue balls and $8$ red balls, so the total number of balls $=x+8$.

Therefore, the probability that a blue ball is drawn $=\frac{x}{x+8}$

The probability that a red ball is drawn $=\frac{8}{x+8}$.

So, by the given condition, we have

$\frac{8}{x+8}:\frac{x}{x+8}=2:5$

$\begin{align} & \Rightarrow 2\left( \frac{x}{x+8} \right)=5\left( \frac{8}{x+8} \right) \\ & \Rightarrow 2x=40 \\ & \Rightarrow x=20 \\ \end{align}$

Thus, the value of $x$ is $20$.

4. A card is drawn from a well shuffled deck of cards.

i) What are the odds in favour of getting spade?

Ans. There are $52$ cards in a deck and among these $13$ are spades.

So, the number of cards remaining is $39$.

Therefore, the odds in favour of getting spades is $\frac{13}{52}:\frac{39}{52}=1:3$.

ii) What are the odds against getting a spade?

Ans. The odds against getting a spade are $39$.

iii) What are the odds in favour of getting a face card?

Ans. The odds of obtaining a face card are $12$ to $52$. The odds of not getting a face card are \[40\] to $52$.

The odds in favour of getting a face card $=\frac{12}{52}:\frac{40}{52}=3:10$.

iv) What are the odds in favour of getting a red king

Ans: The odds of obtaining a red king are $2$ to $52$.

The odds of not getting a red king are $50$ to $52$.

Thus, the odds in favour of getting a red king are $=\frac{2}{52}:\frac{50}{52}=1:25$.

5. A die is thrown repeatedly until a six comes up. What is the sample space for this experiment? HINT: \[{A}=\left\{ {6} \right\},\text{ }{B}=\left\{ {1},{2},{3},{4},{5}, \right\}\].

Ans: The sample space for the given experiment is \[\left\{ \text{A, BA, BBA, BBBA, BBBBA} \right\}\].

6. Why is tossing a coin considered to be a fair wav of deciding which team should get the ball at the beginning of a football match?

Ans: Tossing a coin gives equally likely outcomes since they are mutually exclusive events. That’s why tossing a coin is considered to be a fair wav of deciding which team should get the ball.

7. 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.

Ans: Suppose that the number of blue balls in the bag is $x$.

Therefore, the number of total balls contained in the bag $=x+5$.

Then, the probability that a blue ball is drawn, $=\frac{x}{x+5}$.

Also, the probability that a red ball is drawn, $=\frac{5}{x+5}$.

Now, by the given condition,

$\begin{align} & \Rightarrow \frac{x}{x+5}=2\cdot \frac{5}{x+5} \\ & \Rightarrow x=10 \\ \end{align}$

Thus, the number of blue balls in the bag is $10$.

8. 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 out in the box the probability of drawing a black ball is now double of what it was before. Find ${x}$?

Ans: The total number of possible outcomes is $12$.

Suppose that, the number of favourable outcomes on the event of drawing black ball is $x$.

Thus, the probability of getting a black ball $=\frac{x}{12}$.

If $6$ more black balls are removed from the box, then the number of possible outcomes \[=\text{12+6}=\text{18}\].

Then, the number of black balls $=x+6$.

Thus, the probability of drawing a black ball is $\frac{x+6}{18}$.

$\frac{x+6}{18}=2\left( \frac{x}{12} \right)$

Therefore, $x=3$.

9. If \[{65}%\] of the populations 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 eves

Ans. The number of black eyes $=65$. The number of Brown eyes $=25$

The number of blue eyes $=10$.

Thus, there are a total of $180$ eyes.

Therefore, the probability of having blue eyes $=\frac{10}{100}=\frac{1}{10}$.

(ii) Brown or black eves

Ans. The probability of having brown or black eyes $=\frac{90}{100}=\frac{9}{10}$

(iii) Blue or black eves

Ans. The probability of having blue or black eyes $=\frac{10}{100}=\frac{1}{10}$.

(iv) neither blue nor brown eves

Ans: The probability of having neither blue nor brown eyes $=\frac{65}{100}=\frac{13}{20}$.

10. Find the probability of having ${53}$ Sundays in

i) a non-leap year

Ans. We know that there are $365$ days in an ordinary year containing \[52\] weeks and $1$ day.

This day may be any one of the $7$ days of the week.

Thus, the probability that this day is Sunday $=\frac{1}{7}$.

Hence, the probability that an ordinary year has $53$ Sunday $=\frac{1}{7}$.

ii) a leap year

Ans: It is known that, there are $366$ days in a leap year containing $52$weeks and $2$ days.

These two days can be \[\left( MT \right),\left( TW \right),\left( W\text{ }Th \right),\left( Th\text{ }F \right),\left( FS \right),\left( SS \right),\left( SM \right)\]. Now, since, $52$ weeks contain $52$ Mondays, so the remaining Monday can be obtained if the two days are \[\left( MT \right)\] or \[\left( SM \right)\], that is, $2$ out of $7$ cases. Thus, the probability of having $53$ Mondays in a leap year $=\frac{2}{7}$.

11. Five cards - the ten, Jack, queen, king and ace, 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?

Ans. The number of total possible outcomes is $5$.

Since, the number of Queen is $1$, so

the number of favourable outcomes$=1$.

Thus, the probability of getting a Queen card $=\frac{1}{5}$.

(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?

Ans. Since, one card is put aside, so now, the number of total possible outcomes is $5-1=4$.

(a) Since, there is only one ace card, so the number of favourable outcomes is $1$.

Thus, the probability of getting an ace card $=\frac{1}{4}$.

(b) Since, there is no queen card left after the first pick up, so the number of favourable outcomes is $0$.

Thus, the probability of getting a Queen card $=\frac{0}{4}=0$.

12. A number x is chosen at random from the numbers\[{-3, -2, -1, 0},{ 1, 2, 3}\]. What is the probability that \[\left| {x} \right|<{2}\].

Ans: $x$ can have the value among the $7$ given values.

Now, for \[\left| x \right|<2\], $x$ can be $-3,-2,-1,0,1$.

So, the required probability, $P\left( \left| x \right|<2 \right)=\frac{5}{7}$.

13. A number ${x}$ is selected from the numbers \[{1},{2},{3}\] and then a second number ${y}$ is randomly selected from the numbers \[{1},{4},{9}\], What is the probability that the product ${xy}$ of the two numbers will be less than ${9}$?

Ans: Number $x$ can be chosen in three different ways, and number $y$ can be chosen in three different ways as well. So, the two numbers can be chosen in $9$ ways such that,

\[\left( 1,1 \right),\left( l,4 \right),\left( 2,l \right),\left( 2,4 \right),\left( 3,1 \right)\].

Therefore, the number of favourable occurrences is $5$.

Thus, the probability that the product will be less than $9$ is $=\frac{5}{9}$.

14. In the adjoining figure a dart is thrown at the dart board and lands in the interior of the circle. What is the probability that the dart will land in the shaded region?

(Image will be uploaded soon)

Ans: It is given that,

\[\text{AB = CD = 8}\] and \[\text{AD = BC = 6}\].

Then, by the Pythagorean Theorem on the triangle\[ABC\], gives

\[\text{A}{{\text{C}}^{\text{2}}}=\text{A}{{\text{B}}^{\text{2}}}+\text{B}{{\text{C}}^{\text{2}}}\]

\[\Rightarrow \text{A}{{\text{C}}^{\text{2}}}=\text{82}+\text{62}=\text{100}\]

$\Rightarrow AC=10$

$\Rightarrow OA=OC=5$ (since, the point $O$ is the midpoint of \[AC\])

Therefore, the area of the circle is $=\pi {{\left( OA \right)}^{2}}=25\pi $ sq. units.

Also, the area of $ABCD=AB\times BC=8\times 6=48$ sq. units.

Thus, the area of the shaded region $=$ Area of circle $-$ Area of the rectangle \[ABCD\]

That is, Area of shaded region $=\left( 25\pi -48 \right)$ sq. units.

Thus, the probability dart will land in the shaded region $\begin{align} & =\frac{\text{area of shaded region}}{area\text{ of circle}} \\ & =\frac{25\pi -48}{25\pi } \\ \end{align}$

15. In the fig points ${A},{B},{C},$ and ${D}$ are the centres of four circles. each having a radius of ${1}$ unit. If a point is chosen at random from the interior of a square \[{ABCD}\]. What is the probability that the point will be chosen from the shaded region?

Ans: It is given that the radius of the circle is $1$ unit.

Therefore, area of the circle $=$ Area of $4$ sector.

Thus, the side of the square \[ABCD\] is \[2\] units.

Therefore, the area of square \[=\text{2 }\!\!\times\!\!\text{ 2}=\text{4}\] units.

So, the area of the shaded region

$=$ area of square $-\text{ }4\times $ area of the sectors.

Hence, the required probability $=\frac{4-\pi }{4}$.

16. In the adjoining figure \[{ABCD}\] is a square with sides of length ${6}$ units points ${P}$ & ${Q}$ are the mid points of the sides \[{BC}\] & \[{CD}\]respectively. If a point is selected at random from the interior of the square what is the probability that the point will be chosen from the interior of the triangle \[{APQ}\].

Ans: The area of the $\Delta PQC$$=\frac{1}{2}\times 3\times 3=\frac{9}{2}=4.5$ sq. units.

The area of the $\Delta ABP$$=\frac{1}{2}\times 6\times 3=9$ sq. units.

The area of the $\Delta ADQ=\frac{1}{2}\times 6\times 3=9$ sq. units.

Therefore, the area of the $\Delta APO$$=$ area of the square $-$(Area of the $\Delta PQC$ $+$ area of the $\Delta ABP$$+$ area of the$\Delta ABP$)

\[=36-\left( 18+4.5 \right)\]

$\begin{align} & =36-22.5 \\ & =13.5 \\ \end{align}$

Hence, the probability that the point will be selected from the interior of $\Delta APQ$$=\frac{13.5}{36}=\frac{135}{360}=\frac{3}{8}$.

17. In a musical chair, the person playing the music has been advised to stop playing the music at any time within ${2}$ minutes after she starts playing. What is the probability that the music will stop within the half minute after starting?

Ans: All the numbers between $0$ and $2$ are conceivable outcomes here. As illustrated in the picture, this is the section of the number line from $0$ to $2$.

Let $A$ be the event in which "the music is turned off during the first half minute." Then there are all points on the number line that are favourable to occurrence \[A\]. i.e.

The points on the number line from Q to P represent the total number of outcomes. That is, from $0$ to $2$.

The required probability, $P\left( A \right)=\frac{\text{length of OQ}}{length\text{ of OP}}=\frac{\frac{1}{2}}{2}=\frac{1}{4}$.

18. A jar contains \[{54}\] marbles each of which is blue. green or white. The probability of selecting a blue marble at random from the jar is $\frac{{1}}{{3}}$ and the probability of selecting a green marble at random is $\frac{{4}}{{9}}$. How many white marbles does the jar contain?

Ans: Suppose that $b,g,$ and $w$ denotes the number of blue, green and white marbles respectively in the jar.

Therefore, $b+g+w=54$. …… (i)

So, the probability of choosing a blue marble $=\frac{b}{54}$.

Since, the probability of choosing a blue marble $=\frac{1}{3}$, so

$\begin{align} & \frac{1}{3}=\frac{b}{54} \\ & \Rightarrow b=18 \\ \end{align}$

Again, the probability of choosing a green marble $=\frac{4}{9}$.

Therefore,

$\begin{align} & \frac{g}{54}=\frac{4}{9} \\ & \Rightarrow g=24 \\ \end{align}$

Now, putting the obtained values of $b$ and $g$ into the equation (i), yields \[\text{18 + 24 + w}=\text{54}\]

$\Rightarrow w=12$.

Hence, there are a total of $12$ white marbles in the jar.

1 Marks Questions

1. Complete the statements:

(i) Probability of event ${E}$ $+$ Probability of event “not ${E}$” $=$_________.

Ans. Probability of event $E+$Probability of event “not $E$” $=1$.

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

Ans. 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________. Such an event is called __________.

Ans. 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 _________.

Ans. 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 ________ and less than or equal to __________.

Ans. The probability of an event is greater than or equal to $0$ and less than or equal to $1$.

2. Which of the following cannot be the probability of an event:

(A) $\frac{{2}}{{3}}$

(B) $-{1}.{5}$

(D) \[{0}{.7}\]

Ans. By the definition of probability, the maximum and minimum value of probability are $1$ and $0$ respectively.

Therefore, $-1.5$ cannot be the probability of an event.

Thus, option (B) is the correct answer.

3. If \[{P}\left( {E} \right)={0}.{05}\], what is the probability of ‘not\[{E}\]’?

Ans. It is known that, \[\text{P}\left( \text{E} \right)+\text{P}\left( \text{not E} \right)=\text{1}\]

$\begin{align} & \Rightarrow P\left( not\text{ }E \right)=1P\left( E \right) \\ & =10.05 \\ & =0.95 \\ \end{align}$

4. 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?

Ans. Suppose that $E$ represents the event that the students have the same birthday.

Then, \[\text{P}\left( \text{E} \right)=\text{0}\text{.992}\].

It is known that, $P\left( E \right)+P\left( \overline{E} \right)=1$

$\begin{align} & \Rightarrow P\left( E \right)=1-P\left( \overline{E} \right) \\ & =1-0.992 \\ & =0.008 \\ \end{align}$

Thus, the probability that the $2$ students have the same birthday is $0.008$.

5. ${12}$ defective pens are accidently 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.

Ans. The number of total favourable outcomes \[=\text{132+12}=\text{144}\].

There are $132$ favourable outcomes.

Thus, the probability of getting a good pen $=\frac{132}{144}=\frac{11}{12}$.

6. Which of the following is polynomial?

(a) ${{{x}}^{{2}}}-{6}\sqrt{{x}}+{2}$

(b) $\sqrt{{x}}+\frac{{1}}{\sqrt{{x}}}$

(d) none of these

Ans. (d) none of these.

7. Polynomial ${2}{{{x}}^{{4}}}{+3}{{{x}}^{{3}}}{-5}{{{x}}^{{2}}}-{5}{{{x}}^{{2}}}{+9x+1}$ is a

(a) linear polynomial

(b) quadratic polynomial

(d) bi-quadratic polynomial

Ans. Since, the degree of the polynomial $\text{2}{{\text{x}}^{\text{4}}}\text{+3}{{\text{x}}^{\text{3}}}\text{-5}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{2}}}\text{+9x+1}$ is $4$, so it is a bi-quadratic polynomial.

Thus, option (d) is the correct answer.

8. If ${\alpha }$ and ${\beta }$ are zeros of ${{{x}}^{{2}}}+{5x}+{8}$, then the value of $\left( {\alpha +\beta } \right)$ is

(a) \[{5}\]

(b) \[-{5}\]

(d) \[-{8}\]

Ans. Since, $\alpha $ and $\beta $ are the zeroes of ${{x}^{2}}+5x+8$, so the value of $\alpha +\beta =-5$.

Thus, option (b) is the correct answer.

9. The sum and product of the zeros of a quadratic polynomial are ${2}$ and \[-{15}\] respectively. The quadratic polynomial is

(a) ${{{x}}^{{2}}}{-2x+15}$

(b) ${{{x}}^{{2}}}{-2x}-{15}$

(d) ${{{x}}^{{2}}}+{2x+15}$

Ans. The polynomial with the sum of zeroes $2$ and product of zeroes $-15$ is given by ${{x}^{2}}-\left( 2 \right)x+\left( -15 \right)$, that is, ${{x}^{2}}-2x-15$.

So, option (b) is the correct answer.

10. Cards each marked with one of the numbers \[{4},{5},{6},\ldots ,{20}\] are placed in a box and mixed thoroughly. One card is drawn at random from the box, what is the probability of getting an even prime number?

(b) \[{1}\]

(d) \[{3}\]

Ans. The number of possible outcomes is $17$.

Now, the only number which is even and prime is $2$, but it does not belong to the list $4,5,6,...,20$.

So, the probability of getting an even prime $=\frac{0}{17}=0$.

Thus, option (a) is the correct answer.

11. A bag contains \[{5}\] red and \[{4}\] black balls. A ball is drawn at random from the bag. What is the probability of getting a black ball?

(a) $\frac{{1}}{{3}}$

(b) $\frac{{2}}{{9}}$

(d) None of these

Ans. There are a total of $5+4=9$ bags and from them, $4$ are black.

Therefore, the probability of getting a black ball $=\frac{4}{9}$.

Thus, option (c) is the correct answer.

12. A dice is thrown once, what is the probability of getting a prime number?

(a) ${1}$

(b) $\frac{{1}}{{2}}$

(d) ${0}$

Ans. The number of possible outcomes while throwing one dice is $6$.

Now, the list of $3$ prime numbers is $2,3,5$.

Therefore, the probability of getting a prime number is $=\frac{3}{6}=\frac{1}{2}$.

13. What is the probability that a number selected from the numbers ${1},{2},{3},\ldots ,{15}$ is a multiple of ${4}$?

(a) $\frac{{1}}{{5}}$

Ans. The number of total possible outcomes is $15$.

The list of $3$ numbers from $1$ to $15$,that are multiple of $4$ is given by $4,8,12$.

Therefore, the probability of selecting a number that is multiple of $4$ $=\frac{3}{15}=\frac{1}{5}$.

14. Cards marked with the numbers ${2}$ to \[{51}\] are placed in a box and mixed thoroughly. One card is drawn from this box, find the probability that the number on the card is an even number.

(a) $\frac{{1}}{{2}}$

(d) None of these

Ans. From $2$ to $51$, there are total $50$ numbers and so the total number of outcomes is $50$.

The number of events from $2$ to $51$ is $\frac{50}{2}=25$.

Thus, the probability of getting a card with an even number $=\frac{25}{50}=\frac{1}{2}$.

Hence, option (a) is the correct answer.

15. The king, queen and jack of clubs are removed from a deck of \[{52}\] playing cards and then well shuffled. One card is selected from the remaining cards, find the probability of getting a king.

(a) $\frac{{3}}{{49}}$

(d) none of these Ans. Since, from the deck of $52$ cards, one king, queen, and jack are removed, so the total cards remaining is $52-3=49$.

In $49$ cards, there are a total of $3$ king cards.

Thus, the probability of getting a king $=\frac{3}{49}$.

16. What is the probability of getting a number less than ${7}$ in a single throw of a die?

Ans. The number of total possible outcomes is $6$.

All the numbers less than $7$ are $1,2,3,4,5,6$.

Thus, the probability of getting a number less than $7$ in a single throw is $=\frac{6}{6}=1$.

Hence, option (c) is the correct answer.

17. One card is drawn from a well shuffled deck of ${52}$ cards. Find the probability of drawing ‘${10}$’ of a black suit.

(a) $\frac{{1}}{{26}}$

Ans. The number of possible outcomes is $52$.

The number of cards with $10$ of black suit is $2$.

Thus, the probability of drawing ‘$10$’ of a black suit $=\frac{2}{52}=\frac{1}{26}$

18. Cards each marked with one of the numbers \[{4,5,6,\ldots ,20}\] are placed in a box and mixed thoroughly. One card is drawn at random from the box, what is the probability of getting an even prime number?

(a) \[{0}\]

19. A bag contains ${5}$ red and ${4}$ black balls. A ball is drawn at random from the bag. What is the probability of getting a black ball?

20. A dice is thrown once, what is the probability of getting a prime number?

21. What is the probability that a number selected from the numbers \[{1,2,3,\ldots ,15}\] is a multiple of ${4}$?

Ans. The number of possible outcomes is $15$.

Now, the list of the $3$ numbers between $1$ and $15$, which are multiple of $4$ are $4,8,12$.

Therefore, the required probability $=\frac{3}{15}=\frac{1}{5}$.

22. If \[{E}\] be any event, then value of \[{P}\left( {E} \right)\] lie in between

(a) ${0}<{P}\left( {E} \right)<{1}$

(b) ${0}\le {P}\left( {E} \right)<{1}$

(d) ${0}\le {P}\left( {E} \right)\le {2}$

So, $0\le P\left( E \right)\le 1$.

23. Maximum and minimum value of probability is

(a) $\left( {1},{1} \right)$

(b) \[\left( {1},{0} \right)\]

Thus, the option (b) is the correct answer.

24. An unbiased die is thrown. What is the probability of getting an even number or a multiple of \[{3}\]?

(a) $\frac{{2}}{{3}}$

(b) $\frac{{3}}{{2}}$

Ans. The number of possible outcomes while throwing an unbiased die is $6$ and the number of events is $3$ (which are $2,4,6$).

Also, the numbers, which are multiple of $3$ are $3,6$.

Thus, the $4$ numbers which are even or multiple of $3$ are $2,3,4,6$.

Hence, the probability of getting an even number or a multiple of $3$ is $=\frac{4}{6}=\frac{2}{3}$.

So, option (a) is the correct answer.

25. Let \[{E}\] be any event, then the value of \[{P}\left( {E} \right)+{P}\left( {not}\text{ }{E} \right)\] equals to

(d) $\frac{{1}}{{2}}$

Ans. By the definition of probability,

if $E$ is an event, then the value of \[P\left( E \right)+P\left( \overline{E} \right)=1\], that is,

$P\left( E \right)+P\left( \text{not E} \right)=1$

26. Degree of polynomial ${{{y}}^{{3}}}{-2}{{{y}}^{{2}}}{-}\sqrt{{3y}}{+}\frac{{1}}{{2}}$ is

(d) $\frac{{3}}{{2}}$

Ans. It is known that the degree of a polynomial is the highest power of the variable contained in that polynomial.

Therefore, the degree of the polynomial ${{y}^{3}}-2{{y}^{2}}-\sqrt{3y}+\frac{1}{2}$ is $3$.

27. Zeros of \[{P}\left( {x} \right)={2}{{{x}}^{{2}}}+{9x}{35}\] are

(a) ${7}$ and $\frac{{5}}{{2}}$

(b) \[-{7}\] and $\frac{{5}}{{2}}$

(d) ${7}$ and \[{2}\]

Ans. The zeros of the polynomial $P\left( x \right)=2{{x}^{2}}+9x-35$ are $-7$ and $\frac{5}{2}$, because $P\left( -7 \right)=2{{\left( -7 \right)}^{2}}+9\left( -7 \right)-35=0$ and

$P\left( \frac{5}{2} \right)=2{{\left( \frac{5}{2} \right)}^{2}}+9\left( \frac{5}{2} \right)-35=0$.

28. The quadratic polynomial whose zeros are ${3}$ and \[-{5}\] is

(a) \[{{{x}}^{{2}}}+{2x}{15}\]

(b) \[{{{x}}^{{2}}}+{3x}{8}\]

Ans. The polynomial with the zeroes $3$ and $-5$ is given by

${{x}^{2}}-\left( 3-5 \right)x+3\left( -5 \right)$

$={{x}^{2}}+2x-15$.

29. If ${\alpha }$ and ${\beta }$ are the zeros of the quadratic polynomial\[{P}\left( {x} \right)={{{x}}^{{2}}}{px+q}\], then the value of ${{{\alpha }}^{{2}}}{+}{{{\beta }}^{{2}}}$ is equal to

(a) \[{{{p}}^{{2}}}{2q}\]

(b) $\frac{{p}}{{q}}$

Ans. Since, $\alpha $ and $\beta $ are the roots of the polynomial $P\left( x \right)={{x}^{2}}-px+q$, so $\alpha +\beta =p$ …… (i)

and $\alpha \beta =q$.

$\begin{align} & {{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \\ & ={{p}^{2}}-2q \\ \end{align}$

2 Marks Questions

1. 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.

Ans. “A driver tries to start a car” in the experiment. We cannot presume that each event is equally likely to occur since the car starts or does not start. As a result, there are no equally likely outcomes in the experiment.

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

Ans. “A player tries to shoot a basketball,” says the experiment. We cannot assume that each result is equally likely to occur, whether she shoots or misses the shot. As a result, no equally likely outcomes are possible in the experiment.

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

Ans. During the test, “A true-false question is asked in a trial. The response is correct or incorrect.” We know with certainty that the outcome will be one of two likely outcomes: right or wrong. We can reasonably expect each event, correct or wrong, to occur in the same way.

As a result, the likelihood of doing it right or wrong are equal.

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

Ans. “A baby is born, it is a boy or a girl,” says the experiment. We know with certainty that the outcome will be one of two likely outcomes: a boy or a girl. We have reason to believe that each event, boy or girl, is equally likely to occur. As a result, both boy and female outcomes are equally likely.

2. 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?

Ans. Because we know that a coin toss can only land in one of two ways – head up or tail up – the tossing of a coin is regarded as a fair manner of selecting which team should have the ball at the start of a football game. It is reasonable to infer that either event, whether head or tail, has the same probability of occurring as the other, i.e., the outcomes head and tail are equally likely to occur. So, the outcome of a coin toss is absolutely unexpected.

3. A bag contains lemon flavoured candles only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:

(i) an orange flavoured candy?

Ans. Consider the occurrence with the experiment of removing an orange-flavoured candy from a bag of lemon-flavored candies.

Note that, there are not any outcomes that represent an orange flavoured candy.

Hence, the event is impossible and so its probability is $0$.

(ii) a lemon flavoured candy?

Ans. Consider the event of taking a lemon flavoured candy from a bag that contains only lemon flavoured candies.

This event represents a certain event. So, its probability is $1$.

4. 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:

Ans. The number of total balls in the bag is $3+5=8$. Therefore, there are a total of $8$ possible outcomes.

Since the bag contains 3 red balls, the number of favourable outcomes is $=3$.

Thus, the probability of getting a red ball $=\frac{3}{8}$.

(ii) not red?

Ans. There are $5$ balls, which are not red.

So, the number of favourable outcomes is $5$.

Thus, the probability of obtaining a ball that is not red $=\frac{5}{8}$.

5. 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:

Ans. The number of total marbles in the box \[=\text{5+8+4}=\text{17}\].

So, the number of possible outcomes is $17$.

The number of red marbles in the box is $5$.

Therefore, the number of favourable outcomes is $5$.

Hence, the probability of getting a red marble $=\frac{5}{17}$.

(ii) white?

Ans. The number of white marbles in the box is $8$.

So, the number of favourable outcomes $=8$.

Thus, the probability of getting a white marble $=\frac{8}{17}$.

(iii) not green?

Ans. The number of marbles which are not green is \[\text{5+8}=\text{13}\].

So, the number of favourable outcomes is $13$.

Thus, the probability of not obtaining a green marble $=\frac{13}{17}$.

6. A piggy bank contains hundred ${50}$ p coins, fifty Re. ${1}$ coins, twenty Rs. \[{2}\] coins and ten Rs. \[{5}\] coins. If it is equally likely that 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?

Ans. The number of total coins in a piggy bank \[=\text{100+50+20+10}=\text{180}\].

That is, the number of total possible outcomes is $180$.

Now, since, in the piggy bank, the number of \[50\] coins are $100$, so there are $100$ favourable outcomes.

Thus, the probability of falling out of a \[50\] p coin $=\frac{100}{180}=\frac{5}{9}$.

(ii) will not be a Rs. ${5}$ coin?

Ans. Except the Rs. $5$ coin, there are \[\text{100+50+20}=\text{170}\] coins.

So, the number of favourable outcomes is $170$.

Thus, the probability of falling out of a coin other than Rs. $5$ coin $=\frac{170}{180}=\frac{17}{18}$.

7. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing ${5}$ male fishes and ${8}$ female fishes. What is the probability that the fish taken out is a male fish?

Ans. The total number of fish (male and female) in the tank \[=\text{5 + 8}=\text{13}\].

So, the total number of possible events is $13$.

Now, since, the number of male fishes in the tank is $5$, so the number of favourable outcomes is $5$.

Thus, the probability of taking out a male fish $=\frac{5}{13}$.

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

9. (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?

Ans. The number of total possible outcomes is $20$.

The number of defective bulbs is $4$.

Therefore, the favourable outcomes are $4$.

Thus, the probability of getting a defective bulb $=\frac{4}{20}=\frac{1}{5}$.

(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?

Ans. Since, one defective bulb is replaced, so the number of favourable outcomes \[=\text{20 -- 1}=\text{19}\].

Now, there are total $19-4=15$ non-defective bulbs.

That is, the number of favourable outcomes is $15$.

Thus, the probability of getting a non-defective bulb is $=\frac{15}{19}$.

10. 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

Ans. Since, there are a total of $90$discs, so the number of possible outcomes is $90$. Now, the number of two-digit numbers from $1$ to $90$ is \[\text{90 -- 9 }=\text{81}\].

So, the number of favourable outcomes is $81$.

Thus, the probability of getting a disc with a two-digit number $=\frac{81}{90}=\frac{9}{10}$.

(ii) a perfect square numbers.

Ans. From 1 to 90, the perfect squares are 1, 4, 9, 16, 25, 36, 49, 64 and 81.

Favourable outcomes = 9

Hence P (getting a perfect square) =

(iii) a number divisible by ${5}$.

Ans. The numbers divisible by 5 from 1 to 90 are 18.

Favourable outcomes = 18

Hence P (getting a number divisible by 5) =

11. A child has a die whose six faces show the letters as given below: \[{A},{B},{C},{D},{E},{A}\].

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

Ans. The number of total possible outcomes $=6$.

Since, there are $2$ $A's$, so the number of possible outcomes $=2$.

Thus, the probability of getting a letter $A$ $=\frac{2}{6}=\frac{1}{3}$.

(ii) ${D}$?

Ans. Since, there is only one $D$, so the number of favourable outcomes is $1$.

Therefore, the probability of getting a letter $D$$=\frac{1}{6}$

12. Suppose you drop a die at random on the rectangular region shown in the figure given on the next page. What is the probability that it will land inside the circle with diameter ${1}$ m?

Ans. The area of rectangle (in the given figure) $=3\times 2=6\text{ }{{\text{m}}^{2}}$

Also, the area of the circle inside the rectangle,

$\begin{align} & =\pi {{r}^{2}} \\ & =\pi \left( {{\frac{1}{2}}^{2}} \right) \\ & =\frac{\pi }{4}\text{ }{{\text{m}}^{2}} \\ \end{align}$

Thus, the probability that the die will land inside the circle $=\frac{\frac{\pi }{4}}{6}=\frac{\pi }{24}$.

13. 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 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

Ans. The total number of ball pens is $144$.

There are $20$ defective pens.

So, the number of good pens \[=\text{144 --20}=\text{124}\].

That is, there are $124$ favourable outcomes.

Hence, the probability that she will buy $=\frac{124}{144}=\frac{31}{36}$.

(ii) she will not buy it?

Ans. There are $20$ favourable outcomes.

Thus, the probability that she will not buy $=\frac{20}{144}=\frac{5}{36}$.

14. 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.

Ans. Suppose that the number of blue balls in the bag is $x$.

Then, the probability that a blue ball is drawn, ${{P}_{1}}=\frac{x}{x+5}$.

Also, the probability that a red ball is drawn, ${{P}_{2}}=\frac{5}{x+5}$.

$\begin{align} & {{P}_{1}}=2{{P}_{2}} \\ & \Rightarrow \frac{x}{x+5}=2\cdot \frac{5}{x+5} \\ & \Rightarrow \frac{x}{{x+5}}=2\cdot \frac{5}{{x+5}} \\ & \Rightarrow x=10 \\ \end{align}$

15. 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 ${x}$.

Ans. It is given that the box contains $12$ balls.

So, there are a total of $12$ possible outcomes.

Now, let there are $x$ favourable outcomes while drawing a black ball.

Then, the probability of getting a black ball, ${{P}_{1}}=\frac{x}{12}$.

Now, if \[6\] more balls are put in the box, then the number of possible outcomes becomes \[=\text{12 + 6}=\text{18}\].

Then, there are $x+6$ favourable outcomes.

Therefore, now, the probability of getting a black ball, ${{P}_{2}}=\frac{x+6}{18}$.

So, the given condition,

$\begin{align} & {{P}_{2}}=2{{P}_{1}} \\ & \Rightarrow \frac{x+6}{18}=2\cdot \frac{x}{12} \\ & \Rightarrow \frac{x+6}{18}=\frac{x}{6} \\ & \Rightarrow 6x+36=18x \\ & \Rightarrow 12x=36 \\ & \Rightarrow x=3 \\ \end{align}$

Thus, the value of $x$ is $3$.

16. 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 $\frac{{2}}{{3}}$. Find the number if blue marbles in the jar.

Ans. Since, the jar contains $24$ marbles, so the number of favourable outcomes is $=24$.

Suppose that there are $x$ green marbles.

Therefore, the number of favourable outcomes is $x$.

So, the probability of getting a green marble is $\frac{x}{24}$.

It is given that the probability that the marble drawn is green $=\frac{2}{3}$.

That is, $\frac{x}{24}=\frac{2}{3}$

$\Rightarrow x=16$

So, the number of green marbles is $16$.

Hence, the number of blue marbles \[=\text{24 -- 16}=\text{8}\].

17. Why is tossing a coin considered is the way of deciding which team should get the ball at the beginning of a football match?

Ans. When a coin is tossed, the probability of getting head $=\frac{1}{2}$ and

probability of getting a tail $=\frac{1}{2}$.

So, both probabilities are the same.

That’s why, tossing a coin is considered to be a fair way of deciding which team should get the ball.

18. An unbiased die is thrown, what is the probability of getting an even prime number?

Ans. When an unbiased die is thrown, the list of $6$ outcomes is given by \[\text{1,2,3,4,5,6}\].

Also, the number of favourable outcomes is $1$ (Since, $2$ is the only number which is even and prime.)

Thus, the probability of getting an even prime number is $\frac{1}{6}$.

19. Two unbiased coins are tossed simultaneously, find the probability of getting two heads.

Ans. The list of $4$ outcomes is given by \[\text{HH, HT, TH, TT}\].

The only favourable outcome is \[HH\] [Two heads]

Thus, the probability of getting two heads while tossing two unbiased coins simultaneously $=\frac{1}{4}$.

20. One card is drawn from a well shuffled deck of ${52}$ cards. Find the probability of getting a jack of hearts.

Ans. The number of total outcomes is $52$.

The only favourable case is $1$. (since, there exists only one jack of hearts)

Hence, the probability of getting a jack of hearts $=\frac{1}{52}$.

21. 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 fails and loses otherwise. Calculate the probability that Hanif will lose the game.

Ans. It is given that a coin is tossed thrice.

So, the list of possible outcomes is given by \[\left\{ \text{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} \right\}\].

Therefore, the number of possible outcomes is $8$.

The outcomes with \[3\] heads and \[3\] tails are \[\text{HHH,TTT}\].

So, the number of favourable outcomes is $2$.

Thus, the probability that Hanif will win the game $\frac{2}{8}=\frac{1}{4}$

The probability that Hanif will lose the game $=1-\frac{1}{4}=\frac{3}{4}$.

22. Gopy 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. What is the probability that the fish taken out is a male fish?

Ans. The number of total fishes \[=\text{5+8}=\text{13}\].

There are $5$ male fishes.

Thus, the probability that the fish taken out is a male $=\frac{5}{13}$.

23. A lot consists of \[{144}\] ball pens of which \[{20}\] are defective and the others are food. Arti will buy a pen if it is good but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that

The probability that she will buy $=\frac{124}{144}=\frac{31}{36}$.

Ans. The probability that she will not buy $=\frac{20}{144}=\frac{5}{36}$.

24. Harpreet tosses two different coins simultaneously (say one is of Rs ${1}$ and other is Rs ${2}$), what is the probability that she gets “at least one head”?

Ans. The list of $4$ possible outcomes is given by \[\left\{ \text{HH,TT,TH,HT} \right\}\].

The list of $3$ outcomes when at least one head is obtained is given by $\left\{ \text{TH,HT,HH} \right\}$.

Thus, the probability of getting at least one head $=\frac{3}{4}$.

25. Why is tossing a coin considered is the way of deciding which team should get the ball at the beginning of a football match?

That is, both the probabilities are the same.

26. Two unbiased coins are tossed simultaneously, find the probability of getting two heads.

Ans. The list of $4$ outcomes while tossing two unbiased coins is given by \[\text{HH, HT, TH, TT}\].

Therefore, the only favourable outcome is $HH$.

Hence, the probability of getting two heads while tossing two unbiased coins $=\frac{1}{4}$.

27. One card is drawn from a well shuffled deck of ${52}$ cards. Find the probability of getting a jack of hearts.

The number of favourable cases is $1$. (since, in a deck of playing cards, there is only one jack of hearts)

Hence, the probability of getting a jack of hearts is $=\frac{1}{52}$.

28. If two dice are thrown once, find the probability of getting ${9}$.

Ans. The number of total possible outcomes while throwing two dice is $6\times 6=36$.

The possible outcomes for getting $9$ are \[\left( \text{3+6} \right)\text{,}\left( \text{4+5} \right)\text{,}\left( \text{5+4} \right)\text{,}\left( \text{6+3} \right)\].

That is, the favourable outcomes are $4$.

Hence, the probability of getting $9$ while throwing two dice is $=\frac{4}{36}=\frac{1}{9}$.

29. A card is drawn from a well shuffled deck of playing cards. Find the probability of getting a face card.

Ans. It is known that the deck of playing cards contains $52$ cards. So, the number of total possible outcomes is $52$.

The number of favourable outcomes is \[=\text{4+4+4}=\text{12}\], where $4$ cards are jack, $4$ cards are queen and $4$ cards are king.

Thus, the probability of getting a face card while drawing a random card from the well shuffled deck is $=\frac{12}{52}=\frac{3}{13}$.

30. What is the probability of having ${53}$ Mondays in a leap year?

Ans. It is known that, there is $366$ days in a leap year containing $52$weeks and $2$ days.

31. Cards bearing numbers ${3}$ to \[{20}\] are placed in a bag and mixed thoroughly. A card is taken out from the bag at random, what is the probability that the number on the card taken out is an even number?

Ans. The number of total outcomes is \[=203=17\].

The list of $9$ even numbers marked on the cards is given by\[\text{4,6,8,10,12,14,16,18,20}\].

So, there are a total of $9$ favourable outcomes.

Thus, the probability of getting a card marked with an even number is $=\frac{9}{17}$.

3 Marks Questions

1. 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 figure) and these are equally likely outcomes. What is the probability that it will point at:

(i) ${8}$?

Ans. An arrow can point to any of the $8$ numbers in eight different ways.

Therefore, there are $8$ possible outcomes.

So, the number $8$ can appear only one way.

Thus, the probability that the arrow points at $8$ is $=\frac{1}{8}$.

(ii) an odd number?

Ans. In the circle, there are only $4$ odd numbers, viz., $1,3,5,7$.

So, there are $4$ favourable outcomes.

Thus, the probability that the arrow points at an odd number is $=\frac{4}{8}=\frac{1}{2}$.

(iii) a number greater than ${2}$?

Ans. In the circle, the numbers greater than $2$ are $3,4,5,6,7,8$.

So, there are a total of $6$ favourable outcomes.

Thus, the probability that the arrow points at a number greater than $2$ is $=\frac{6}{8}=\frac{3}{4}$.

(iv) a number less than ${9}$?

Ans. In the circle, the numbers that are less than $9$ are $1,2,3,4,5,6,7,8$.

So, there are $8$ possible outcomes.

Thus, the probability that the arrow points at a number less than $9$, is $=\frac{8}{8}=1$.

2. A dice is thrown once. Find the probability of getting:

(i) a prime number.

Ans. The total number of favourable outcomes from a dice roll is $6$.

The prime numbers that can appear in the dice are \[\text{2, 3 and 5}\].

Thus, the number of possible outcomes is $3$.

So, the probability of getting a prime number is $=\frac{3}{6}=\frac{1}{2}$

(ii) a number lying between ${2}$ and ${6}$.

Ans. On a dice, the numbers which lie between $2$ and $6$ are \[\text{3, 4, 5}\].

So, there are a total of $3$ favourable outcomes.

Thus, the probability of getting a number lying between $2$ and $6$ is $=\frac{3}{6}=\frac{1}{2}$.

(iii) an odd number.

Ans. The odd numbers on a dice are \[\text{1, 3 and 5}\].

Thus, the probability of obtaining an odd number is $=\frac{3}{6}=\frac{1}{2}$.

3. 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.

Ans. When a coin is tossed three times, then the outcomes are: \[\text{HHH, HHT, HTH, THH, TTH, HTT, THT, TTT}\].

So, there are a total of $8$ possible outcomes.

The outcomes when he loses the game are: $HHT,\text{ HTH, THH, TTH, HTT, THT}$.

Therefore, there are a total of $6$ favourable outcomes.

Thus, the probability of losing the game is $=\frac{6}{8}=\frac{3}{4}$.

4. 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:

Ans. The following describes the complete table:

There are total $6$ rows and $6$ columns in the table, so $6\times 6=36$ possible outcomes can be obtained.

Now, the number of favourable outcomes of obtaining a total score that is even is $18$.

Thus, the probability of obtaining a total score even is $=\frac{18}{36}=\frac{1}{2}$.

Ans. The favourable outcomes of obtaining a total score $6$ are $4$.

Thus, the probability of obtaining a total score of $6$ is $=\frac{4}{36}=\frac{1}{9}$.

(iii) at least ${6}$?

Ans. The favourable outcomes of obtaining a total score at least $6$ are \[15\]. Thus, the probability of obtaining the total score at least $6$ is $=\frac{15}{36}=\frac{5}{12}$.

5. \[{18}\] cards numbered \[{1,2,3,\ldots ,18}\] are put in a box and mixed thoroughly. A card is drawn at random from the box. Find the probabilities that the card bears

(i) an even number.

Ans. From $1$ to $18$, there are $18$ cards.

So, the number of possible outcomes is $18$.

Now, the list of the $9$ even numbers marked on the cards is given by \[\text{2,4,6,8,10,12,14,16,18}\].

Thus, the probability of drawing a card marked with an even number $=\frac{9}{18}=\frac{1}{2}$.

(ii) a number divisible by ${2}$ or ${3}$.

Ans. The list of $12$ numbers that are divisible by $2$ or $3$ are \[\text{2,3,4,6,8,9,10,12,14,15,16,18}\].

So, the number of favourable outcomes is $12$.

Thus, the probability of drawing a card marked with a number divisible by $2$ or $3$ is $=\frac{12}{18}=\frac{2}{3}$.

6. A bag contains \[{5}\] red balls, \[{4}\] green balls and \[{7}\] white balls. A ball is drawn at random from the box. Find the probability that the ball drawn is

Ans. The bag contains a total of $5+4+7=16$ balls.

So, there are a total of $16$ possible outcomes.

Now, the number of white balls in the bag is $7$, that is, the number of favourable outcomes is $7$.

Thus, the probability of drawing a white ball is $=\frac{7}{16}$.

(b) neither red nor white

Ans. The number of balls that are neither red nor white $=$ the number of green balls $=4$.

So, the number of possible outcomes is $4$.

Hence, the probability of drawing a ball that is neither red nor white $=\frac{4}{16}=\frac{1}{4}$.

7. A box contains \[{20}\] balls bearing numbers \[{1,2,3,4,\ldots ,20}\]. A ball is drawn at random from the box, what is the probability that the number on the ball is

(i) an odd number

Ans. From $1$ to $20$, there are a total $20$ balls, that is, the number of total outcomes is $20$.

Now, the list of $10$ odd numbers from $1$ to $20$ is given by \[\text{1,3,5,7,9,11,13,15,17,19}\].

Thus, the probability of drawing a ball bearing an odd number is $=\frac{10}{20}=\frac{1}{2}$.

(ii) divisible by \[{2}\] or \[{3}\]

Ans. The list of $10$ numbers that are divisible by $2$ is given by

\[\text{2,4,}\underline{\text{6}}\text{,8,10,}\underline{\text{12}}\text{,14,16,}\underline{\text{18}}\text{,20}\].

The list of $6$ numbers that are divisible by $3$ is given by

\[\text{3,}\underline{\text{6}}\text{,9,}\underline{\text{12}}\text{,15,}\underline{\text{18}}\].

Now, the $3$ numbers that are divisible by both $2$ and $3$ are \[\text{6,12,18}\].

So, the numbers divisible by $2$ or $3$ are $=\text{10+6 --3}=\text{13}$.

Thus, there are a total of $13$ possible favourable outcomes.

Hence, the probability of drawing a ball bearing a number that is divisible by $2$ or $3$ is $=\frac{13}{20}$.

(iii) prime number

Ans. The list of $8$ prime numbers between $1$ and $20$ is given by \[\text{2,3,5,7,11,13,17,19}\].

So, the number of favourable outcomes is $8$.

Thus, the probability of drawing a ball bearing a prime number is $=\frac{8}{20}=\frac{2}{5}$.

8. A bag contains ${5}$ red and some blue balls,

(i) if probability of drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag.

Ans. Suppose that there are $x$ blue balls in the bag.

So, there are a total of $x+5$ balls in the bag.

Therefore, the probability of drawing a blue ball $=\frac{x}{x+5}$ and

the probability of drawing a red ball $=\frac{5}{x+5}$.

$\frac{x}{{x+5}}=2\cdot \frac{5}{{5+x}}$ $\Rightarrow x=10$

Thus, there are a total of $10$ blue balls in the bag.

(ii) if probability of drawing a blue ball from the bag is four times that of a red ball, find the number of blue balls in the bag.

Ans. According to the given condition,

$\frac{x}{5+x}=4\cdot \frac{5}{5+x}$

$\Rightarrow \frac{x}{{5+x}}=4\cdot \frac{5}{{5+x}}$

$\Rightarrow x=20$

Thus, there are a total of $20$ blue balls in the bag.

9. A box contains ${3}$ blue marbles, ${2}$ white marbles. If a marble is taken out at random from the box, what is the probability that it will be a white one? Blue one? Red one?

Ans. The number of total possible outcomes is $3=2+4=9$.

There are a total of $2$ white marbles.

So, the probability of taking a white marble $=\frac{2}{9}$.

There are a total of $3$ blue marbles.

Therefore, the probability of taking a blue marble $=\frac{3}{9}=\frac{1}{3}$.

Also, there are a total of $4$ red marbles.

Thus, the probability of getting a red marble $=\frac{4}{9}$.

10. The integers from 1 to 30 inclusive are written on cards (one number on one card). These card one put in a box and well mixed. Joseph picked up one card. What is the probability that his card has

(i) number ${7}$

Ans. From $1$ to $30$, there are a total of $30$ integers.

So, there are a total of $30$ possible outcomes.

Thus, the probability of getting the number $7$ is $=\frac{1}{30}$

(ii) an even number

Ans. The list of $15$ even numbers from $1$ to $30$ is given by \[\text{2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}\].

So, there are a total of $15$ favourable outcomes.

Hence, the probability of picking up an even number $=\frac{15}{30}=\frac{1}{2}$.

(iii) a prime number

Ans. The list of $10$ prime numbers from $1$ to $30$ is given by \[\text{2,3,5,7,11,13,17,19,23,29}\].

So, there are a total of $10$ favourable outcomes.

Hence, the probability of picking up a prime number $=\frac{10}{30}=\frac{1}{3}$.

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

Ans. It is given that the bag has only lemon flavoured candies.

So, the probability of getting an orange flavoured candy $=\frac{0}{1}=0$.

Ans. It is provided that the bag contains only the lemon flavoured candies.

So, the probability of getting an orange flavoured candy $=\frac{1}{1}=1$.

12. A bag contains ${6}$ red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red, find the number of blue balls in the bag.

Ans. Let there be $x$ blue balls in the bag.

So, the bag contains a total of $x+6$ balls.

Therefore, the probability of drawing a blue ball $=\frac{x}{x+6}$ and

the probability of drawing a red ball $=\frac{6}{x+6}$.

$\frac{x}{{x+6}}=2\cdot \frac{6}{{6+x}}$

$\Rightarrow x=12$

Thus, there are a total of $12$ blue balls in the bag.

13. A bag contains \[{5}\] red, \[{4}\] black and \[{3}\] green balls. A ball is taken out of the bag at random, find the probability that the selected ball is

(i) of red colour.

Ans. The bag contains total \[5+4+3=12\] balls, that is, there are a total of $12$ outcomes.

Also, the number of red balls contained in the bag is $5$.

Hence, the probability of selecting a red ball is $=\frac{5}{12}$.

(ii) not of green colour.

Ans. The number of balls except the green ones \[=\text{12--3}=\text{9}\].

So, the number of favourable outcomes is $9$.

Thus, the probability of selecting a ball that is not green is $=\frac{9}{12}=\frac{3}{4}$.

14. From a well shuffled pack of \[{52}\] cards, black aces and black queens are removed. From the remaining cards a card is drawn at random, find the probability of drawing a king or a queen.

Ans. A pack of cards contains $52$ cards. From them, there are only $2$ black ace cards and $2$ black queen cards.

Therefore, the number of cards left $=52-2-2=48$.

So, there are $48$ equally likely cases available.

Also, in $48$ cards, there are a total \[4+2=6\] queens and kings left. Thus, the number of favourable cases is $6$.

Hence, the probability of drawing a king or a queen is $=\frac{6}{48}=\frac{1}{6}$.

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

Ans. When a driver tries to start a car, the likelihood of the car starting or not starting are not equal.

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

Ans. When a player tries to shoot a basketball, the likelihood of hitting the target or missing the shot are not equal.

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

Ans. A baby is born; whether it is a boy or a girl, is an equally likely event.

16. Find the probability that a number selected at random from the numbers \[{1,2,3,\ldots ,35}\] is a

(i) prime number

Ans. The list of $11$ prime numbers between $1$ and $35$ is given by\[\text{2,3,5,7,11,13,17,19,23,29,31}\].

Between $1$ and $35$, there are total $35$ numbers, so the number of total outcomes is $35$.

Thus, the probability that the chosen number is a prime $=\frac{11}{35}$.

(ii) multiple of ${7}$

Ans. The list of $5$ numbers between $1$ and $35$ that are multiple of $7$ is given by \[\text{7,14,21,28,35}\].

Thus, the probability of choosing a number that is multiple of $7$ $=\frac{5}{35}=\frac{1}{7}$

(iii) multiple of ${3}$ or ${5}$.

Ans. The list of $11$ numbers between $1$ and $35$ which are multiple of $3$ is given by \[3,6,9,12,\underline{15},18,21,24,27,\underline{30},33\].

Also, the list of $7$ numbers between $1$ and $35$ which are multiple of $5$ is given by \[5,10,\underline{15},20,25,\underline{30},35\].

Thus, the numbers that are multiple of both $3$ and $5$ are \[15,30\].

So, the numbers which are multiple of $3$ or $5$ is given by

$\begin{align} & =11+7-2 \\ & =16 \\ \end{align}$

Thus, the probability of getting a number that is multiple of $3$ or $5$ is

$=\frac{16}{35}$.

4 Marks Questions

1. One card is drawn from a well-shuffled deck of ${52}$ cards. Find the probability of getting:

(i) a king of red colour.

Ans. Total number of outcomes \[=52\].

There are two suits of red cards, that is, diamond and heart.

Each of the suits has one king.

So, there is only one favourable outcome.

Thus, the probability of getting a king of red colour $=\frac{2}{52}=\frac{1}{26}$.

(ii) a face card

Ans. The number of face cards in a pack is \[12\].

So, the number of favourable outcomes is $12$.

Thus, the probability of getting a face card is $=\frac{12}{52}=\frac{3}{13}$.

(iii) a red face card

Ans. It is known that there are $2$ suits of red cards, that is, diamond and heart and each of the suits contain $3$ face cards.

Thus, the probability of getting a red face card $=\frac{6}{52}=\frac{3}{26}$.

(iv) the jack of hearts

Ans. Recall that, in a deck of $52$ cards, there is only one jack of heart.

So, the number of favourable outcomes is $1$.

Thus, the probability of getting the jack of hearts $=\frac{1}{52}$.

(v) a spade

Ans. It is known that there are \[13\] cards of spades.

Thus, the probability of getting a spade is $=\frac{13}{52}=\frac{1}{4}$.

(vi) the queen of diamonds.

Ans. Note that, there is only one queen of diamonds.

So, the number of favourable outcomes is $1$.

Thus, the probability of getting the queen of diamonds $=\frac{1}{52}$.

2. A die is thrown twice. What is the probability that:

(i) ${5}$ will not come up either time?

Ans. The following are the outcomes of the experiment where a dice is thrown twice:

\[\left( \text{1, 1} \right)\text{ }\left( \text{1, 2} \right)\text{ }\left( \text{1, 3} \right)\text{ }\left( \text{1, 4} \right)\text{ }\left( \text{1, 5} \right)\text{ }\left( \text{1, 6} \right)\]

\[\left( \text{2, 1} \right)\text{ }\left( \text{2, 2} \right)\text{ }\left( \text{2, 3} \right)\text{ }\left( \text{2, 4} \right)\text{ }\left( \text{2, 5} \right)\text{ }\left( \text{2, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{3, 1} \right)\text{ }\left( \text{3, 2} \right)\text{ }\left( \text{3, 3} \right)\text{ }\left( \text{3, 4} \right)\text{ }\left( \text{3, 5} \right)\text{ }\left( \text{3, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{4, 1} \right)\text{ }\left( \text{4, 2} \right)\text{ }\left( \text{4, 3} \right)\text{ }\left( \text{4, 4} \right)\text{ }\left( \text{4, 5} \right)\text{ }\left( \text{4, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{5, 1} \right)\text{ }\left( \text{5, 2} \right)\text{ }\left( \text{5, 3} \right)\text{ }\left( \text{5, 4} \right)\text{ }\left( \text{5, 5} \right)\text{ }\left( \text{5, }\!\!~\!\!\text{ 6} \right)\]

\[\left( \text{6, 1} \right)\text{ }\left( \text{6, 2} \right)\text{ }\left( \text{6, 3} \right)\text{ }\left( \text{6, 4} \right)\text{ }\left( \text{6, 5} \right)\text{ }\left( \text{6, }\!\!~\!\!\text{ 6} \right)\]

So, there are a total $36$ outcomes.

Now let $A$ denotes the event that first throw shows $5$ and

let $B$ denote the event that the second throw shows $5$.

Thus, there are a total $6$ favourable outcomes in each of the cases.

Therefore, $P\left( A \right)=\frac{6}{36}$ and $P\left( B \right)=\frac{6}{36}$

$\Rightarrow P\left( \overline{A} \right)=1-\frac{6}{36}=\frac{30}{36}=\frac{5}{6}$ and $P\left( \overline{B} \right)=\frac{5}{6}$.

Hence, the probability that $5$ will not come in either time

$\begin{align} & =\frac{5}{6}\times \frac{5}{6} \\ & =\frac{25}{36} \\ \end{align}$

(ii) \[{5}\] will come up at least once?

Ans. Suppose that, \[S\] is the sample space related to the random experiment of throwing a die two times. So, \[n\left( S \right)=36\].

Therefore, \[A\cap B=\]first and second throw shoe $5$, that is, obtaining $5$ in each throw.

Then, \[\text{A=}\left\{ \left( \text{5,1} \right),\left( \text{5,2} \right),\left( \text{5,3} \right),\left( \text{5,4} \right),\left( \text{5,5} \right),\left( \text{5,6} \right) \right\}\]

and \[\text{B=}\left\{ \left( \text{1,5} \right),\left( \text{2,5} \right),\left( \text{3,5} \right),\left( \text{4,5} \right),\left( \text{5,5} \right),\left( \text{6,5} \right) \right\}\].

\[P\left( A \right)=\frac{6}{36}\],\[P\left( B \right)=\frac{6}{36}\] and \[P\left( A~\cap B \right)=\frac{1}{36}\].

Thus, the probability that $5$ will come at least once in throwing a dice twice, $P\left( A\cap B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cup B \right)$

$\begin{align} & =\frac{6}{36}+\frac{6}{36}-\frac{1}{36} \\ & =\frac{11}{36} \\ \end{align}$

3. 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 the following days?

(i) the same day?

Ans. The list of favourable outcomes related with two consumers visiting a specific shop in the same week (Tuesday to Saturday) is:

\[\left( \text{T,T} \right)\text{ }\left( \text{T,W} \right)\text{ }\left( \text{T,TH} \right)\text{ }\left( \text{T,F} \right)\text{ }\left( \text{T,S} \right)\]

\[\left( \text{W,T} \right)\text{ }\left( \text{W,W} \right)\text{ }\left( \text{W,TH} \right)\text{ }\left( \text{W,F} \right)\text{ }\left( \text{W,S} \right)\]

\[\left( \text{TH, T} \right)\text{ }\left( \text{TH, W} \right)\text{ }\left( \text{TH, TH} \right)\text{ }\left( \text{TH, F} \right)\text{ }\left( \text{TH, S} \right)\]

\[\left( \text{F, T} \right)\text{ }\left( \text{F, W} \right)\text{ }\left( \text{F, TH} \right)\text{ }\left( \text{F, F} \right)\text{ }\left( \text{F, S} \right)\]

\[\left( \text{S,T} \right)\text{ }\left( \text{S, W} \right)\text{ }\left( \text{S, TH} \right)\text{ }\left( \text{S, F} \right)\text{ }\left( \text{S S} \right)\]

Therefore, there are a total of $25$ favourable outcomes.

Now, the outcomes of visiting on the same day can be listed as

\[\left( \text{T,T} \right)\text{,}\left( \text{W,W} \right)\text{,}\left( \text{TH,TH} \right)\text{,}\left( \text{F,F} \right)\] and \[\left( S,S \right)\].

Thus, the probability that both Shyam and Ekta will visit the shop on the same day is $=\frac{5}{25}=\frac{1}{5}$.

(ii) consecutive days?

Ans. The list of favourable outcomes of visiting the shop on consecutive days are \[\left( \text{T, W} \right)\text{,}\left( \text{W, T} \right)\text{,}\left( \text{W, TH} \right)\text{,}\left( \text{TH, W} \right)\text{,}\left( \text{TH, F} \right)\text{,}\left( \text{F, TH} \right)\text{,}\left( \text{S, F} \right)\] and \[\left( F,\text{ }S \right)\].

So, there are a total of $8$ favourable outcomes.

Thus, the probability that both Shyam and Ekta will visit the shop on consecutive days $=\frac{8}{25}$.

(iii) different days?

Ans. There are total \[255=20\] favourable outcomes of visiting on different days.

So, there are a total of $20$ favourable outcomes.

Thus, probability that both Shyam and Ekta will visit the shop on different days $=\frac{20}{25}=\frac{4}{5}$.

4. A card is drawn at random from a well shuffled deck of playing cards. Find the probability that the card drawn is

(i) a card of spades of an ace

Ans. The number of cards in a deck \[=52\].

There are a total of $13$ spades and $4$ aces in a deck of playing cards.

There is one card which is common [i.e., ace of spade]

So, the number of favourable outcomes \[=13+41=16\].

Thus, the probability of getting a card of spades of an ace $=\frac{16}{52}=\frac{4}{13}$.

(ii) a red king

Ans. The number of red kings \[=2\].

Thus, the probability of getting a red king is $=\frac{2}{52}=\frac{1}{26}$.

(iii) neither a king nor a queen.

Ans. The total number of kings and queens \[=4+4=8\].

The number of cards that are neither king nor a queen \[=528=44\].

Thus, the probability of getting neither a king nor a queen is $=\frac{44}{52}=\frac{11}{13}$.

(iv) either a king or a queen

Thus, the probability of getting either a king or a queen is $=\frac{8}{52}=\frac{2}{13}$.

(v) a face card.

Ans. The number of face cards \[=4+4+4=12\] (Jack, queen and king are the face cards)

Therefore, the probability of getting a face card is $=\frac{12}{52}=\frac{3}{13}$.

(vi) cards which is neither king nor a red card.

Ans. The number of cards that are neither red card nor king \[=\text{ }52\left( 26+42 \right)\]

\[\begin{align} & =5228 \\ & =24 \\ \end{align}\]

Thus, the probability of getting a card that is neither a king nor a red card $\begin{align} & =\frac{24}{52} \\ & =\frac{6}{13} \\ \end{align}$

5. Cards marked with numbers \[{1},{2},{3},\ldots ,{25}\] are placed in a box and mixed thoroughly and one card is drawn at random from the box, what is the probability that the number on the card is

(i) a prime number?

Ans. Total number of outcomes that are possible $=25$

There are $9$ favourable cases, which can be listed as \[2,3,5,7,11,13,17,19,23\].

Thus, the probability of getting a prime number $=\frac{9}{25}$.

(ii) a multiple of \[{3}\] or \[{5}\]?

Ans. The numbers that are multiple of $3$ or $5$ can be listed as\[3,5,6,9,10,12,15,18,20,21,24,25\].

So, there are a total of $12$ favourable cases.

Therefore, the probability of getting a multiple of $3$ and $5$ is $=\frac{12}{25}$.

(iii) an odd number?

Ans. The odd numbers can be listed as \[1,3,5,7,9,11,13,15,17,19,21,23,25\].

Therefore, there are a total $13$ favourable cases.

Thus, the probability of getting an odd number $=\frac{13}{25}$.

(iv) neither divisible by \[{5}\] nor by \[{10}\]?

Ans. The numbers that are neither divisible by $5$ nor by $10$ can be listed as \[1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24\].

So, there are a total of $20$ favourable cases.

Thus, the probability of getting a number that is neither divisible by $5$ nor by $10$ is $=\frac{20}{25}=\frac{4}{5}$.

(v) perfect square?

Ans. The perfect square numbers can be listed as \[1,4,9,16,25\].

Therefore, there are a total of $5$ favourable cases.

Thus, the probability of getting a perfect square number $=\frac{5}{25}=\frac{1}{5}$.

(vi) a two-digit number?

Ans. All two-digit numbers can be listed as \[10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25\].

So, there are a total $16$ favourable cases.

Thus, the probability of getting a two-digit number $=\frac{16}{25}$.

6. From a pack of ${52}$ playing cards, jacks, queens, kings and aces of red colour are removed. From the remaining a card is drawn at random. Find the probability that the card drawn is

(i) a black queen

Ans. Total number of cards $=52$.

Number of cards that are removed $=2+2+2+2=8$ ($2$ jack, $2$ queen, $2$ king and $2$ aces of red colour)

The number of cards remains $=52-8=44$

Therefore, the total number of outcomes $=44$

The number of favourable outcomes $=2$ (Since, the number of black queens is $2$)

Thus, the probability of getting a black queen $=\frac{2}{44}=\frac{1}{22}$

(ii) a red card

Ans. The number of favourable outcomes $=$ the number of remaining red cards$=26-8=18$.

Thus, the probability of getting a red card is $=\frac{18}{44}=\frac{9}{22}$.

(iii) a black jack

Ans. The number of favourable outcomes $=$ The number of black jacks$=2$

Therefore, the probability of getting a black jack $=\frac{2}{44}=\frac{1}{22}$.

(iv) a picture cards

Ans. The number of picture cards remains $=2+2+2=6$ [King, queen, jack are picture cards]

Therefore, the probability of getting a picture card is $=\frac{6}{44}=\frac{3}{22}$.

(v) a honourable card

Ans. The number of Honourable cards remains $=2+2+2+2=8$ (ace, jack, queen and king are honourable cards)

Thus, the probability of getting a honourable card $=\frac{8}{44}=\frac{2}{11}$.

CBSE Class 10 Maths Probability Important Questions

Explain the concept of probability mentioned in important questions of probability class 10.

Probability is a branch of mathematics that deals with numerical explanations, determining the outcome of whether an event is true or not. This possibility explains the concept of probability. The numerical value of probability is expressed from zero to one. Probability has been introduced in mathematics to explain the likelihood of an event that is about to take place. This is also helpful in the probability distribution, where one can know the outcome of a random event. The probability of all events in a given sample space adds up to 1. Class 10 probability questions are crucial for students who are appearing for their Class 10th boards.

The Probability Formula is Depicted as Follows

Probability of an event about to occur P(E) = Number of favourable outcomes/Total number of outcomes

Class 10 Maths Chapter 15 Important Questions Based on Exercise?

Q. Which of the given experiments have equally likely outcomes? Explain.

(i) A driver tries to start a car. The car starts or may not start.

(ii) A player strives to shoot a football. He misses or shoots the football.

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

(iv) A child is born. Whether it is a girl or a boy.

Solution:

(i) This statement doesn’t have a likely outcome, as the car may begin to operate or it may not.

(ii) This statement also doesn’t have a likely outcome as the player may shoot or miss the ball.

(iii) This statement has a likely outcome, as the solution has to be either right or wrong.

(iv) This statement has a likely outcome because the newborn child has to be a girl or a boy.

Q. Pull a random card from a pack of cards. What is the probability that the card pulled has a feminine face?

A standard deck of cards has 52 cards.

Total number of outcomes = 52

Number of favorable events = 4 x 1 = 4 (considering Queen only)

Therefore Probability, P = Number of Favourable Outcomes/Total Number of Outcomes = 4/52= 1/13.

Q. If P(E) = 0.25, what is the probability of ‘not E’?

We already know that

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

It is provided that, P(E) = 0.25

So, P(not E) = 1 – P(E)

P(not E) = 1 – 0.25

Hence, P(not E) = 0.75

Q. If 10 defective balls are accidentally mixed with 144 good ones. It is not possible to just look at a ball and tell whether or not it is defective. One ball is drawn out at random from this set of balls. Determine the probability that the ball pulled out is a good one.

Numbers of balls = Numbers of defective balls + Numbers of good balls

∴ Total number of pens = 144 + 10 = 154 pens

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

P(picking a good ball ) = 144/154 = 72/77 = 0.935

Q. One card is taken out from a well-shuffled deck of 52 cards. State the probability that the card will

(i) be a king,

(ii) not be a king.

Well-shuffling of cards ensures fairly possible outcomes.

(i) Card drawn is a king

There are a total of four kings in a deck of cards.

Let A be the event ‘the card is a king’.

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

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

Hence, P(A) = n(A)/n(S) = 4/52 = 1/13

(ii) Card drawn is not a king

Let B be the event ‘card drawn is not a king’.

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

Hence, P(B) = n(B)/n(S) = 48/52 = 12/13

Probability Important Questions Class 10- State the Different Types of Probability?

Here is the answer to probability class 10 most important questions. There are mainly three types of probability.

Theoretical Probability

Experimental probability, axiomatic probability.

It depends on the possible chances of something that is about to occur. The theoretical probability concept originates from the reasoning behind probability. For example, if a dice is rolled, the theoretical probability of getting a sixer will be ⅙.

It depends on the basis of observation of an experiment. The experimental probability is generally calculated by the total number of possible outcomes by the total number of trials taken. For example, if a coin is tossed 10 times and out of that head comes 4 times, then the experimental probability of a head is 4/10 or ⅖.

A set of rules or axioms are established, which are applied to all types in an axiomatic probability. These axioms are set by Kolmogorov and are popularly called the Kolmogorov’s three axioms. In Kolmogorov's axiomatic approach to probability, the chances of non-occurrence and occurrence of an event can be quantified.

Explain Events and Outcomes in Probability Important Questions Class 10?

An outcome is a result obtained through a random experiment. For example, when we toss a coin, the tail is the outcome. An event refers to the set of outcomes. For example, when we roll a dice, the probability of getting a result less than four is an event. This is the answer to important questions of probability class 10.

What are Various Types of Events Based on Probability Class 10 Important Questions?

Here are the following types of events in probability, elementary events .

An event that has only one outcome of an event is referred to as an elementary event. For example: take a coin and toss it in the air for ‘n’ number of times. After the trial of this experiment, it will possibly have two outcomes- Heads and Tails. So, for any individual toss in a coin, the outcome has to be between the head and tail.

In elementary events, the sum of probabilities of all events in an experiment is one. For example- the tossing of a coin experiment P(Heads) + P (Tails)

= (1/2)+ (1/2) =1

Impossible Events

The event that has no chance of happening or occurring is termed as an impossible event, is called as an impossible event, i.e P(E)=0. For example, the probability of getting an eight on a dice is zero. This is because the number 8 can never appear on a dice.

The event where there is a 100% chance of occurrence or happening is termed as a sure event. The probability of occurrence in a sure event is one. For example, the probability of throwing a number less than 7 in dice is 1. So, P(E) = P(Obtaining a number less than 7) = 6/6= 1.

Complementary Event

In complementary events, there are only two outcomes of an event and these are the only two possible outcomes. A simple example is that of a coin where there are only two possible outcomes, the heads, and tails. P(E) + P (È) = 1, where E and È are two complementary events. The event È represents the complement of the event E.

What is Probability Density Function?

The probability density function is an important question of probability class 10 . It is represented as a density of a continuous random variable that comes between a certain range of values. It explains the normal distribution and how the mean distribution occurs. The standard normal distribution helps create the statistics and database, which are applicable in science for representing the real-valued variable, whose distribution is unknown.

How Class 10 Probability Questions are Crucial for Preparation?

By referring to CBSE class 10 maths probability important questions , students can stay ahead of the competition.

They can practice important questions and recognize their mistakes.

Students can make notes and highlight the vital information based on these questions.

By solving the important questions, the students can revise the probability topic very well.

These crucial questions are also valuable to help solve their homework problems.

The questions will allow students to understand various sums given in this probability chapter.

Overall, these important questions serve as a resourceful study guide for the students.

Important Related Links for CBSE Class 10 Maths

Vedantu's Probability Class 10 Notes for CBSE Maths Chapter 15 offer a comprehensive and well-structured resource for students seeking to master this fundamental concept. Through their free PDF download, Vedantu provides a valuable aid to learners, simplifying complex probability principles with clarity and precision. These notes enable students to understand probability's theoretical foundations and practical applications, enhancing their problem-solving skills. The user-friendly format and engaging examples ensure an enjoyable learning experience. Vedantu's commitment to providing high-quality educational materials empowers students to excel in their studies, instilling confidence and proficiency in probability, a skill that extends beyond mathematics into various real-world scenarios. By making these notes freely accessible, Vedantu exemplifies its dedication to fostering academic excellence for all.

FAQs on Important Questions for CBSE Class 10 Maths Chapter 15 - Probability

1. What is the difference between theoretical probability and experimental probability?

Theoretical probability is the probability of an event occurring based on the assumptions that the events are equally likely and that the experiment is repeated an infinite number of times. Experimental probability is the probability of an event occurring based on the results of a finite number of trials.

2. What are the addition and multiplication theorems of probability?

The addition theorem of probability states that the probability of two events occurring is equal to the sum of the probabilities of each event occurring, if the events are mutually exclusive. Mutually exclusive events are events that cannot occur at the same time. The multiplication theorem of probability states that the probability of two events occurring is equal to the product of the probabilities of each event occurring, if the events are not mutually exclusive. Not mutually exclusive events are events that can occur at the same time.

3. What are some applications of probability in real life?

Probability is used in a variety of fields, such as gambling, insurance, weather forecasting, marketing, and medicine.

Gambling : Probability is used to calculate the odds of winning a particular bet. For example, in a game of roulette, the probability of winning a bet on red is 1/2.

Insurance: Probability is used to calculate the risk of a particular event occurring and to set premiums accordingly. For example, an insurance company might charge a higher premium for a house in a flood-prone area than for a house in a low-risk area.

Weather forecasting: Probability is used to predict the likelihood of different weather conditions occurring. For example, a weather forecaster might say that there is a 60% chance of rain tomorrow.

Marketing: Probability is used to target advertising campaigns to specific groups of people. For example, an advertiser might target their ads to people who are likely to be interested in their product based on their age, gender, location, and other factors.

Medicine: Probability is used to diagnose diseases and to assess the risks and benefits of different treatments. For example, a doctor might use a probability model to determine the likelihood that a patient has a particular disease based on their symptoms.

4. What are some tips for solving probability problems?

Here are some tips for solving probability problems:

Read the problem carefully and identify the relevant information.

List the possible outcomes of the experiment.

Identify the favorable outcomes.

Calculate the probability of the event by dividing the number of favorable outcomes by the total number of possible outcomes.

Check your answer to make sure it makes sense.

5. What is the meaning of "sample space"?

The sample space is the set of all possible outcomes of an experiment. For example, if you roll a die, the sample space is the set of numbers 1, 2, 3, 4, 5, and 6.

6. What is an elementary event?

An elementary event is a single outcome of an experiment. For example, if you roll a die, one elementary event is the outcome of getting a 1.

7. What is a favorable outcome?

A favorable outcome is an outcome of an experiment that satisfies a particular condition. For example, if you roll a die and you want to get a number greater than 3, the favorable outcomes are 4, 5, and 6.

8. What is the probability of an event?

The probability of an event is the likelihood of that event occurring. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For example, the probability of getting a head when a coin is tossed is 1/2.

11. 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 which is a perfect square.

Total number of cards = 50 – 6 + 1 = 45 Perfect square numbers are 9, 16, 25, 36, 49, i.e., 5 numbers ∴ P(a prefect square) = $\dfrac{5}{45}$ = $\dfrac{1}{9}$

12. From a pack of 52 playing cards, Jacks, Queens and Kings of red colour are removed. From the remaining, a card is drawn at random. Find the probability that drawn card is:

1. A black king.

2. A card of red colour.

3. A card of black colour.

Removed red colour cards = 3×2 = 6 Remaining cards = 52 – 6 = 46

1. Number of black kings = 2 P(a black king) = $\dfrac{2}{46}$=$\dfrac{1}{23}$

2. Number of red colour cards = 26 Remaining red colour cards = 26 – 6 = 20 P(a card of red colour) =$\dfrac{20}{46}$=$\dfrac{10}{23}$

3. Number of black cards = 26 P(a black colour card) = $\dfrac{26}{46}$=$\dfrac{13}{23}$

13. Three distinct coins are tossed together. Find the probability of getting:

I. at least 2 heads

II. at most 2 heads.

When three coins are tossed following is the sample space: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

I. P(at least 2 heads) = $\dfrac{4}{8}$ =$\dfrac{1}{2}$

II. P(at most 2 heads) = $\dfrac{7}{8}$

14. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. Find the probability that the selected ticket has a number which is a multiple of 5.

Total number of tickets = 40 Multiples of 5 are 5,10,15,20,25,30,35,40 Number of favourable tickets = 8 $\therefore$ P(a number which is a multiple of 5) =$\dfrac{8}{40}$=$\dfrac{1}{5}$

15. How students can score good marks in Mathematics class 10.

Students can score good marks in mathematics class 10 by revising each chapter 2-3 times. Also, they need to solve sample papers to improve their time management and accuracy. Students can solve sample papers on Vedantu site.

CBSE Class 10 Maths Important Questions

Cbse study materials.

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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CBSE Class 10 Maths Case Study Questions for Chapter 14 - Statistics (Published By CBSE)

Case study question bank for cbse class 10 maths chapter 14 - statistics is available here. practice this new format of questions to score good marks in your board exam..

CBSE Class 10 Maths Case Study Questions for Chapter 14 - Statistics are published by the CBSE board itself. These questions are perfect to acquaint with the new format of the questions and make your board exam preparations. Questions are based on the real life situations. You can easily understand how concepts and logic are used in the case study questions. All the questions are provided with answers.

Check Case Study Questions for Class 10 Maths Chapter 14 - Statistics

CASE STUDY 1:

COVID-19 Pandemic The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans.

The following tables shows the age distribution of case admitted during a day in two different hospitals

Refer to table 1

1. The average age for which maximum cases occurred is

Answer: c) 36.82

2. The upper limit of modal class is

Answer: d) 45

3. The mean of the given data is

Answer: d) 35.4

Refer to table 2

4. The mode of the given data is

Answer: a) 41.4

5. The median of the given data is

Answer: b) 40.2

Electricity Energy Consumption

CASE STUDY 2:

Electricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption).

Refer to data received from Colony A

1. The median weekly consumption is

a) 12 units

b) 16 units

c) 20 units

d) None of these

Answer: c) 20 units

2. The mean weekly consumption is

a) 19.64 units

b) 22.5 units

c) 26 units

Answer: a) 19.64 units

3. The modal class of the above data is I

Answer: c) 20-30

Refer to data received from Colony B

4. The modal weekly consumption is

a) 38.2 units

b) 43.6 units

d) 32 units

Answer: b) 43.6 units

5. The mean weekly consumption is

a) 15.65 units

b) 32.8 units

c) 38.75 units

d) 48 units

Answer: c) 38.75 units

Also Check:

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CBSE Class 10 Maths Case Study Questions for Chapter 14
Case study question bank for CBSE Class 10 Maths Chapter 14 - Statistics is available here. Practice this new format of questions to score good marks in your board exam. By Gurmeet Kaur