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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

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

Probability Case Study Questions With Answers

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

Case Study/Passage-Based Questions

Question 1:

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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Case Study Class 10 Maths Questions

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

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

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

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

Class 10 Maths has the following chapters.

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

Format of Maths Case-Based Questions

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

Maths Case Study Question Paper 2023

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

Case Study Question in Mathematics

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

Case Study Question – 1

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

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

Case Study Question – 2

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

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

Case Study Question – 3

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

Case Study Question – 4

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

Case Study Question – 5

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

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

Case Study Question – 6

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

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

More Case Study Questions

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

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

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

How to Solve Case-Based Questions?

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

What are Class 10 competency-based questions?

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

Case Study Questions in Maths Question Paper

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

Case Studies on CBSE’s Official Website

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

10 Maths Case Studies in myCBSEguide App

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

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

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

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

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

CASE STUDY 1:

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

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

Answer: a) 1/26

2. Find the probability of getting a face card.

Answer: d) 3/13

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

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

Answer: a) 3/13

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

Answer: d) 1/4

CASE STUDY 2:

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

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

Answer: b) 5/36

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

Answer: d) 0

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

Answer: a) 1

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

Answer: c) 1/6

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

Answer: d) 5/18

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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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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Computational Probability and Statistics

7 case study, 7.1 objectives.

• Use R to simulate a probabilistic model.
• Use basic counting methods.

7.2 Introduction to probability models

In this second block of material we will focus on probability models. We will take two approaches, one is mathematical and the other is computational. In some cases we can use both methods on a problem and in others only the computational approach is feasible. The mathematical approach to probability modeling allows us insight into the problem and the ability to understand the process. Simulation has a much greater ability to generalize but can be time intensive to run and often requires the writing of custom functions.

This case study is extensive and may seem overwhelming, do not worry we will discuss these ideas again in the many lessons we have coming up this block.

7.3 Probability models

Probability models are an important tool for data analysts. They are used to explain variation in outcomes that cannot be explained by other variables. We will use these ideas later in this text to help us make decisions about our statistical models.

Often probability models are used to answer a question of the form “What is the chance that …..?” This means that we typically have an experiment or trial where multiple outcomes are possible and we only have an idea of the frequency of those outcomes. We use this frequency as a measure of the probability of a particular outcome.

For this block we will focus just on probability models. To apply a probability model we will need to

• Select the experiment and its possible outcomes.
• Have probability values for the outcomes which may include parameters that determine the probabilities.
• Understand the assumptions behind the model.

7.4 Case study

There is a famous example of a probability question that we will attack in this case study. The question we want to answer is “In a room of \(n\) people what is the chance that at least two people have the same birthday?”

Exercise : The typical classroom at USAFA has 18 students in it. What do you think the chance that at least two students have the same birthday? 62

7.4.1 Break down the question

The first action we should take is to understand what is being asked.

• What is the experiment or trial?
• What does it mean to have the same birthday?
• What about leap years?
• What about the frequency of births? Are some days less likely than others?

Exercise : Discuss these questions and others that you think are relevant. 63

The best first step is to make a simple model, often these are the only ones that will have a mathematical solution. For our problem this means we answer the above questions.

• We have a room of 18 people and we look at their birthdays. We either have two or more birthdays matching or not; thus there are two outcomes.
• We don’t care about the year, only the day and month. Thus two people born on May 16th are a match.
• We will ignore leap years.
• We will assume that a person has equal probability of being born on any of the 365 days of the year.
• At least two means we could have multiple matches on the same day or several different days where multiple people have matching birthdays.

7.4.2 Simulate (computational)

Now that we have an idea about the structure of the problem, we next need to think about how we would simulate a single classroom. We have 18 students in the classroom and they all could have any of the 365 days of the year as a birthday. What we need to do is sample birthdays for each of the 18 students. But how do we code the days of the year?

An easy solution is to just label the days from 1 to 365. The function seq() does this for us.

Next we need to pick one of the days using the sample function. Note that we set the seed to get repeatable results, this is not required.

The first person was born on the 228th day of the year.

Since R works on vectors, we don’t have to write a loop to select 18 days, we just have sample() do it for us.

What do we want R to do? Sample from the numbers 1 to 365 with replacement, which means a number can be picked more than once.

Notice in our sample we have at least one match, although it is difficult to look at this list and see the match. Let’s sort them to make it easier for us to see.

The next step is to find a way in R for the code to detect that there is a match.

Exercise : What idea(s) can we use to determine if a match exists?

We could sort the data and look at differences in sequential values and then check if the set of differences contains a zero. This seems to be computationally expensive. Instead we will use the function unique() which gives a vector of unique values in an object. The function length() gives the number of elements in the vector.

Since we only have 17 unique values in a vector of size 18, we have a match. Now let’s put this all together to generate another classroom of size 18.

The next problem that needs to be solved is how to repeat the classrooms and keep track of those that have a match. There are several functions we could use to include replicate() but we will use do() from the mosaic package because it returns a data frame so we can use tidyverse verbs to wrangle the data.

The do() function allows us to repeat an operation many times. The following template

where {stuff to do} is typically a single R command, but may be something more complicated.

Load the libraries.

Let’s repeat for a larger number of simulated classroom, remember you should be asking yourself:

What do I want R to do? What does R need to do this?

This is within 2 decimal places of the mathematical solution we develop shortly.

How many classrooms do we need to simulate to get an accurate estimate of the probability of a match? That is a statistical modeling question and it depends on how much variability we can accept. We will discuss these ideas later in the semester. For now, you can run the code multiple times and see how the estimate varies. If computational power is cheap, you can increase the number of simulations.

7.4.3 Plotting

By the way, the method we have used to create the data allows us to summarize the number of unique birthdays using a table or bar chart. Let’s do that now. Note that since the first argument in tally() is not data then the pipe operator will not work without some extra effort. We must tell R that the data is the previous argument in the pipeline and thus use the symbol . to denote this.

Figure 7.1 is a plot of the number of unique birthdays in our sample.

Figure 7.1: Bar chart of the number of unique birthdays in the sample.

Exercise : What does it mean if the length of unique birthdays is 16, in terms of matches? 64

7.4.4 Mathematical solution

To solve this problem mathematically, we will step through the logic one step at a time. One of the key ideas that we will see many times is the idea of the multiplication rule. This idea is the foundation for permutation and combinations which are counting methods frequently used in probability calculations.

The first step that we take is to understand the idea of 2 or more people with the same birthday. With 18 people, there are a great deal of possibilities for 2 or more birthdays. We could have exactly 2 people with the same birthday. We could have 18 people with the same birthday, We could have 3 people with the same birthday and another 2 people with the same birthday but different from the other 3. Accounting for all these possibilities is too large a counting process. Instead, we will take the approach of finding the probability of no one having a matching birthday. Then the probability of at least 2 people having a matching birthday is 1 minus the probability that no one has a matching birthday. This is known as a complementary probability. A simpler example is to think about rolling a single die. The probability of rolling a 6 is equivalent to 1 minus the probability of not rolling a 6.

We first need to think about all the different ways we could get 18 birthdays. This is going to be our denominator in the probability calculation. First let’s just look at 2 people. The first person could have 365 different days for their birthday. The second person could also have 365 different birthdays. So for each birthday of the first person there could be 365 birthdays for the second. Thus for 2 people there are \(365^2\) possible sets of birthdays. This is an example of the multiplication rule . For 18 people there are \(365^{18}\) sets of birthdays. That is a large number. Again, this will be our denominator in calculating the probability.

The numerator is the number of sets of birthdays with no matches. Again, let’s consider 2 people. The first person can have a birthday on any day of the year, so 365 possibilities. Since we don’t want a match, the second person can only have 364 possibilities for a birthday. Thus we have \(365 \times 364\) possibilities for two people to have different birthdays.

Exercise: What is the number of possibilities for 18 people so that no one has the same birthday.

The answer for 18 people is \(365 \times 364 \times 363 ... \times 349 \times 348\) . This looks like a truncated factorial. Remember a factorial, written as \(n!\) with an explanation point, is the product of successive positive integers. As an example \(3!\) is \(3 \times 2 \times 1\) or 6. We could write the multiplication for the numerator as \[\frac{365!}{(365-n)!}\] As we will learn, the multiplication rule for the numerator is known as a permutation .

We are ready to put it all together. For 18 people, the probability of 2 or more people with the same birthday is 1 minus the probability that no one has the same birthday, which is

\[1 - \frac{\frac{365!}{(365-18)!}}{365^{18}}\] or

\[1 - \frac{\frac{365!}{347!}}{365^{18}}\]

In R there is a function called factorial() but factorials get large fast and we will overflow the memory. Try factorial(365) in R to see what happens.

It is returning infinity because the number is too large for the buffer. As is often the case we will have when using a computational method, we must be clever about our approach. Instead of using factorials we can make use of R s ability to work on vectors. If we provide R with a vector of values, the prod() will perform a product of all the elements.

7.4.5 General solution

We now have the mathematics to understand the problem. We can easily generalize this to any number of people. To do this, we have to write a function in R . As with everything in R , we save a function as an object. The general format for creating a function is

For this problem we will call the function birthday_prob() . The only parameter we need is the number of people in the room, n . Let’s write this function.

Notice we assigned the function to the name birthday_prob , we told R to expect one argument to the function, which we are calling n , and then we provide R with the code to find the probability. We set a default value for n in case one is not provided to prevent an error when the function is run. We will learn more about writing functions over this and the next semester.

Test the code with a know answer.

Now we can determine the probability for any size room. You may have heard that it only takes about 23 people in a room to have a 50% probability of at least 2 people matching birthdays.

Let’s create a plot of the probability versus number of people in the room. To do this, we need to apply the function to a vector of values. The function sapply() will work or we can also use Vectorize() to alter our existing function. We choose the latter option.

First notice what happens if we input a vector into our function.

It only uses the first value. There are several ways to solve this problem. We can use the map() function in the purrr package. This idea of mapping a function to a vector is important in data science. It is used in scenarios where there is a lot of data. In this case the idea of map-reduce is used to make the analysis amenable to parallel computing.

We could also just vectorize the function.

Now notice what happens.

We are good to go. Let’s create our line plot, Figure 7.2 .

Figure 7.2: The probability of at least 2 people having mathcing birthdays

Is this what you expected the curve to look like? We, the authors, did not expect this. It has a sigmodial shape with a large increase in the middle range and flatten in the tails.

7.4.6 Data science approach

The final approach we will take is one based on data, a data science approach. In the mosaicData package is a data set called Births that contains the number of births in the US from 1969 to 1988. This data will allow us to estimate the number of births on any day of the year. This allows us to eliminate the reliance on the assumption that each day is equally likely. Let’s first inspect() the data object.

It could be argued that we could randomly pick one year and use it. Let’s see what happens if we just used 1969. Figure 7.3 is a scatter plot of the number of births in 1969 for each day of the year.

Figure 7.3: The number of births for each day of the year in 1969

Exercise : What patterns do you see in Figure 7.3 ? What might explain them?

There are definitely bands appearing in the data which could be the day of the week; there are less birthdays on the weekend. There is also seasonality with more birthdays in the summer and fall. There is also probably an impact from holidays.

Quickly, let’s look at the impact of day of the week by using color for day of the week. Figure 7.4 makes it clear that the weekends have less number of births as compared to the work week.

Figure 7.4: The number of births for each day of the year in 1969 broken down by day of the week

By only using one year, this data might give poor results since holidays will fall on certain days of the week and the weekends will also be impacted. Note that we also still have the problem of leap years.

The years 1972, 1976, 1980, 1984, and 1988 are all leap years. At this point, to make the analysis easier, we will drop those years.

Notice in filter() we used the %in% argument. This is a logical argument checking if year is one of the values. The ! at the front negates this in a sense requiring year not to be one of those values.`

We are almost ready to simulate. We need to get the count of births on each day of the year for the non-leap years.

Let’s look at a plot of the number of births versus day of the year. We combined years in Figure 7.5 .

Figure 7.5: Number of births by day of the year for all years.

This curve has the seasonal cycling we would expect. The smaller scale cycling is unexpected. Maybe because we are dropping the leap years, we are getting some days appearing in our time interval more frequently on weekends. We leave it to you to investigate this phenomenon.

We use these counts as weights in a sampling process. Days with more births will have a higher probability of being selected. Days such as Christmas and Christmas Eve have a lower probability of being selected. Let’s save the weights in an object to use in the sample() function.

The pull() function pulls the vectors of values out of the data frame format into a vector format which the sample() needs.

Now let’s simulate the problem. The probability of a match should change slightly, maybe go down slightly?, but not much since most of the days have about the same probability or number of occurrences.

We could not solve this problem of varying frequency of birth days using mathematics, at least as far as we know.

Cool stuff, let’s get to learning more about probability models in the next chapters.

7.5 Homework Problems

• Exactly 2 people with the same birthday - Simulation . Complete a similar analysis for case where exactly 2 people in a room of 23 people have the same birthday. In this exercise you will use a computational simulation.
• Create a new R Markdown file and create a report. Yes, we know you could use this file but we want you to practice generating your own report.
• Simulate having 23 people in the class with each day of the year equally likely. Find the cases where exactly 2 people have the same birthday, you will have to alter the code from the Notes more than changing 18 to 23.
• Plot the frequency of occurrences as a bar chart.
• Estimate the probability of exactly two people having the same birthday.
• Exactly 2 people with the same birthday - Mathematical . Repeat problem 1 but do it mathematically. As a big hint, you will need to use the choose() function. The idea is that with 23 people we need to choose 2 of them to match. We thus need to multiply, the multiplication rule again, by choose(23,2) . If you are having trouble, work with a total of 3 people in the room first.
• Find a formula to determine the exact probability of exactly 2 people in a room of 23 having the same birthday.
• Generalize your solution to any number n people in the room and create a function.
• Vectorize the function.
• Plot the probability of exactly 2 people having the same birthday versus number of people in the room.
• Comment on the shape of the curve and explain it.

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Case Study on Probability Class 12 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Probability Class 12 Maths can use this page to download the PDF file. 

The case study questions on Probability are based on the CBSE Class 12 Maths Syllabus, and therefore, referring to the Probability case study questions enable students to gain the appropriate knowledge and prepare better for the Class 12 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

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There are three major reasons why one should solve Probability case study questions on Class 12 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 12 Maths students, therefore, it is important to solve Probability Case study questions as it will help better prepare for the Class 12 board exam preparation.
• Develop Problem-Solving Skills: Class 12 Maths Probability case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 12 students develop their problem-solving skills, which are essential for success in any profession rather than Class 12 board exam preparation.
• Understand Real-Life Applications: Several Probability Class 12 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Probability as well as real-life implications of those learnings too.

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Students can choose their own way to answer Case Study on Probability Class 12 Maths, however, we believe following these three steps would help a lot in answering Class 12 Maths Probability Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Probability questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Probability Class 12 Maths case study questions make sure you are approaching each question in a step-wise manner.

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 A few essential things to know to solve Case Study Questions on Class 12 Probability are -

• Basic Formulas of Probability: One of the most important things to know to solve Case Study Questions on Class 12 Probability is to learn about the basic formulas or revise them before solving the case-based questions on Probability.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 12 Maths Probability case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

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Case Study Questions for Class 12 Maths Chapter 13 Probability

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[PDF] Download Case Study Questions for Class 12 Maths Chapter 13 Probability

Here we are providing case study questions for class 12 maths. In this article, we are sharing Class 12 Maths Chapter 13 Probability case study questions. All case study questions of class 12 maths are solved so that students can check their solutions after attempting questions.

Case Study Questions for Class 12 Maths

Case study questions are a type of question that is commonly used in academic and professional settings to evaluate a person’s ability to analyze, interpret, and solve problems based on a given scenario or case study.

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Solving case study questions for Class 12 Maths is extremely important as it provides students with an opportunity to apply the mathematical concepts they have learned to real-world scenarios. These questions present a situation or problem that requires students to use their problem-solving skills and critical thinking abilities to arrive at a solution.

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• Enhances problem-solving skills: Case study questions challenge students to think beyond textbook examples and apply their knowledge to real-world situations. This enhances their problem-solving skills and helps them develop a deeper understanding of the mathematical concepts.
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In conclusion, solving case study questions for Class 12 Maths is important as it helps students develop problem-solving and critical thinking skills, retain concepts better, and prepare for exams.

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CBSE 10th Standard Maths Subject Probability Case Study Questions With Solution 2021

By QB365 on 22 May, 2021

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

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

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

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

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

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

(ii) a monkey

(iii) a teddy bear

(iv) not a monkey 

(v) not a pokemon 

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

(iii) The probability of distributing white chocolates is

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

(v) The probability of distributing all the chocolates is

(ii) The probability of getting exactly 1 head is

(iii) The probability of getting exactly 3 tails is 

(iv) The probability of getting atmost 3 heads is 

(v) The probability of getting atleast two heads is

(ii) not of yellow colour

(iii) of green colour

(iv) of yellow colour 

(v) not of blue colour 

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

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

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

Total number of puppets in claw crane = 58 + 42 + 36 + 64 = 200 (i) (b): P(picking a tiger) =  \(\begin{equation} \frac{36}{200}=\frac{9}{50} \end{equation}\) (ii) (a): P(picking a monkey) =  \(\begin{equation} \frac{64}{200}=\frac{8}{25} \end{equation}\) (iii) (c) : P(picking a teddy bear) =  \(\begin{equation} \frac{58}{200}=\frac{29}{100} \end{equation}\) (iv) (d) : P(not picking a monkey) = 1 - P(picking a monkey) \(\begin{equation} =1-\frac{8}{25}=\frac{17}{25} \end{equation}\) (v) (d): P(picking a pokemon) =  \(\begin{equation} \frac{42}{200}=\frac{21}{100} \end{equation}\) P(not picking a pokemon) = 1 - P(picking a pokemon) \(\begin{equation} =1-\frac{21}{100}=\frac{79}{100} \end{equation}\)

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

Sample space (5) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} \(\Rightarrow\) n(5) = 8 (i) (c): Let A be the event of getting atmost one tail.  A = {HHH, HHT, HTH, THH} \(\Rightarrow\) n(A) = 4 Required probability =  \(\begin{equation} \frac{4}{8}=\frac{1}{2} \end{equation}\) (ii) (d ): Let B be the event of getting exactly 1 head. B = {HTT, THT, TTH} \(\Rightarrow\)  n(B) = 3 Required probability =  \(\begin{equation} \frac{3}{8} \end{equation}\) (iii) (d) : Let C be the event of getting exactly 3 tails. C = {TTT}  \(\Rightarrow\)  n( C) = 1 Required probability =  \(\begin{equation} \frac{1}{8} \end{equation}\) \(\begin{equation} \frac{1}{8} \end{equation}\) \(\begin{equation} \frac{1}{8} \end{equation}\) \(\begin{equation} \frac{1}{8} \end{equation}\) (iv) (b) : Let D be the event of getting atmost 3 heads. D = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} \(\Rightarrow\) n(D) = 8 Required probability =  \(\begin{equation} \frac{8}{8}=1 \end{equation}\) (v) (c): Let E be the event of getting atleast two heads. E = {HHT, HTH, THH, HHH} \(\Rightarrow\) n(E) = 4 Required probability =  \(\begin{equation} \frac{n(E)}{n(S)}=\frac{4}{8}=\frac{1}{2} \end{equation}\)

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

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Chapter 14 Class 10 Probability

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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!

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CBSE Case Study Questions for Class 10 Maths Probability PDF

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