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CBSE 12th Standard Maths Subject Probability Case Study Questions With Solution 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

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Cbse 12th standard maths probability case study questions with solution 2021.

12th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) What is the probability that the first ball is yellow and the second ball is red?

(iii) What is the probability that both the balls are red?

(iv) What is the probability that the first ball is green and the second ball is not yellow?

(v) What is the probability that both the balls are not blue?

(ii) When the doctor arrives late, what is the probability that he comes by cab?

(iii) When the doctor arrives late, what is the probability that he comes by bike?

(iv) When the doctor arrives late, what is the probability that he comes by other means of transport?

(v) What is the probability that the doctor is late by any means?

(ii) Probability that puzzle is solved by Ravi but not by Priya, is

(iii) Find the probability that puzzle is solved.

(iv) Probability that exactly one of them solved the puzzle, is

(v) Probability that none of them solved the puzzle, is

(ii) The probability that it does not rain on chosen day is

(iii) The probability that the weatherman predicts correctly is

(iv) The probability that it will rain on the chosen day, if weatherman predict rain for that day, is

(v) The probability that it will not rain on the chosen day, if weatherman predict rain for that day, is

(ii) Teacher ask Mivaan, what is the probability that both tickets drawn by Aadya shows odd number?

(iii) Teacher ask Deepak, what is the probability that tickets drawn by Mivaan, shows a multiple of 4 on one ticket and a multiple 5 on other ticket?

(iv) Teacher ask Archit, what is the probability that tickets are drawn by Deepak, shows a prime number on one ticket and a multiple of 4 on other ticket?

(v) Teacher ask Aadya, what is the probability that tickets drawn by Vrinda, shows an even number on first ticket and an odd' number on second ticket?

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Cbse 12th standard maths probability case study questions with solution 2021 answer keys.

Let B, R, Y and G denote the events that ball drawn is blue, red, yellow and green respectively. \(\therefore P(B)=\frac{12}{35}, P(R)=\frac{8}{35}, P(Y)=\frac{10}{35} \text { and } P(G)=\frac{5}{35}\) \((i) \ (c): P(G \cap B)=P(B) \cdot P(G \mid B)=\frac{12}{35} \cdot \frac{5}{34}=\frac{6}{119}\) \((ii) \ (\mathbf{b}): P(R \cap Y)=P(Y) \cdot P(R \mid Y)=\frac{10}{35} \cdot \frac{8}{34}=\frac{8}{119}\) (iii) (a): Let E = event of drawing a first red ball and F = event of drawing a second red ball Here,  \(P(E)=\frac{8}{35} \text { and } P(E)=\frac{7}{34}\) \(\therefore \P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{8}{35} \cdot \frac{7}{34}=\frac{4}{85}\) \(\text {(iv) }(c): P\left(Y^{\prime} \cap G\right)=P(G) \cdot\left(Y^{\prime} \mid G\right)=\frac{5}{35} \cdot \frac{24}{34}=\frac{12}{119}\) (v) (d): Let E = event of drawing a first non-blue ball and F = event of drawing a second non-blue ball Here, \(P(E)=\frac{23}{35} \text { and } P(F)=\frac{22}{34}\) \(\therefore \ P(F \cap E)=P(E) \cdot P(F \mid E)=\frac{23}{35} \cdot \frac{22}{34}=\frac{253}{595}\)

Let E be the event that the doctor visit the patient late and let A 1 , A 2 , A 3 , A 4 be the events that the doctor comes by cab, metro, bike and other means of transport respectively. \(P\left(A_{1}\right)=0.3, P\left(A_{2}\right)=0.2, P\left(A_{3}\right)=0.1, P\left(A_{4}\right)=0.4\) P(E I A 1 ) = Probability that the doctor arriving late when he comes by cab = 0.25 Similarly, P ( E I A 2 ) = 0.3, P (E I A 3 ) = 0.35 and P ( E I A 3 ) = 0.1 (i) (b): P(A 2  | E) = Probability that the doctor arriving late and he comes by metro \(=\frac{P\left(A_{2}\right) P\left(E \mid A_{2}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\) \(=\frac{(0.2)(0.3)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\) \(=\frac{0.06}{0.21}=\frac{2}{7}\) (ii) (c): P(A 1 | E) = Probability that the doctor arriving late and he comes by cab \(=\frac{P\left(A_{1}\right) P\left(E \mid A_{1}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\) \(=\frac{(0.3)(0,25)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\) \(=\frac{0.075}{0.21}=\frac{5}{14}\) (iii) (d): P(A 3 | E) = Probability that the doctor arriving late and he comes by bike \(=\frac{P\left(A_{3}\right) P\left(E \mid A_{3}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\) \(=\frac{(0.1)(0.35)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\) \(=\frac{0.035}{0.21}=\frac{1}{6}\) (iv) (c): P(A 4 | E) = Probability that the doctor arriving late and he comes by other means of transport \(=\frac{P\left(A_{4}\right) P\left(E \mid A_{4}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\) \(=\frac{(0.4)(0.1)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\) \(=\frac{0.04}{0.21}=\frac{4}{21}\) (v) (a): Probability that the doctor is late by any means \(=\frac{2}{7}+\frac{5}{14}+\frac{1}{6}+\frac{4}{21}=1\)

Let E 1 be the event that Ravi solved the puzzle and E 2 be the event that Priya solved the puzzle Then, p(E 1 ) = 1/4 and P(E2) = 1/5 (i) (b): Since, E 1 and E 2 are independent events. ∴ P(both solved the puzzle) \(=P\left(E_{1} \cap E_{2}\right)\) \(=P\left(E_{1}\right) \cdot P\left(E_{2}\right)=\frac{1}{4} \times \frac{1}{5}=\frac{1}{20}=\frac{1}{20} \times 100 \%=5 \%\) (ii) (b): P(puzzle is solved by Ravi but not by Priya) \(=P\left(\bar{E}_{2}\right) P\left(E_{1}\right)=\left(1-\frac{1}{5}\right) \cdot \frac{1}{4}=\frac{4}{5} \cdot \frac{1}{4}=\frac{1}{5}\) (iii) (c) : P(puzzle is solved) = P(E 1 or E 2 ) \(=P\left(E_{1} \cup E_{2}\right)=P\left(E_{1}\right)+P\left(E_{2}\right)-P\left(E_{1} \cap E_{2}\right)\) \(=\frac{1}{4}+\frac{1}{5}-\frac{1}{20}=\frac{8}{20}=\frac{2}{5}\) (iv) (c): P(Exactly one of them solved the puzzle) \(=P\left[\left(E_{1} \text { and } \bar{E}_{2}\right) \operatorname{or}\left(E_{2} \text { and } \bar{E}_{1}\right)\right] \) \(=P\left(E_{1} \cap \bar{E}_{2}\right)+P\left(E_{2} \cap \bar{E}_{1}\right) \) \(=P\left(E_{1}\right) \times P\left(\bar{E}_{2}\right)+P\left(E_{2}\right) \times P\left(\bar{E}_{1}\right) \) \(=\frac{1}{4} \times \frac{4}{5}+\frac{1}{5} \times \frac{3}{4} \quad\left[\because P\left(\bar{E}_{1}\right)=1-P\left(E_{1}\right)\right] \) \(=\frac{4}{20}+\frac{3}{20}=\frac{7}{20}\) (v) (b): P(none of them solved the puzzle) \(=P\left(\bar{E}_{1} \cap \bar{E}_{2}\right)=P\left(\bar{E}_{1}\right) \cdot P\left(\bar{E}_{2}\right)=\frac{3}{4} \times \frac{4}{5}=\frac{3}{5}\)

(i) (d): Since, it rained only 6 days each year, therefore, probability that it rains on chosen day is  \(\frac{6}{366}=\frac{1}{61}\) (ii) (c): The probability that it does not rain on chosen day =  \(1-\frac{1}{61}=\frac{60}{61}=\frac{360}{366}\) (iii) (c): It is given that, when it actually rains, the weatherman correctly forecasts rain 80% of the time. Required probability \( =\frac{80}{100}=\frac{8}{10}=\frac{4}{5}\) (iv) (a): Let A 1 be the event that it rains on chosen day, A 2 be the event that it does not rain on chosen day and E be the event the weatherman predict rain. Then we have, P(A 1 ) \(=\frac{6}{366}, P\left(A_{2}\right)=\frac{360}{366},\) \(P\left(E \mid A_{1}\right)=\frac{8}{10} \text { and } P\left(E \mid A_{2}\right)=\frac{2}{10}\) Required probability = P(A 1 I E) \(=\frac{P\left(A_{1}\right) \cdot P\left(E \mid A_{1}\right)}{P\left(A_{1}\right) \cdot P\left(E \mid A_{1}\right)+P\left(A_{2}\right) \cdot P\left(E \mid A_{2}\right)}\) \(=\frac{\frac{6}{366} \times \frac{8}{10}}{\frac{6}{366} \times \frac{8}{10}+\frac{360}{366} \times \frac{2}{10}}=\frac{48}{768}=0.0625\) (v) (a): Required probability = 1 - P(A 1 I E) = 1 - 0.0625 = 0.9375 ≈ 0.94.

(i) (b): Total number of tickets = 50 Let event A = First ticket shows even number and B = Second ticket shows even number Now, P(Both tickets show even number) = P(A)・P(B | A) \(=\frac{25}{50} \cdot \frac{24}{49}=\frac{12}{49}\) (ii) (c): Let the event A = First ticket shows odd number and B = Second ticket shows odd number P (Both tickets show odd number) \(=\frac{25}{50} \times \frac{24}{49}=\frac{12}{49}\) (iii) (c): Required probability = P(one number is a multiple of 4 and other is a multiple of 5) = P(multiple of 5 on first ticket and multiple of 4 on second ticket) + P(multiple of 4 on first ticket and multiple of 5 on second ticket) \(=\frac{10}{50} \cdot \frac{12}{49}+\frac{12}{50} \times \frac{10}{49}=\frac{12}{245}+\frac{12}{245}=\frac{24}{245}\) (iv) (d): Required probability = P(one ticket with prime number and other ticket with a multiple of 4) \(=2\left(\frac{15}{50} \times \frac{12}{49}\right)=\frac{36}{245}\) (v) (b): Let the event A = First ticket shows even number and B = Second ticket shows odd number. Now, P(First ticket shows an even number and second ticket shows an odd number) = P(A)・P(B | A) \(=\frac{25}{50} \times \frac{25}{49}=\frac{25}{98}\)

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Case Study Questions for Class 12 Maths Chapter 13 Probability

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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.

Typically, a case study question presents a real-world situation or problem that requires the individual to apply their knowledge and skills to identify the issues, consider various solutions, and recommend a course of action.

Case study questions are designed to test critical thinking skills, problem-solving abilities, and the capacity to work through complex and ambiguous situations.

Preparing for case study questions involves developing a deep understanding of the subject matter, being able to analyze and synthesize information quickly, and being able to communicate ideas clearly and effectively.

Importance of Solving Case Study Questions for Class 12 Maths

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.

The importance of solving case study questions for Class 12 Maths can be summarized as follows:

• 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.
• Improves critical thinking abilities: Case study questions require students to analyze and evaluate information, and draw conclusions based on their understanding of the situation. This helps them develop their critical thinking abilities, which are essential for success in many areas of life.
• Helps in retaining concepts: Solving case study questions helps students retain the concepts they have learned for a longer period of time. This is because they are more likely to remember the concepts when they have applied them to a real-world situation.
• Better preparation for exams: Many competitive exams, including the Class 12 Maths board exam, contain case study questions. Solving these questions helps students become familiar with the format of the questions and the skills required to solve them, which can improve their performance in exams.

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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Class 12 Maths: Case Study Based Questions PDF Download

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You must practice some good Case Study questions of Class 12 Maths to boost your preparation to score 95+% on Boards. In this post, you will get Case Study Questions of All Chapters which will come in CBSE Class 12 Maths Board Exams.

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

We have provided here Case Study questions for the Class 12 Maths exams. You can read these chapter-wise Case Study questions. Prepared by subject experts and experienced teachers. The answer key is also provided so that you can check the correct answer for each question. Practice these questions to score well in your Board Final exams.

We are providing Case Study questions for class 12 Biology based on the latest syllabi. There is a total of 13 chapters included in CBSE class 12 Maths exams. Students can practice these questions for concept clarity and score better marks in their exams.

Table of Contents

CBSE Class 12th – MATHS : Chapterwise Case Study Question & Solution

CBSE will ask two Case Study Questions in the CBSE class 12 maths questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them.

Case Study Based Questions for Class 12 Maths

Class 12 Physics Case Study Questions Class 12 Chemistry Case Study Questions Class 12 Biology Case Study Questions Class 12 Maths Case Study Questions

Books for Class 12 Maths

Strictly as per the new term-wise syllabus for Board Examinations to be held in the academic session 2022-23 for class 12 Multiple Choice Questions based on new typologies introduced by the board- Stand-Alone MCQs, MCQs based on Assertion-Reason Case-based MCQs. Include Questions from CBSE official Question Bank released in April 2022 Answer key with Explanations What are the updates in the book: Strictly as per the Term wise syllabus for Board Examinations to be held in the academic session 2022-23. Chapter-wise -Topic-wise Multiple choice questions based on the special scheme of assessment for Board Examination for Class 12th.

Class 12 Maths Syllabus 2022-23

Unit-i: relations and functions.

1. Relations and Functions (15 Periods)

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions.

2. Inverse Trigonometric Functions (15 Periods)

Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit-II: Algebra

1. Matrices (25 Periods)

Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operation on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Oncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants 25 Periods

Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit-III: Calculus

1. Continuity and Differentiability (20 Periods)

Continuity and differentiability, chain rule, derivative of inverse trigonometric functions, 𝑙𝑖𝑘𝑒 sin −1  𝑥 , cos −1  𝑥 and tan −1  𝑥, derivative of implicit functions. Concept of exponential and logarithmic functions. Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.

2. Applications of Derivatives (10 Periods)

Applications of derivatives: rate of change of bodies, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as reallife situations).

3. Integrals (20 Periods)

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals (15 Periods)

Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only)

5. Differential Equations (15 Periods)

Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

Unit-IV: Vectors and Three-Dimensional Geometry

1. Vectors (15 Periods)

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors.

2. Three – dimensional Geometry (15 Periods)

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines.

Unit-V: Linear Programming

1. Linear Programming (20 Periods)

Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Unit-VI: Probability

1. Probability 30 (Periods)

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

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Chapter 13 Class 12 Probability

Click on any of the links below to start learning from Teachoo ...

Updated according to new NCERT - 2023-24 NCERT Books.

Get NCERT solutions of all examples, exercises and Miscellaneous questions of Chapter 13 Class 12 Probability with detailed explanation. Formula sheet also available.

We started learning about Probability from Class 6,

we learned that Probability is Number of outcomes by Total Number of Outcomes.

In Class 11, we learned about Sample Space, Events, using Sets . 

In this chapter, we will learn about

• Conditional Probability - Finding probability of something when an event has already occurred. For example - finding probability of 4 coming in second throw of die if 6 has come in first throw. We also discuss its formula, properties and questions
• Independent events - What is an independent event, and where is it used?
• Multiplication rule of probability - We learn about dependent and independent events, and the multiplication rule for 2, or more than two events
• Basic Probability - We solve questions using basic formula - Number of outcomes/Total Outcomes to find Probability, set theory , and permutation and combinations  to find probability.
• Theorem of total probability - We use the formula P(A) = P(B) P(A|B) + P(B') P(A|B')
• Bayes theorem - Finding probability when an event has already happened
• Random Variable - Writing random variable 
• Probability distribution - Finding probability distribution of random variable, and finding its mean (or expectation)
• Variance and Standard Deviation of a Random Variable - Finding variance and standard deviation using probability distribution
• Bernoulli Trials - Checking if an event is a Bernoulli trial
• Binomial Distribution - For Bernoulli Trial, finding probability using Binomial Distribution

Check the chapter from different Concepts, starting from Basic to Advanced, or you can also refer to the exercises mentioned in the NCERT Book. Click on a topic below to start

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CBSE Class 12 Maths Exam 2024 Important Case Study Based Questions

Case study questions for class 12 maths: check here the important case study based questions of section e in the cbse class 12 maths exam 2024 for last minute preparation..

CBSE Class 12 Maths Exam 2024 Important Questions: The Central Board of Secondary Education is the largest and one of the most famed school boards in India, and lakhs of students are currently enrolled in it. The CBSE conducts the Class 12 board exams annually. The next paper is arguably the most important for science and commerce stream students: Maths on 9 March. Maths is essential for non-medical science and commerce aspirants and is also required in subjects like physics, statistics and accounts. CBSE Class 12 Maths requires extensive practice, especially important topics like Calculus and Algebra. There will be five sections in the 2024 CBSE Class 12 Maths exam, and the last section E will comprise two case study-based questions of 4 marks. These questions are quite important from the exam point of view, and you can check and practice the solved versions here.

CBSE Class 12   Maths   Unit Wise Marks Distribution 2024

CBSE Maths Previous Year Question Paper Class 12

CBSE Class 12 Maths Case Study Questions 2023

Q uestion 1:  Ramesh is   elder brother of Suresh. Ramesh wants to help his younger brother Suresh to solve the following problems of integrals. Write the suitable substitution by which Ramesh can help him.

Question 2: Mr Shashi, who is an architect, designs a building for a small company. The design of window on the ground floor is proposed to be different than other floors. The window   is in the shape of a rectangle which is surmounted by a semi-circular opening. This window is having a perimeter of 10 m as shown below :

Based on the above information answer the following :

(i) If 2x and 2y represents the length and breadth of the rectangular portion of the windows, then the relation between the variables is:      

(ii) The combined area (A) of the rectangular region and semi-circular region of the window expressed as a function of x is:

(iii) The maximum value of area A, of the whole window is

The owner of this small company is interested in maximizing the area of the whole window so that maximum light input is possible.

For this to happen, the length of rectangular portion of the window should be

(i) 4y = 10 - (2 + π)x

(ii) A = 10x - (2 + 12π)x 2

(iii) 50/4 + π

20/4 + π  

Question 3:  Read the following and answer the questions given below 

The front gate of a building is in the shape of a trapezium as shown below. Its three sides other than base are of 10 m each. The height of the gate is h meter. On the basis of below figure, answer the following questions:

(i)  Write the Area (A) of the gate in terms of   .

(ii) Write the value of  when Area (A)  is maximum.

(iii) Write the value of h when Area (A) is maximum .

      Write the Maximum value of Area (A) .

(i) (10 + x)√100 - x 2

(iii) 5√3m OR 75√3/m.m2

Question 4 : Read the following and answer the questions given below 

Given three identical boxes 1 st,  2 nd  and 3 rd  each containing two coins. In 1 st  box both coins are gold coins, in 2 nd  box both are silver coins and in 3 rd  box there is one gold and one silver coin. A person chooses a box at random and takes out a coin.

On the basis of above information, answer the following questions:    

(i)   What is the probability of choosing 1 st  box  ?

(ii)  What is the probability of getting gold coin from 3 rd  box ?

(iii)  What is the total probability of drawing gold coin ?

       If drawn coin is of gold the probability that other coin in the box is also of gold?

(iii) 1/2  Or  2/3            

Question 5:  Read the following and answer the questions given below

Sand is pouring from a pipe at the rate of 12 cm 3 / second the falling sand forms a cone on the ground in such a way that the height of the cone is always 1/6 th of the radius of the base. Based on above information answer the following:

(i)  Write the expression for volume in terms of height only.

(ii)   What is the rate of Change of height, when height is 4 cm?

i)  12πh 3

(ii) 1/48 cm/s

Question 6: There are two antiaircraft guns, named as A and B. The probabilities that the shell fired from them hits an airplane are 0.3 and 0.2 respectively. Both of them fired one shell at an airplane at the same time.

(i) What is the probability that the shell fired from exactly one of them hit the plane?

(ii) If it is known that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B?

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• Important Questions for CBSE Class 12 Maths Chapter 13 - Probability 2024-25

CBSE Class 12 Maths Chapter-13 Important Questions - Free PDF Download

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Also, check CBSE Class 12 Maths Important Questions for other chapters:

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Study Important Questions for Class 12 Mathematics Chapter 13 – Probability

Very Short Answer Question (1 Mark)

1. Find $P\left( {A}/{B}\; \right)$ if $P\left( A \right)=0.4$, $P\left( B \right)=0.8$ and $P\left( {B}/{A}\; \right)=0.6$.

Ans: It is given that,

$P\left( A \right)=0.4$

$P\left( B \right)=0.8$

$P\left( {B}/{A}\; \right)=0.6$

It is known that,

$P\left( {B}/{A}\; \right)=\frac{P\left( A\cap B \right)}{P\left( A \right)}$

$\Rightarrow P\left( A\cap B \right)=0.4\times 0.6=0.24$

$P\left( {A}/{B}\; \right)=\frac{P\left( A\cap B \right)}{P\left( B \right)}$

$\Rightarrow P\left( {A}/{B}\; \right)=\frac{0.24}{0.8}=0.3$

Therefore, $P\left( {A}/{B}\; \right)=0.3$

2. Find $P\left( A\cap B \right)$ if $A$ and $B$ are two events such that $P\left( A \right)=0.5$, $P\left( B \right)=0.6$ and $P\left( A\cup B \right)=0.8$.

$P\left( A \right)=0.5$

$P\left( B \right)=0.6$

$P\left( A\cup B \right)=0.8$

$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ 

$\Rightarrow 0.8=0.5+0.6-P\left( A\cap B \right)$

$\Rightarrow P\left( A\cap B \right)=1.1-0.8=0.3$

Therefore, $P\left( A\cap B \right)=0.3$

3. A soldier fires three bullets at the enemy. The probability that the enemy will be killed by one bullet is $0.7$. What is the probability that the enemy is still alive?

Probability that the enemy will be killed by one bullet is $0.7$

Probability that the enemy will be alive after one bullet is fired $=1-0.7=0.3$

Probability that the enemy is still alive after three bullets is fired$=0.3\times 0.3\times 0.3=0.027$ 

Therefore, the probability that the enemy is still alive is $0.027$.

4. What is the probability that a leap year has $53$ Sundays?

Ans: It is known that a leap year has $366$ days.

It means that there are $52$ weeks and $2$ extra days.

The $52$ weeks have $52$ Sundays.

The possibilities for two extra days are  $\left\{ Monday,Tuesday \right\}$, $\left\{ Tuesday,Wednesday \right\}$, $\left\{ Wednesday,Thursday \right\}$, $\left\{ Thursday,Friday \right\}$, $\left\{ Friday,Saturday \right\}$, $\left\{ Saturday,Sunday \right\}$and $\left\{ Saturday,Sunday \right\}$

The possibility of having Sunday in the other days is $\left\{ Saturday,Sunday \right\}$ and$\left\{ Saturday,Sunday \right\}$

The number of favourable outcomes will be $2$ 

Total number of outcomes is $7$ 

Probability of an event $A$: $P\left( A \right)=\frac{Favourable\text{ }Outcomes}{\text{Total }Outcomes}$ 

$\Rightarrow P\left( A \right)=\frac{2}{7}$ 

Probability that a leap year has $53$ Sundays is $\frac{2}{7}$ .

5. $20$ cards are numbered $1$ to $20$. One card is drawn at random. What is the probability that the number on the card will be a multiple of $4$?

Sample space, $S=1,2,3,.........,20$ 

Total number of outcomes in the sample space, $n\left( S \right)=20$

Let $A$ be the event of getting multiple of $4$,$A=4,8,12,16,20$ 

Total number of favourable outcomes, $n\left( A \right)=5$

Probability of an event $A$: $P\left( A \right)=\frac{n\left( A \right)}{n\left( S \right)}$ 

$\Rightarrow P\left( A \right)=\frac{5}{20}=0.25$ 

The probability that the number on the card will be a multiple of $4$is $0.25$.

6. Three coins are tossed once. Find the probability of getting at least one head.

Ans: Sample space, $S=\left\{ HHH,HHT,HTH,THH,HTT,THT,TTH,TTT \right\}$ 

Total number of outcomes in the sample space, $n\left( S \right)=8$

Let $A$ be the event of getting at least one head,$A=\left\{ HHH,HHT,HTH,THH,HTT,THT,TTH \right\}$ 

Total number of favourable outcomes, $n\left( A \right)=7$

$\Rightarrow P\left( A \right)=\frac{7}{8}$ 

Therefore, the probability of getting at least one head is $\frac{7}{8}$.

7. The probability that a student is not a swimmer is $\frac{1}{5}$. Find the probability that out of $5$ students, $4$ are swimmers.

The probability that a student is not a swimmer $=\frac{1}{5}$

The probability that a student is a swimmer $=1-\frac{1}{5}=\frac{4}{5}$

Probability that out of $5$ students, $4$ are swimmers:

From Binomial distribution:$P\left( X=x \right)={}^{n}{{C}_{x}}{{p}^{x}}{{q}^{n-x}}$ 

Here, $x=4$, $n=5$, $p=\frac{4}{5}$ and $q=\frac{1}{5}$ 

$\Rightarrow P\left( X=4 \right)={}^{5}{{C}_{4}}{{\left( \frac{4}{5} \right)}^{4}}{{\left( \frac{1}{5} \right)}^{5-4}}$ 

$\Rightarrow P\left( X=4 \right)=\frac{5!}{4!\left( 5-4 \right)!}{{\left( \frac{4}{5} \right)}^{4}}{{\left( \frac{1}{5} \right)}^{1}}$

$\Rightarrow P\left( X=4 \right)=5{{\left( \frac{4}{5} \right)}^{4}}\left( \frac{1}{5} \right)$

$\Rightarrow P\left( X=4 \right)={{\left( \frac{4}{5} \right)}^{4}}$

The probability that out of $5$ students, $4$ are swimmers is ${{\left( \frac{4}{5} \right)}^{4}}$.

8. Find $P\left( {A}/{B}\; \right)$ if $P\left( B \right)=0.5$ and $P\left( A\cap B \right)=0.32$.

$P\left( B \right)=0.5$

$P\left( A\cap B \right)=0.32$

$\Rightarrow P\left( {A}/{B}\; \right)=\frac{P\left( A\cap B \right)}{P\left( B \right)}$

$\Rightarrow P\left( {A}/{B}\; \right)=\frac{0.32}{0.50}=\frac{16}{25}$

Therefore, $P\left( {A}/{B}\; \right)=\frac{16}{25}$

9. A random variable $X$ has the following probability distribution.

Find the value of $k$.

Ans: It is known that the sum of all the probabilities of a random variable is equal to $1$ i.e. $\sum{P\left( X \right)=1}$.

$\Rightarrow \frac{1}{15}\times 0+k\times 1+\frac{15k-2}{15}\times 2+k\times 3+\frac{15k-1}{15}\times 4+\frac{1}{15}\times 5=1$ 

$\Rightarrow 9k-\frac{3}{15}=1$

$\Rightarrow 4k=1+\frac{1}{15}$

$\Rightarrow 4k=\frac{16}{15}$

$\Rightarrow k=\frac{4}{15}$

10. A random variable $X$ taking values $0$, $1$, $2$ has the following probability distribution for some number $k$.

P(X)=\begin{cases}k &if & X = 0\\2k &if & X = 1,find &k.\\3k &if & X = 2\end{cases}

Ans: It is known that the sum of probabilities of a random variable is equal to $1$.

$\Rightarrow k+2k+3k=1$ 

$\Rightarrow 6k=1$ 

$\Rightarrow k=\frac{1}{6}$ 

Therefore, the value of $k$ is $\frac{1}{6}$.

Long Answer Questions (4 Marks)

11. A problem in Mathematics is given to three students whose chance of solving it are $\frac{1}{2}$, $\frac{1}{3}$ and $\frac{1}{4}$. What is the probability that the problem is solved?

Ans: Let $A,B,C$ be the events of solving the problem.

$\bar{A},\bar{B},\bar{C}$ be the event  of not solving the problem.

It is known that, $\bar{A},\bar{B},\bar{C}$ are independent events.

From the given data,

$P\left( A \right)=\frac{1}{2}\Rightarrow P\left( {\bar{A}} \right)=\frac{1}{2}$ 

$P\left( B \right)=\frac{1}{3}\Rightarrow P\left( {\bar{B}} \right)=\frac{2}{3}$

\[P\left( C \right)=\frac{1}{4}\Rightarrow P\left( {\bar{C}} \right)=\frac{3}{4}\]

$P\left( None\text{ }solving\text{ }the\text{ }problem \right)=P(\bar{A}\text{ }and\text{ }\bar{B}\text{ }and\text{ }\bar{C})$

$\Rightarrow P(None\text{ }solving\text{ }the\text{ }problem)=P(\bar{A}\cap \bar{B}\cap \bar{C})$ 

$\Rightarrow P(None\text{ }solving\text{ }the\text{ }problem)=P(\bar{A})P(\bar{B})P(\bar{C})$

$\Rightarrow P(None\text{ }solving\text{ }the\text{ }problem)=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}=\frac{1}{4}$

$P(The\text{ problem will b}e\text{ solved})=1-P(None\text{ }solving\text{ }the\text{ }problem)$ 

$P(The\text{ problem will b}e\text{ solved})=1-\frac{1}{4}=\frac{3}{4}$

Therefore, the probability that the problem is solved is $\frac{3}{4}$.

12. A die is rolled. If the outcome is an even number, what is the probability that it is a prime?

Ans: Sample space when a die is rolled, $S=\left\{ 1,2,3,4,5,6 \right\}$ 

Let $A$ be the event of getting an even number.

Let $B$ be the event of getting a prime number.

$A=\left\{ 1,3,5 \right\}$ $B=\left\{ 2,3,5 \right\}$ $A\cap B=\left\{ 3,5 \right\}$ 

Conditional probability, $P\left( {B}/{A}\; \right)=\frac{P\left( A\cap B \right)}{P\left( A \right)}$ 

$\Rightarrow P\left( {B}/{A}\; \right)=\frac{\frac{2}{6}}{\frac{3}{6}}=\frac{2}{3}$

Therefore, when a die is rolled if the outcome is an even number the probability that it is prime is $\frac{2}{3}$.

13. If $A$ and $B$ are two events such that $P\left( A \right)=\frac{1}{4}$, $P\left( B \right)=\frac{1}{2}$ and $P\left( A\cap B \right)=\frac{1}{8}$. Find $P$(not $A$ and not $B$).

Ans: Given,

$P\left( A \right)=\frac{1}{4}$

$P\left( B \right)=\frac{1}{2}$

$P\left( A\cap B \right)=\frac{1}{8}$

\[P\left( not~A~\And \text{ n}ot~\text{B} \right)=P\left( A'\cap B' \right)\] 

\[\Rightarrow P\left( not~A~\And \text{ n}ot~\text{B} \right)=P\left( A\cup B \right)'\]

\[\Rightarrow P\left( not~A~\And \text{ n}ot~\text{B} \right)=1-P\left( A\cup B \right)\]

\[\Rightarrow P\left( not~A~\And \text{ n}ot~\text{B} \right)=1-\left[ P\left( A \right)+P\left( B \right)-P\left( A\cap B \right) \right]\]

\[\Rightarrow P\left( not~A~\And \text{ n}ot~\text{B} \right)=1-\left[ \frac{1}{4}+\frac{1}{2}-\frac{1}{8} \right]\]

\[\Rightarrow P\left( not~A~\And \text{ n}ot~\text{B} \right)=1-\left[ \frac{2}{8}+\frac{4}{8}-\frac{1}{8} \right]\]

\[\Rightarrow P\left( not~A~\And \text{ n}ot~\text{B} \right)=1-\frac{5}{8}\]

\[\Rightarrow P\left( not~A~\And \text{ n}ot~\text{B} \right)=\frac{3}{8}\]

Therefore, \[P\left( not~A~\And \text{ n}ot~\text{B} \right)=\frac{3}{8}\]

14. In a class of $25$ students with roll numbers $1$ to $25$, a student is picked up at random to answer a question. Find the probability that the roll number of the selected student is either a multiple of $5$ or of $7$.

Ans: Sample space, $S=\left\{ 1,2,3,4,........25 \right\}$

Total number of outcomes in the sample space, $n\left( S \right)=25$

Let $A$ be the event that the roll number of selected student is either a multiple of $5$ or of $7$,$A=\left\{ 5,7,10,14,15,20,21,25 \right\}$ 

Total number of favourable outcomes, $n\left( A \right)=8$

$\Rightarrow P\left( A \right)=\frac{8}{25}$ 

Therefore, the probability that the roll number of the selected student is either a multiple of $5$ or of $7$is $\frac{8}{25}$.

15. A can hit a target $4$ times in $5$ shots $B$ three times in $4$ shots and $C$ twice in $3$ shots. They fire a volley. What is the probability that at least two shots hit.

Ans: Fire a volley means that $A$, $B$ and $C$ all try to hit the target simultaneously.

At least two shots hit the target:

$A$ and $B$ hit and $C$ fails to hit.

$A$ and $C$ hit and $B$ fails to hit.

$B$ and $C$ hit and $A$ fails to hit.

$A$, $B$ ,$C$ all three hit the target.

The chance of hitting by $A,P\left( A \right)=\frac{4}{5}$ and of not hitting by $A,P\left( A' \right)=1-\frac{4}{5}=\frac{1}{5}$

The chance of hitting by $B,P\left( B \right)=\frac{3}{4}$ and of not hitting by $B,P\left( B' \right)=1-\frac{3}{4}=\frac{1}{4}$

The chance of hitting by $C,P\left( C \right)=\frac{2}{3}$ and of not hitting by $C,P\left( C' \right)=1-\frac{2}{3}=\frac{1}{3}$

Probability of $\left( I \right)$ is $P\left( ABC' \right)=\frac{4}{5}\times \frac{3}{4}\times \frac{1}{3}=\frac{1}{5}$ 

Probability of $\left( II \right)$ is $P\left( ACB' \right)=\frac{4}{5}\times \frac{2}{3}\times \frac{1}{4}=\frac{2}{15}$ 

Probability of $\left( III \right)$ is $P\left( BC'A \right)=\frac{3}{4}\times \frac{2}{3}\times \frac{1}{5}=\frac{1}{10}$

Probability of $\left( IV \right)$ is $P\left( ABC \right)=\frac{4}{5}\times \frac{3}{4}\times \frac{2}{3}=\frac{2}{5}$

$Total\text{ }probability=\frac{1}{5}+\frac{2}{15}+\frac{1}{10}+\frac{2}{5}$  

$Total\text{ }probability=\frac{6}{30}+\frac{4}{30}+\frac{3}{30}+\frac{12}{30}$

$Total\text{ }probability=\frac{25}{30}=\frac{5}{6}$

Therefore, the probability that at least two shots hit the target is $\frac{5}{6}$.

16. Two dice are thrown once. Find the probability of getting an even number on the first die or a total of $8$ .

Ans: Sample space, $S=\left\{ 1,2,3,4,........36 \right\}$

Total number of outcomes in the sample space, $n\left( S \right)=36$

Let $A$ be the event of getting even number on the first die$A=\left\{ \left( 2,1 \right),\left( 2,2 \right),...,\left( 2,6 \right),\left( 4,1 \right),\left( 4,2 \right),...,\left( 4,6 \right),\left( 6,1 \right),\left( 6,2 \right),...,\left( 6,6 \right) \right\}$ 

Total number of favourable outcomes, $n\left( A \right)=18$

Let $B$ be the event of getting a total of $8$, $B=\left\{ \left( 2,6 \right),\left( 3,5 \right),\left( 4,4 \right),\left( 5,3 \right),\left( 6,2 \right) \right\}$ 

Total number of favourable outcomes, $n\left( B \right)=5$

Let $A\cap B$ be the event of getting an even number on first die and a total of $8$,$A\cap B=\left\{ \left( 2,6 \right),\left( 4,4 \right),\left( 6,2 \right) \right\}$ 

Total number of favourable outcomes, $n\left( A\cap B \right)=3$

Probability of an event $A$: $P\left( A \right)=\frac{n\left( A \right)}{n\left( S \right)}=\frac{18}{36}$ 

Probability of an event $B$: $P\left( A \right)=\frac{n\left( B \right)}{n\left( S \right)}=\frac{5}{36}$ 

Probability of an event $A\cap B$: $P\left( A\cap B \right)=\frac{n\left( A\cap B \right)}{n\left( S \right)}=\frac{3}{36}$ 

$P\left( A\cup B \right)=\frac{18}{36}+\frac{5}{36}-\frac{3}{36}$

$P\left( A\cup B \right)=\frac{20}{36}=\frac{5}{9}$

Therefore, the probability of getting an even number on the first die or a total of $8$ is $\frac{5}{9}$.

17. $A$ and $B$ throw a die alternatively till one of them throws a $'6'$ and wins the game. Find their respective probabilities of winning, if $A$ starts the game.

Ans: Let $S$ be the success of getting $6$ and $F$ be the failure of not getting $6$.

$\Rightarrow P\left( S \right)=p=\frac{1}{6}$ 

$\Rightarrow P\left( F \right)=q=\frac{5}{6}$

\[P\left( A\text{ }wins\text{ }in\text{ }first\text{ }throw \right)=P\left( S \right)=p\] 

\[P\left( A\text{ }wins\text{ }in\text{ third }throw \right)=P\left( FFS \right)=qqp\]

\[P\left( A\text{ }wins\text{ }in\text{ fifth }throw \right)=P\left( FFFFS \right)=qqqqp\]

\[P\left( A\text{ }wins \right)=p+qqp+qqqqp+......\]

\[\Rightarrow P\left( A\text{ }wins \right)=p\left( 1+{{q}^{2}}+{{q}^{4}}+...... \right)\]

\[\Rightarrow P\left( A\text{ }wins \right)=\frac{p}{1-{{q}^{2}}}\]

\[\Rightarrow P\left( A\text{ }wins \right)=\frac{\frac{1}{6}}{1-\frac{25}{36}}=\frac{36}{6\times 11}=\frac{6}{11}\] 

\[P\left( \text{B }wins \right)=1-P\left( \text{A }wins \right)\]

\[P\left( \text{B }wins \right)=1-\frac{6}{11}=\frac{5}{11}\]

Therefore, \[P\left( \text{A }wins \right)=\frac{6}{11}\]and \[P\left( \text{B }wins \right)=\frac{5}{11}\].

18. If $A$and $B$ are events such that $P\left( A \right)=\frac{1}{2}$ , $P\left( A\bigcup B \right)=\frac{3}{5}$ and $P\left( B \right)=p$ find p if events are

a. mutually exclusive

Ans: It is given that, 

$P\left( A \right)=\frac{1}{2}$

 $P\left( B \right)=p$ 

$P\left( A\cup B \right)=\frac{3}{5}$

It is known that, if two events $A$ and $B$ are mutually exclusive then$P\left( A\cap B \right)=0$.

$P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-0$

$\Rightarrow p=\frac{3}{5}-\frac{1}{2}$

$\Rightarrow p=\frac{1}{10}$

Therefore, the value of $p$ is $\frac{1}{10}$.

b. independent

 $P\left( A \right)=\frac{1}{2}$

 $P\left( B \right)=p$

 $P\left( A\cup B \right)=\frac{3}{5}$

We know that if two events $A$ and $B$ are independent then $P\left( A\cap B \right)=P\left( A \right)P\left( B \right)$

$\Rightarrow \frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}$ $\left[ P\left( A \right)P\left( B \right)=\frac{p}{2} \right]$

$\Rightarrow \frac{p}{2}=\frac{3}{5}-\frac{1}{2}$

$\Rightarrow p=\frac{2}{10}=\frac{1}{5}$

Therefore, the value of $p$ is $\frac{1}{5}$.

19. A man takes a step forward with probability $0.4$ and backward with probability $0.6$ . Find the probability that at the end of eleven steps he is one step away from the starting point.

Ans: Let, a step forward be the success $S$ and a step backward be the failure $F$ .

$\Rightarrow P\left( S \right)=p=0.4$ 

$\Rightarrow P\left( F \right)=q=0.6$

In eleven steps he will be one step away from the starting point if the number of successes and failures differ by $1$.

Therefore there are two chances i.e,

Number of successes$=6$ 

Number of failures$=5$ OR

Number of successes$=5$ 

Number of failures$=6$

Required probability$={}^{11}{{C}_{6}}{{p}^{6}}{{q}^{5}}+{}^{11}{{C}_{5}}{{p}^{5}}{{q}^{6}}$ 

$\Rightarrow \frac{11!}{6!5!}{{\left( 0.4 \right)}^{6}}{{\left( 0.6 \right)}^{5}}+\frac{11!}{6!5!}{{\left( 0.4 \right)}^{5}}{{\left( 0.6 \right)}^{6}}$

$\Rightarrow \frac{11!}{6!5!}{{\left( 0.4 \right)}^{5}}{{\left( 0.6 \right)}^{5}}\left[ 0.4+0.5 \right]$

$\Rightarrow \frac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2\times 1}{{\left( 0.4 \right)}^{5}}{{\left( 0.6 \right)}^{5}}\left( 1 \right)=0.3678$

20. Two cards are drawn from a pack of well shuffled $52$ cards one by one with replacement. Getting an ace or a spade is considered a success. Find the probability distribution for the number of successes.

Ans: Let $A$ be the event of a number of aces and spades.

$\Rightarrow n(A)=16$ 

Number of outcomes in sample space,$n(S)=52$ 

Probability of picking an Ace or a spade, $P(A)=\frac{4}{13}$ 

Probability of not picking an Ace or a spade, $P(A')=1-\frac{4}{13}=\frac{9}{13}$ 

Let $X$ be the event of number of aces or spades

When $X=0$ 

$P\left( X=0 \right)=P(A')\times P(A')$ 

$\Rightarrow P\left( X=0 \right)=\frac{9}{13}\times \frac{9}{13}=\frac{81}{169}$

When $X=1$ 

$P\left( X=1 \right)=P(A)\times P(A')$ 

$\Rightarrow P\left( X=1 \right)=\frac{4}{13}\times \frac{9}{13}=\frac{36}{169}$

When $X=2$ 

$P\left( X=2 \right)=P(A)\times P(A)$ 

$\Rightarrow P\left( X=0 \right)=\frac{4}{13}\times \frac{4}{13}=\frac{16}{169}$

Therefore, the probability distribution for the number of successes is:

21. In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/losses.

Ans: The following cases will be obtained from the given scenario:

Man gets $6$in the first throw

Man doesn’t get $6$ in the first throw, gets $6$ in the second throw

Man doesn’t get $6$ in the first and second throws, gets $6$ in the third throw

Man doesn’t get $6$ in the first, second and third throws

Case I: Man gets $6$in the first throw

Probability of getting $6$$=\frac{1}{6}$ 

Amount won$=1$ 

Case II: Man doesn’t get $6$ in the first throw, gets $6$ in the second throw

Probability of getting $6$$=\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}$ 

Amount won$=-1+1=0$ 

Case III: Man doesn’t get $6$ in the first and second throws, gets $6$ in the third throw

Probability of getting $6$$=\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{25}{216}$ 

Amount won$=-1-1+1=-1$ 

Case IV: Man doesn’t get $6$ in the first, second and third throws

Probability of getting $6$$=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}=\frac{125}{216}$ 

Amount won$=-1-1-1=-3$ 

\[Expected\text{ }value=Amount\text{ }won\text{ }\times \text{ }Probability\text{ }for\text{ }all\text{ }four\text{ }throws\] 

\[Expected\text{ }value=\left( 1\times \frac{1}{6} \right)+\left( 0\times \frac{5}{36} \right)+\left( -1\times \frac{25}{216} \right)+\left( -3\times \frac{125}{216} \right)\]

\[Expected\text{ }value=\frac{1}{6}+0-\frac{25}{216}-\frac{125}{216}\]

\[Expected\text{ }value=\frac{36-25-125}{216}\]

\[Expected\text{ }value=\frac{36-400}{216}\]

\[Expected\text{ }value=\frac{-364}{216}=\frac{-91}{54}\]

Therefore, the expected value of the amount he wins or loses is $\frac{-91}{54}$.

22. Suppose that $10\%$ of men and $5\%$ women have grey hair. A grey haired person is selected at random. What is the probability that the selected person is male assuming that there are $60\%$ males and $40\%$  females?

Ans: Let $A$ be the event that the person selected has grey hair

Let ${{E}_{1}}$ be the event that the person selected is male

Let ${{E}_{2}}$ be the event that the person selected is female

$\Rightarrow P\left( {{E}_{1}} \right)=\frac{60}{100}=\frac{3}{5}$ 

$\Rightarrow P\left( {{E}_{2}} \right)=\frac{40}{100}=\frac{2}{5}$

Probability that the selected is male with grey hair:$P\left( {A}/{{{E}_{1}}}\; \right)=\frac{10}{100}=\frac{1}{10}$ 

Probability that the selected is male with grey hair:$P\left( {A}/{{{E}_{2}}}\; \right)=\frac{5}{100}=\frac{1}{20}$ 

From Bayes theorem:\[P\left( {{{E}_{1}}}/{A}\; \right)=\frac{P({{E}_{1}})P\left( {A}/{{{E}_{1}}}\; \right)}{P({{E}_{1}})P\left( {A}/{{{E}_{1}}}\; \right)+P({{E}_{2}})P\left( {A}/{{{E}_{2}}}\; \right)}\] 

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{\frac{3}{5}\times \frac{1}{10}}{\frac{3}{5}\times \frac{1}{10}+\frac{2}{5}\times \frac{1}{20}}\] 

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{3}{3+1}=\frac{3}{4}\]

Therefore, the probability that the selected person is male is $\frac{3}{4}$.

23. A card from a pack of $52$ cards are lost. From the remaining cards of the pack, two cards are drawn. What is the probability that they both are diamonds?

Ans: Let ${{E}_{1}}$ be the event that the lost cad is diamond

In a pack of $52$ there are $13$ diamonds$\Rightarrow P({{E}_{1}})=\frac{13}{52}=\frac{1}{4}$ 

Let ${{E}_{2}}$ be the event that the lost cad is not a diamond

$\Rightarrow P({{E}_{2}})=1-P({{E}_{1}})=1-\frac{1}{4}=\frac{3}{4}$ 

Let $A$ be the event that the two cards drawn are both diamonds

Number of ways of drawing 2 diamond cards

\[\Rightarrow P\left( {A}/{{{E}_{1}}}\; \right)=\frac{Number\text{ }of\text{ }ways\text{ }of\text{ }drawing\text{ }2\text{ }diamond\text{ }cards}{Total\text{ }Number\text{ }of\text{ }ways\text{ }of\text{ }drawing\text{ }2\text{ }cards}\]

$\Rightarrow P\left( {A}/{{{E}_{1}}}\; \right)=\frac{{}^{12}{{C}_{2}}}{{}^{51}{{C}_{2}}}=\frac{66}{1275}$ 

$\Rightarrow P\left( {A}/{{{E}_{2}}}\; \right)=\frac{{}^{13}{{C}_{2}}}{{}^{51}{{C}_{2}}}=\frac{78}{1275}$

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{\frac{66}{1275}\times \frac{1}{4}}{\frac{66}{1275}\times \frac{1}{4}+\frac{78}{1275}\times \frac{3}{4}}=\frac{66}{66+224}\]

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{\frac{66}{1275}\times \frac{1}{4}}{\frac{66}{1275}\times \frac{1}{4}+\frac{78}{1275}\times \frac{3}{4}}=\frac{66}{66+234}\]

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{66}{300}=\frac{11}{50}\]

Therefore, the probability that they both are diamonds is $\frac{11}{50}$.

24. Ten eggs are drawn successively with replacement from a lot containing $10\%$  defective eggs. Find the probability that there is at least one defective egg.

Probability of defective eggs$=10%$ 

$\Rightarrow p=\frac{10}{100}=\frac{1}{10}$ 

Probability of good eggs$=q$ 

$\Rightarrow q=1-\frac{1}{10}=\frac{9}{10}$ 

Probability that at least one egg is defective out of $=p(1)+p(2)+p(3)+.......$ 

$\Rightarrow \left[ p(0)+p(1)+p(2)+.....+p(10) \right]-p(0)$

$\Rightarrow 1-p(0)=1-{{\left( \frac{9}{10} \right)}^{10}}$ 

Therefore, the probability that there is at least one defective egg is $1-{{\left( \frac{9}{10} \right)}^{10}}$.

25. Find the variance of the number obtained on a throw of an unbiased die.

Ans: Let the number obtained on an unbiased die is $X$.

The probability of getting each number is equal in an unbiased die. Here, it is $\frac{1}{6}$

Probability distribution in an unbiased die:

Mean Expectation value is $E(X)=\sum\limits_{i=1}^{n}{{{x}_{i}}{{p}_{i}}}$ 

$\Rightarrow E(X)=1\times \frac{1}{6}+2\times \frac{1}{6}+3\times \frac{1}{6}+4\times \frac{1}{6}+5\times \frac{1}{6}+6\times \frac{1}{6}$

$\Rightarrow E(X)=\frac{21}{6}$

Variance: $Var(X)=E\left( {{X}^{2}} \right)-{{\left[ E\left( X \right) \right]}^{2}}$ 

$E({{X}^{2}})=\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}{{p}_{i}}}$

$\Rightarrow E({{X}^{2}})={{1}^{2}}\times \frac{1}{6}+{{2}^{2}}\times \frac{1}{6}+{{3}^{2}}\times \frac{1}{6}+{{4}^{2}}\times \frac{1}{6}+{{5}^{2}}\times \frac{1}{6}+{{6}^{2}}\times \frac{1}{6}$

$\Rightarrow E({{X}^{2}})=\frac{1+4+9+16+25+36}{6}$

\[\Rightarrow E({{X}^{2}})=\frac{91}{6}\]

$Var(X)=E\left( {{X}^{2}} \right)-{{\left[ E\left( X \right) \right]}^{2}}$

$\Rightarrow Var(X)=\frac{91}{6}-{{\left[ \frac{21}{6} \right]}^{2}}$

$\Rightarrow Var(X)=\frac{91}{6}-\frac{441}{36}$

$\Rightarrow Var(X)=\frac{546-441}{36}=\frac{105}{36}$

$\Rightarrow Var(X)=\frac{35}{12}$

Therefore, the variance of the number obtained on a throw of an unbiased die is $\frac{35}{12}$.

Long Answer Questions (6 Marks)

26. In a hurdle race, a player has to cross $8$ hurdles. The probability that he will clear a hurdle is $\frac{4}{5}$, what is the probability that he will knock down in fewer than $2$  hurdles?

Ans: Let $X$ be the event of a number of hurdles that the player knocks down.

From Binomial distribution,

$P\left( X=x \right)={}^{n}{{C}_{x}}{{q}^{n-x}}{{p}^{x}}$ 

It is given that,

Number of hurdles, $n=8$ 

Probability that he will clear the hurdle, $q=\frac{4}{5}$ 

Probability that he will not clear the hurdle, $p=1-\frac{4}{5}=\frac{1}{5}$

$P\left( X=x \right)={}^{8}{{C}_{x}}{{\left( \frac{4}{5} \right)}^{8-x}}{{\left( \frac{1}{5} \right)}^{x}}$

Probability that he will knock down fewer than $2$ hurdles:

$\Pr obability=P\left( Knock\text{ }0\text{ }hurdles \right)+P\left( Knock\text{ 1 }hurdles \right)$ 

$\Rightarrow \Pr obability=P\left( X=0 \right)+P\left( X=1 \right)$

$\Rightarrow \Pr obability={}^{8}{{C}_{0}}{{\left( \frac{4}{5} \right)}^{8-0}}{{\left( \frac{1}{5} \right)}^{0}}+{}^{8}{{C}_{1}}{{\left( \frac{4}{5} \right)}^{8-1}}{{\left( \frac{1}{5} \right)}^{1}}$

$\Rightarrow \Pr obability=1\times {{\left( \frac{4}{5} \right)}^{8}}\times 1+8\times {{\left( \frac{4}{5} \right)}^{7}}\times \left( \frac{1}{5} \right)$

$\Rightarrow \Pr obability={{\left( \frac{4}{5} \right)}^{7}}\left[ \left( \frac{4}{5} \right)+8\times \left( \frac{1}{5} \right) \right]$

$\Rightarrow \Pr obability={{\left( \frac{4}{5} \right)}^{7}}\left[ \frac{12}{5} \right]$

Therefore, the probability that he will knock down in fewer than $2$  hurdles is $\frac{12}{5}{{\left( \frac{4}{5} \right)}^{7}}$.

27. Bag $A$ contains $4$ red, $3$ white and $2$ black balls. Bag $B$ contains $3$ red, $2$ white and $3$ black balls. One ball is transferred from bag $A$ to bag $B$  and then a ball is drawn from bag $B$. The ball so drawn is found to be red. Find the probability that the transferred ball is black.

Ans: Consider the following notations:

${{A}_{1}}$ : One red ball is transferred from bag $A$ to bag $B$ 

${{A}_{2}}$ : One white ball is transferred from bag $A$ to bag $B$

${{A}_{3}}$ : One black ball is transferred from bag $A$ to bag $B$

$E$ : Drawing a red ball from bag $B$

\[P\left( {{A}_{1}} \right)=\frac{Number\text{ }of\text{ }red\text{ }balls\text{ }in\text{ }bag\text{ }A}{Total\text{ }number\text{ }of\text{ }balls\text{ }in\text{ }bag\text{ }A}\] 

\[\Rightarrow P\left( {{A}_{1}} \right)=\frac{4}{4+3+2}=\frac{4}{9}\]

\[P\left( {{A}_{2}} \right)=\frac{Number\text{ }of\text{ white }balls\text{ }in\text{ }bag\text{ A}}{Total\text{ }number\text{ }of\text{ }balls\text{ }in\text{ }bag\text{ }A}\] 

\[\Rightarrow P\left( {{A}_{2}} \right)=\frac{3}{4+3+2}=\frac{3}{9}\]

\[P\left( {{A}_{3}} \right)=\frac{Number\text{ }of\text{ black }balls\text{ }in\text{ }bag\text{ A}}{Total\text{ }number\text{ }of\text{ }balls\text{ }in\text{ }bag\text{ }A}\] 

\[\Rightarrow P\left( {{A}_{3}} \right)=\frac{3}{4+3+2}=\frac{2}{9}\]

If event ${{A}_{1}}$ occurs, number of red balls in bag $B$ $=(3+1)=4$ 

Total number of balls in bag $B=(3+2+3)+1=9$

$P\left( {E}/{{{A}_{1}}}\; \right)=\frac{Number\text{ }of\text{ red }balls\text{ }in\text{ }bag\text{ B}}{Total\text{ }number\text{ }of\text{ }balls\text{ }in\text{ }bag\text{ B}}$ 

$\Rightarrow P\left( {E}/{{{A}_{1}}}\; \right)=\frac{4}{9}$

If event ${{A}_{2}}$ occurs, number of red balls in bag $B$ $=3$ 

$P\left( {E}/{{{A}_{2}}}\; \right)=\frac{Number\text{ }of\text{ black }balls\text{ }in\text{ }bag\text{ B}}{Total\text{ }number\text{ }of\text{ }balls\text{ }in\text{ }bag\text{ B}}$ 

$\Rightarrow P\left( {E}/{{{A}_{2}}}\; \right)=\frac{3}{9}$

If event ${{A}_{3}}$ occurs, number of red balls in bag $B$ $=3$ 

$P\left( {E}/{{{A}_{3}}}\; \right)=\frac{Number\text{ }of\text{ black }balls\text{ }in\text{ }bag\text{ B}}{Total\text{ }number\text{ }of\text{ }balls\text{ }in\text{ }bag\text{ B}}$ 

$\Rightarrow P\left( {E}/{{{A}_{3}}}\; \right)=\frac{3}{9}$

From Bayes theorem: \[P\left( {{{A}_{3}}}/{E}\; \right)=\frac{P({{A}_{3}})P\left( {E}/{{{A}_{3}}}\; \right)}{P({{A}_{1}})P\left( {E}/{{{A}_{1}}}\; \right)+P({{A}_{2}})P\left( {E}/{{{A}_{2}}}\; \right)+P({{A}_{3}})P\left( {E}/{{{A}_{3}}}\; \right)}\] 

\[\Rightarrow P\left( {{{A}_{3}}}/{E}\; \right)=\frac{\frac{2}{9}\times \frac{3}{9}}{\frac{4}{9}\times \frac{4}{9}+\frac{3}{9}\times \frac{3}{9}+\frac{2}{9}\times \frac{3}{9}}\]

\[\Rightarrow P\left( {{{A}_{3}}}/{E}\; \right)=\frac{6}{16+9+6}=\frac{6}{31}\]

28. If a fair coin is tossed $10$ times, find the probability of getting.

a. exactly six heads, 

Ans: Let $X$ be the number of heads obtained.

Number of tosses, $n=10$ 

Probability of head, $p=\frac{1}{2}$ 

Probability of no getting head,$q=1-p=\frac{1}{2}$ 

From binomial distribution: $P\left( X=x \right)={}^{n}{{C}_{x}}{{q}^{n-x}}{{p}^{x}}$ 

Here, $x=6$ 

\[\Rightarrow P\left( X=6 \right)={}^{10}{{C}_{6}}{{\left( \frac{1}{2} \right)}^{10-6}}{{\left( \frac{1}{2} \right)}^{6}}\] 

$\Rightarrow P\left( X=6 \right)=\frac{10!}{\left( 10-6 \right)!6!}{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P\left( X=6 \right)=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P\left( X=6 \right)=\frac{105}{512}$

b. at least six heads,

Ans: At least six heads means that$\Rightarrow P(x\ge 6)$ 

$P(x\ge 6)=P(x=6)+P(x=7)+P(x=8)+P(x=9)+P(x=10)$ 

$\Rightarrow P(x\ge 6)={}^{10}{{C}_{6}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{7}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{8}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{9}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{10}}{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\ge 6)=\left[ {}^{10}{{C}_{6}}+{}^{10}{{C}_{7}}+{}^{10}{{C}_{8}}+{}^{10}{{C}_{9}}+{}^{10}{{C}_{10}} \right]{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\ge 6)=\left[ \frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}+\frac{10\times 9\times 8}{3\times 2\times 1}+\frac{10\times 9}{2\times 1}+10+1 \right]{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\ge 6)=\left[ 210+120+45+10+1 \right]{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\ge 6)=\frac{193}{512}$

c. at most six heads.

Ans: At most six heads means that$\Rightarrow P(x\le 6)$ 

$P(x\le 6)=P(x=6)+P(x=5)+P(x=4)+P(x=3)+P(x=2)+P(x=1)+P(x=0)$

$P(x\le 6)={}^{10}{{C}_{6}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{5}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{4}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{3}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{1}}{{\left( \frac{1}{2} \right)}^{10}}+{}^{10}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\le 6)=\left[ {}^{10}{{C}_{6}}+{}^{10}{{C}_{5}}+{}^{10}{{C}_{4}}+{}^{10}{{C}_{3}}+{}^{10}{{C}_{2}}+{}^{10}{{C}_{1}}+{}^{10}{{C}_{0}} \right]{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\le 6)=\left[ \frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}+\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}+\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}+\frac{10\times 9\times 8}{3\times 2\times 1}+\frac{10\times 9}{2\times 1}+10+1 \right]{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\le 6)=\left[ 210+252+210+120+45+10+1 \right]{{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\le 6)=\left( 848 \right){{\left( \frac{1}{2} \right)}^{10}}$

$\Rightarrow P(x\le 6)=\frac{53}{64}$

29. A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter by other means of transport are respectively $\frac{3}{13}$, $\frac{1}{5}$, $\frac{1}{10}$ and $\frac{2}{5}$. The probabilities that he will be late are $\frac{1}{4}$, $\frac{1}{3}$and $\frac{1}{12}$ if he comes by train, bus and scooter respectively but if comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?

 $E$ : Event that the doctor visits the patient late

${{T}_{1}}$ : Event that the doctor comes by train, $\Rightarrow P\left( {{T}_{1}} \right)=\frac{3}{10}$

${{T}_{2}}$ : Event that the doctor comes by bus, $\Rightarrow P\left( {{T}_{2}} \right)=\frac{1}{5}$

${{T}_{3}}$ : Event that the doctor comes by scooter, $\Rightarrow P\left( {{T}_{3}} \right)=\frac{1}{10}$

${{T}_{4}}$ : Event that the doctor comes by other means of transport, $\Rightarrow P\left( {{T}_{4}} \right)=\frac{2}{5}$

Probability that the doctor arriving late comes by train, $P\left( {E}/{{{T}_{1}}}\; \right)=\frac{1}{4}$ 

Probability that the doctor arriving late comes by bus, $P\left( {E}/{{{T}_{2}}}\; \right)=\frac{1}{3}$ 

Probability that the doctor arriving late comes by scooter, $P\left( {E}/{{{T}_{3}}}\; \right)=\frac{1}{12}$ 

Probability that the doctor arriving late comes by other transport, $P\left( {E}/{{{T}_{4}}}\; \right)=0$ 

From Baye’s theorem: \[P\left( {{{T}_{1}}}/{E}\; \right)=\frac{P({{T}_{1}})P\left( {E}/{{{T}_{1}}}\; \right)}{P({{T}_{1}})P\left( {E}/{{{T}_{1}}}\; \right)+P({{T}_{2}})P\left( {E}/{{{T}_{2}}}\; \right)+P({{T}_{3}})P\left( {E}/{{{T}_{3}}}\; \right)+P({{T}_{4}})P\left( {E}/{{{T}_{4}}}\; \right)}\]

\[\Rightarrow P\left( {{{T}_{1}}}/{E}\; \right)=\frac{\frac{3}{10}\times \frac{1}{4}}{\frac{3}{10}\times \frac{1}{4}+\frac{1}{5}\times \frac{1}{3}+\frac{1}{10}\times \frac{1}{12}+\frac{2}{5}\times 0}\]

\[\Rightarrow P\left( {{{T}_{1}}}/{E}\; \right)=\frac{\frac{3}{40}}{\frac{3}{40}+\frac{1}{15}+\frac{1}{120}+0}\]

\[\Rightarrow P\left( {{{T}_{1}}}/{E}\; \right)=\frac{\frac{3}{40}}{\frac{18}{120}}=\frac{1}{2}\]

Therefore, the probability that the doctor arrives late by train is $\frac{1}{2}$. 

30. A man is known to speak the truth $3$ out of $4$ times. He throws a die and reports that it is six. Find the probability that it is actually a six.

Ans: Let $A$ be the event that the number on die is $6$.

Probability that the man speaks truth, $P(T)=\frac{3}{4}$ 

Probability that the man lies, $P(F)=1-\frac{3}{4}=\frac{1}{4}$ 

Probability of getting $6,P\left( {A}/{T}\; \right)=\frac{1}{6}$ 

Probability of not getting $6,P\left( {A}/{F}\; \right)=1-\frac{1}{6}=\frac{5}{6}$ 

From Bayes theorem: \[P\left( {T}/{A}\; \right)=\frac{P(T)P\left( {A}/{T}\; \right)}{P(T)P\left( {A}/{T}\; \right)+P(F)P\left( {A}/{F}\; \right)}\]

\[\Rightarrow P\left( {T}/{A}\; \right)=\frac{\frac{3}{4}\times \frac{1}{6}}{\frac{3}{4}\times \frac{1}{6}+\frac{1}{4}\times \frac{5}{6}}\]

\[\Rightarrow P\left( {T}/{A}\; \right)=\frac{3}{3+5}=\frac{3}{8}\]

Therefore, Probability that its actually $6$ is $\frac{3}{8}$.

31. An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of an accidents are $0.01$, $0.03$and$0.15$respectively if one of the insured persons meets with an accident. What is the probability that he is a scooter driver?

$E$ : The event when the driver drives scooter 

$F$ :  The event when the driver drives car

$G$ : The event when the driver drives truck

$K$ : The event that the driver meets accident

From the given data:

$P\left( E \right)=\frac{2000}{12000}$

$P\left( F \right)=\frac{4000}{12000}$

$P\left( G \right)=\frac{6000}{12000}$

$P\left( K\left| E \right. \right)=\frac{1}{100}$ $\left( K\left| E \right. \right)$: The driver meets accident provided he drives scooter

$P\left( K\left| F \right. \right)=\frac{3}{100}$ $\left( K\left| F \right. \right)$: The driver meets accident provided he drives car

$P\left( K\left| G \right. \right)=\frac{15}{100}$ $\left( K\left| G \right. \right)$: The driver meets accident provided he drives truck

From Bayes theorem:

$P\left( E\left| K \right. \right)=\frac{P\left( E \right)P\left( K\left| E \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)+P\left( G \right)P\left( K\left| G \right. \right)}$

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{1}{6}\times \frac{1}{100}}{\frac{1}{6}\times \frac{1}{100}+\frac{1}{3}\times \frac{3}{100}+\frac{1}{2}\times \frac{15}{100}}\]

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}}\]

\[\Rightarrow P\left( E\left| K \right. \right)=\frac{1}{52}\]

Therefore, the probability that the accidental person is a scooter driver is $\frac{1}{52}$.

32. Two cards from a pack of $52$ cards are lost. One card is drawn from the remaining cards. If a drawn card is a heart, find the probability that the lost cards were both hearts.

Ans: Consider the following events:

${{E}_{1}}$ be the event both lost cards are hearts.

${{E}_{2}}$ be the event both lost cards are non hearts.

${{E}_{3}}$ be the event one lost card is non heart and one is heart.

$A$ be the event of picking a heart from remaining  $50$ cards.

$P\left( {{E}_{1}} \right)=\frac{{}^{13}{{C}_{2}}}{{}^{52}{{C}_{2}}}=\frac{13\times 12}{52\times 51}$ 

$\Rightarrow P\left( {{E}_{1}} \right)=\frac{1}{17}$

$P\left( \frac{A}{{{E}_{1}}} \right)=\frac{11}{50}$ 

$P\left( {{E}_{2}} \right)=\frac{{}^{39}{{C}_{2}}}{{}^{52}{{C}_{2}}}=\frac{39\times 38}{52\times 51}$ 

$\Rightarrow P\left( {{E}_{2}} \right)=\frac{19}{34}$

\[P\left( \frac{A}{{{E}_{2}}} \right)=\frac{13}{50}\] 

$P\left( {{E}_{3}} \right)=\frac{(One~~heart)~and\left( One~~non~heart \right)}{{}^{52}{{C}_{2}}}$

$P\left( {{E}_{3}} \right)=\frac{13\times 39}{{}^{52}{{C}_{2}}}=\frac{13\times 39\times 2}{52\times 51}$ 

$\Rightarrow P\left( {{E}_{3}} \right)=\frac{39}{102}$

$P\left( \frac{A}{{{E}_{3}}} \right)=\frac{12}{50}$ 

From Baye’s theorem:\[P\left( {{{E}_{1}}}/{A}\; \right)=\frac{P({{E}_{1}})P\left( {A}/{{{E}_{1}}}\; \right)}{P({{E}_{1}})P\left( {A}/{{{E}_{1}}}\; \right)+P({{E}_{2}})P\left( {A}/{{{E}_{2}}}\; \right)+P({{E}_{3}})P\left( {A}/{{{E}_{3}}}\; \right)}\] 

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{\frac{1}{17}\times \frac{11}{50}}{\frac{1}{17}\times \frac{11}{50}+\frac{19}{34}\times \frac{13}{50}+\frac{39}{102}\times \frac{12}{50}}\]

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{11}{11+\frac{247}{2}+78}\]

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{22}{22+247+156}\]

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{22}{425}\]

Therefore, the probability that the lost cards were both hearts is $\frac{22}{425}$.

33. A box $X$ contains $2$ white and $3$ red balls and a bag $Y$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag $Y$.

Ans: Let the event of selecting a red ball be $R$.

Let the event of selecting the bag $X$ be $X$.

Let the event of selecting the bag $Y$ be $Y$.

Bag $X$ has $2$ white and $3$ red balls.

Bag $Y$ has $4$ white and $5$ red balls.

$\Rightarrow P\left( X \right)=P\left( Y \right)=\frac{1}{2}$ 

$P\left( {R}/{X}\; \right)=\frac{3}{5}$ 

$P\left( {R}/{Y}\; \right)=\frac{5}{9}$

From Bayes theorem: \[P\left( {Y}/{R}\; \right)=\frac{P(Y)P\left( {R}/{Y}\; \right)}{P(X)P\left( {R}/{X}\; \right)+P(Y)P\left( {R}/{Y}\; \right)}\]

\[\Rightarrow P\left( {Y}/{R}\; \right)=\frac{\frac{5}{9}\times \frac{1}{2}}{\frac{3}{5}\times \frac{1}{2}+\frac{5}{9}\times \frac{1}{2}}\]

\[\Rightarrow P\left( {Y}/{R}\; \right)=\frac{25}{27+25}=\frac{25}{52}\]

Therefore, the probability that the ball is red and drawn from bag $Y$is$\frac{25}{52}$.

34. In answering a question on a multiple choice, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be incorrect with probability $\frac{1}{4}$. What is the probability that the student knows the answer, given that he answered correctly?

Ans: Let ${{E}_{1}}$ and ${{E}_{2}}$ be the respective events in which the student knows the answer and guesses the answer.

Let $A$ be the event that the answer is correct.

$P\left( {{E}_{1}} \right)=\frac{3}{4}$ 

$P\left( {{E}_{2}} \right)=\frac{1}{4}$

The probability that the student answered correctly by knowing the answer is $1$.

$\Rightarrow P\left( {A}/{{{E}_{1}}}\; \right)=1$  

The probability that the student answered correctly by guessing the answer is $\frac{1}{4}$

$\Rightarrow P\left( {A}/{{{E}_{2}}}\; \right)=\frac{1}{4}$  

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{\frac{3}{4}\times 1}{\frac{3}{4}\times 1+\frac{1}{4}\times \frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}}\]

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{\frac{3}{4}}{\frac{13}{16}}\]

\[\Rightarrow P\left( {{{E}_{1}}}/{A}\; \right)=\frac{12}{13}\]

Therefore, the probability that the student answers correctly by knowing the answer is $\frac{12}{13}$.

35. Suppose a girl throws a die. If she gets $5$ or $6$She tosses a coin three times and notes the number of heads. If she gets $1$, $2$, $3$ or $4$ She tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head. What is the probability that she throws $1$, $2$, $3$ or $4$ with the die?

$E$ : The event when $5\text{ or 6}$ come

$F$ :  The event when $\text{1,2,3 or 4}$ come

$K$ : The event that the coin shows exactly one head

$P\left( E \right)=\frac{2}{6}$

$P\left( F \right)=\frac{4}{6}$

$P\left( K\left| E \right. \right)=\frac{3}{8}$ $\left( K\left| E \right. \right)$: The event that the coin shows exactly one head provided $5\text{ or 6}$come

$P\left( K\left| F \right. \right)=\frac{1}{2}$ $\left( K\left| F \right. \right)$: The event that the single throw of coin shows exactly one head provided $\text{1,2,3 or 4}$come

From Bayes theorem 

$P\left( F\left| K \right. \right)=\frac{P\left( F \right)P\left( K\left| F \right. \right)}{P\left( E \right)P\left( K\left| E \right. \right)+P\left( F \right)P\left( K\left| F \right. \right)}$

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{\frac{4}{6}\times \frac{1}{2}}{\frac{2}{6}\times \frac{3}{8}+\frac{4}{6}\times \frac{1}{2}}\]

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{2}{\frac{3}{4}+2}=\frac{8}{11}\]

Therefore, the probability that the item is produced from the second group is $\frac{8}{11}$.

36. In a bolt factory machines $A$, $B$ and $C$ manufacture $60%$, $30%$ and $10%$ of the total bolts respectively, $2%$, $5%$ and $10%$ of the bolts produced by them respectively are defective. A bolt is picked up at random from the product and is found to be defective. What is the probability that it has been manufactured by machine $A$?

In a bolt factory machines $A$, $B$ and $C$ manufacture $60%$, $30%$ and $10%$ of the total bolts.

Let total bolts be $x$.

Number of bolts in Machine $A=x\times \frac{60}{100}=0.6x$ 

Number of bolts in Machine $B=x\times \frac{30}{100}=0.3x$ 

Number of bolts in Machine $C=x\times \frac{10}{100}=0.1x$ 

Probability of picking a bolt from $A$, $P\left( A \right)=\frac{0.6x}{0.6x+0.3x+0.1x}=0.6x$ 

Probability of picking a bolt from $B$, $P\left( B \right)=\frac{0.3x}{0.6x+0.3x+0.1x}=0.3x$ 

Probability of picking a bolt from $C$, $P\left( C \right)=\frac{0.1x}{0.6x+0.3x+0.1x}=0.1x$ 

Let $D$ be the event of a bolt picked up at random from the product and is found to be defective.

$P\left( {D}/{A}\; \right)=\frac{2}{100}=0.02$ 

$P\left( {D}/{B}\; \right)=\frac{5}{100}=0.05$

\[P\left( {D}/{C}\; \right)=\frac{10}{100}=0.1\]

From Baye’s theorem:\[P\left( {A}/{D}\; \right)=\frac{P(A)P\left( {D}/{A}\; \right)}{P(A)P\left( {D}/{A}\; \right)+P(B)P\left( {D}/{B}\; \right)+P(C)P\left( {D}/{C}\; \right)}\]

\[\Rightarrow P\left( {A}/{D}\; \right)=\frac{0.6\times 0.02}{0.6\times 0.02+0.3\times 0.05+0.1\times 0.1}\]

\[\Rightarrow P\left( {A}/{D}\; \right)=\frac{12}{12+15+10}\]

\[\Rightarrow P\left( {A}/{D}\; \right)=\frac{12}{37}\]

Therefore, probability of a bolt picked up at random from the product and is found to be defective and is from machine $A$is $\frac{12}{37}$.

37. Two urns $A$ and $B$ contain $6$ black and $4$ white, $4$ black and $6$ white balls respectively. Two balls are drawn from one of the urns. If both the balls drawn are white, find the probability that the balls are drawn from urn $B$.

Ans: Let $E$be the event of drawing two white balls.

Urn $A$contains $6$ black and $4$ white balls.

Urn $B$contains $4$ black and $6$ white balls.

Probability of choosing Urn $A$or Urn $B$ is $P(A)=P(B)=\frac{1}{2}$

Probability of choosing two white balls from \[A=P\left( {E}/{A}\; \right)\] 

\[\Rightarrow P\left( {E}/{A}\; \right)=\frac{{}^{4}{{C}_{2}}}{{}^{10}{{C}_{2}}}=\frac{6}{45}\]

Probability of choosing two white balls from \[B=P\left( {E}/{B}\; \right)\] 

\[\Rightarrow P\left( {E}/{B}\; \right)=\frac{{}^{6}{{C}_{2}}}{{}^{10}{{C}_{2}}}=\frac{15}{45}\]

From Bayes theorem:\[P\left( {B}/{E}\; \right)=\frac{P(B)P\left( {E}/{B}\; \right)}{P(A)P\left( {E}/{A}\; \right)+P(B)P\left( {E}/{B}\; \right)}\]

\[\Rightarrow P\left( {B}/{E}\; \right)=\frac{\frac{1}{2}\times \frac{15}{45}}{\frac{1}{2}\times \frac{6}{45}+\frac{1}{2}\times \frac{15}{45}}=\frac{15}{6+15}\]

\[\Rightarrow P\left( {B}/{E}\; \right)=\frac{15}{21}=\frac{5}{7}\]

Therefore, the probability that the balls are drawn from urn $B$ is $\frac{5}{7}$.

38. Two cards are drawn from a well shuffled pack of $52$ cards. Find the mean and variance for the number of face cards obtained.

Ans: Let $X$ be the event of the number of face cards. 

When two cards are picked with replacement there are chances of getting either \[0,1or\text{ 2}\]face cards.

Number of face cards in a deck$=12$ 

Number of events in sample space, $n\left( S \right)=52$

$P\left( N \right)$ be the probability of not picking face card , $P\left( N \right)=\frac{40}{52}$

$P\left( F \right)$ be the probability of picking a face card , $P\left( F \right)=\frac{12}{52}$

For $X=0$: 

Probability, $P\left\{ NN \right\}=\frac{40}{52}\times \frac{40}{52}$

$\Rightarrow P\left\{ NN \right\}=\frac{40}{52}\times \frac{40}{52}=\frac{100}{169}$

For $X=1$:  

Probability, \[P\left( NF\text{ or }FN \right)=P(N)P(F)+P(F)P(N)\] 

\[\Rightarrow P\left( NF\text{ or }FN \right)=2\times \frac{40}{52}\times \frac{12}{52}=\frac{60}{169}\]

For $X=2$: 

Probability, \[P\left( FF \right)=\frac{12}{52}\times \frac{12}{52}\] 

\[\Rightarrow P\left( FF \right)=\frac{9}{169}\]

Probability distribution:

Mean Expectation value is $\mu =E(X)=\sum\limits_{i=1}^{n}{{{x}_{i}}{{p}_{i}}}$ 

$\Rightarrow \mu =0\times \frac{100}{169}+1\times \frac{60}{169}+2\times \frac{9}{169}$

$\Rightarrow \mu =0+\frac{100}{169}+\frac{60}{169}+\frac{18}{169}$

$\Rightarrow \mu =\frac{78}{169}=\frac{6}{13}$

Therefore, the mean number of heads is $\frac{6}{13}$.

$\Rightarrow E({{X}^{2}})={{0}^{2}}\times \frac{100}{169}+{{1}^{2}}\times \frac{60}{169}+{{2}^{2}}\times \frac{9}{169}$

$\Rightarrow E({{X}^{2}})=\frac{0+60+36}{169}$

\[\Rightarrow E({{X}^{2}})=\frac{96}{169}\]

$\Rightarrow Var(X)=\frac{96}{169}-{{\left[ \frac{6}{13} \right]}^{2}}$

$\Rightarrow Var(X)=\frac{96-36}{169}$

$\Rightarrow Var(X)=\frac{60}{169}$

Therefore, the variance of the number of heads is $\frac{60}{169}$.

39. Write the probability distribution for the number of heads obtained when three coins are tossed together. Also, find the mean and variance of the number of heads.

Ans: Let $X$ be the event of the number of heads.

When three coins are tossed simultaneously there are chances of getting either \[0,1,2~or\text{ 3}\]heads.

Sample space, $S=\left\{ TTT,HTT,THT,TTH,HHT,HTH,THH,HHH \right\}$ 

Number of events in sample space, $n\left( S \right)=8$ 

For $X=0$: Possible outcome$\left\{ TTT \right\}$ 

Number of possible outcomes$=1$ 

Probability, \[P\left( X \right)=\frac{1}{8}\] 

For $X=1$: Possible outcome$\left\{ HTT,THT,TTH \right\}$ 

Number of possible outcomes$=3$ 

Probability, \[P\left( X \right)=\frac{3}{8}\] 

For $X=2$: Possible outcome$\left\{ HHT,HTH,THH \right\}$ 

For $X=3$: Possible outcome$\left\{ HHH \right\}$ 

$\Rightarrow \mu =0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$

$\Rightarrow \mu =0+\frac{3}{8}+\frac{6}{8}+\frac{3}{8}$

$\Rightarrow \mu =\frac{12}{8}=\frac{3}{2}$

Therefore, the mean number of heads is $\frac{3}{2}$.

$\Rightarrow E({{X}^{2}})={{0}^{2}}\times \frac{1}{8}+{{1}^{2}}\times \frac{3}{8}+{{2}^{2}}\times \frac{3}{8}+{{3}^{2}}\times \frac{1}{8}$

$\Rightarrow E({{X}^{2}})=\frac{0+3+12+9}{8}$

\[\Rightarrow E({{X}^{2}})=\frac{24}{8}=3\]

$\Rightarrow Var(X)=3-{{\left[ \frac{3}{2} \right]}^{2}}$

$\Rightarrow Var(X)=\frac{12-9}{4}$

$\Rightarrow Var(X)=\frac{3}{4}$

Therefore, the variance of the number of heads is $\frac{3}{4}$.

40. Two groups are competing for the position on the Board of Directors of a corporation. The probabilities that the first and the second groups will win are $0.6$ and $0.4$ respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$ and the corresponding probability is $0.3$ if the second group wins. Find the probability that the new product introduced was by the second group.

$E$ : The event when the first group wins

$F$ :  The event when the second group wins

$K$ : The event that the new item is produced

$P\left( E \right)=0.6$

$P\left( F \right)=0.4$

$P\left( K\left| E \right. \right)=0.7$ $\left( K\left| E \right. \right)$: The event that the item is introduced by first group 

$P\left( K\left| F \right. \right)=0.3$ $\left( K\left| F \right. \right)$: The event that item is introduced by second group

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{0.4\times 0.3}{0.6\times 0.7+0.4\times 0.3}\]

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{12}{42+12}\]

\[\Rightarrow P\left( F\left| K \right. \right)=\frac{12}{54}=\frac{2}{9}\]

Thus, the probability that the item is produced from the second group is $\frac{2}{9}$.

Important Related Links for CBSE Class 12 Maths 

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NCERT Solutions for Class 12 Maths Chapter 13 – Probability

Ncert solutions for class 12 maths chapter 13 – probability pdf.

NCERT Solutions for Class 12 Maths Chapter 13 – Probability includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 13 – Probability Maths NCERT Solutions for Class 12 to help you to score more marks in your board exams and as well as competitive exams.

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Class 12 Maths Case Study Questions

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Class 12 Maths question paper will have 1-2 Case Study Questions. These questions will carry 5 MCQs and students will attempt any four of them. As all of these are only MCQs, it is easy to score good marks with a little practice. Class 12 Maths Case Study Questions are available on the myCBSEguide App and Student Dashboard .

Why Case Studies in CBSE Syllabus?

CBSE has introduced case study questions in the CBSE curriculum recently. The purpose was to make students ready to face real-life challenges with the knowledge acquired in their classrooms. It means, there was a need to connect theories with practicals. Whatsoever the students are learning, they must know how to apply it in their day-to-day life. That’s why CBSE is emphasizing case studies and competency-based education .

Case Study Questions in Maths

Let’s have a look over the class 12 Mathematics sample question paper issued by CBSE, New Delhi. Question numbers 17 and 18 are case study questions.

Focus on concepts

If you go through each MCQ there, you will find that the theme/case study is common but the questions are based on different concepts related to the theme. It means, that if you have done ample practice on the various concepts, you can solve all these MCQs in minutes.

Easy Questions with a Practical Approach

The difficulty level of the questions is average or say easy in some cases. On the other hand, you get four options to choose from. So, you get two levels of support to get full marks with very little effort.

Practice Questions Regularly

Most of the time we feel that it’s easy and neglect it. But in the end, we have to pay for this negligence. This may happen here too. Although it’s easy to score good marks on the case study questions if you don’t practice such questions, you may lose your marks. So, we suggest students should practice at least 30-40 such questions before writing the board exam.

12 Maths Case-Based Questions

We are giving you some examples of case study questions here. We have arranged hundreds of such questions chapter-wise on the myCBSEguide App. It is the complete guide for CBSE students. You can download the myCBSEguide App and get more case study questions there.

Case Study Question – 1

• A is a diagonal matrix
• A is a scalar matrix
• A is a zero matrix
• A is a square matrix
• If A and B are two matrices such that AB = B and BA = A, then B 2 is equal to

Case Study Question – 2

• 4(x 3  – 24x 2   + 144x)
• 4(x 3 – 34x 2   + 244x)
• x 3  – 24x 2   + 144x
• 4x 3  – 24x 2   + 144x
• Local maxima at x = c 1
• Local minima at x = c 1
• Neither maxima nor minima at x = c 1
• None of these

Case Study Questions Matrices -1

Answer Key:

Case Study Questions Matrices – 2

Read the case study carefully and answer any four out of the following questions: Once a mathematics teacher drew a triangle ABC on the blackboard. Now he asked Jose,” If I increase AB by 11 cm and decrease the side BC by 11 cm, then what type of triangle it would be?” Jose said, “It will become an equilateral triangle.”

Again teacher asked Suraj,” If I multiply the side AB by 4 then what will be the relation of this with side AC?” Suraj said it will be 10 cm more than the three times AC.

Find the sides of the triangle using the matrix method and  answer the following questions:

• (a) 3  ×  3

Case Study Questions Determinants – 01

DETERMINANTS:  A determinant is a square array of numbers (written within a pair of vertical lines) that represents a certain sum of products. We can solve a system of equations using determinants, but it becomes very tedious for large systems. We will only do 2 × 2 and 3 × 3 systems using determinants. Using the properties of determinants solve the problem given below and answer the questions that follow:

Three shopkeepers Ram Lal, Shyam Lal, and Ghansham are using polythene bags, handmade bags (prepared by prisoners), and newspaper envelopes as carrying bags. It is found that the shopkeepers Ram Lal, Shyam Lal, and Ghansham are using (20,30,40), (30,40,20), and (40,20,30) polythene bags, handmade bags, and newspapers envelopes respectively. The shopkeeper’s Ram Lal, Shyam Lal, and Ghansham spent ₹250, ₹270, and ₹200 on these carry bags respectively.

• (b) Shyam Lal
• (a) Ram Lal

Case Study Questions Determinants – 02

Case study questions application of derivatives.

• R(x) = -x 2  + 200x + 150000
• R(x) = x 2  – 200x – 140000
• R(x) = 200x 2  + x + 150000
• R(x) = -x 2  + 100 x + 100000
• R'(x) > 0
• R'(x) < 0
• R”(x) = 0
• (a) -x 2  + 200x + 150000
• (a) R'(x) = 0
• (c) 257, -63

Case Study Questions Vector Algebra

• tan−1⁡(5/12)
• tan−1⁡(12/3)
• (b) 130 m/s
• (a)  tan−1⁡(5/12)
• (b) 170 m/s

More Case Study Questions

These are only some samples. If you wish to get more case study questions for CBSE class 12 maths, install the myCBSEguide App. It has class 12 Maths chapter-wise case studies with solutions.

12 Maths Exam pattern

Question Paper Design of CBSE class 12 maths is as below. It clearly shows that 20% weightage will be given to HOTS questions. Whereas 55% of questions will be easy to solve.

• No. chapter-wise weightage. Care to be taken to cover all the chapters
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections

12 Maths Prescribed Books

• Mathematics Part I – Textbook for Class XII, NCERT Publication
• Mathematics Part II – Textbook for Class XII, NCERT Publication
• Mathematics Exemplar Problem for Class XII, Published by NCERT
• Mathematics Lab Manual class XII, published by NCERT

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CBSE Class 12 Maths –Chapter 13 Probability- Study Materials

NCERT Solutions Class 12 All Subjects Sample Papers Past Years Papers

Notes and Study Materials

• Concepts of Probability
• Master File for Probability
• Probability Note
• NCERT Solutions for – Probability
• NCERT Exemplar Solutions for – Probability
• R D Sharma Solution of Mean Random Variable
• R D Sharma Solution of Binomial Distribution
• R D Sharma Solution of Probability
• Past Many Years CBSE Questions and Answer Of Relation and Function
• Probability Mind Map

Examples and Exercise

• Probability : Practice Paper 1
• Probability : Practice Paper 2
• Probability : Practice Paper 3
• Probability : Practice Paper 4

CBSE Expert

Case Study Questions for Class 12 Maths PDF Download

We have provided here Case Study questions for the Class 12 Maths for board exams. You can read these chapter-wise Case Study questions. These questions are prepared by subject experts and experienced teachers. The answer key is also provided so that you can check the correct answer for each question. Practice these questions to score well in your final exams.

CBSE 12th Standard CBSE Maths question papers, important notes, study materials, Previous Year questions, Syllabus, and exam patterns. Free 12th Standard CBSE Maths books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 12 Maths Case Study Questions

CBSE Class 12 Maths question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 12 Maths students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 12 Maths

Class 12 students should go through important Case Study problems for Maths before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 12 Maths examinations. Our expert faculty for standard 12 Maths have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 12 students understand the concepts and also easy to learn solutions.

Books for Class 12 Maths

Strictly as per the new term-wise syllabus for Board Examinations to be held in the academic session 2022-23 for class 12 Multiple Choice Questions based on new typologies introduced by the board- Stand-Alone MCQs, MCQs based on Assertion-Reason Case-based MCQs. Include Questions from CBSE official Question Bank released in April 2022 Answer key with Explanations What are the updates in the book: Strictly as per the Term wise syllabus for Board Examinations to be held in the academic session 2022-23. Chapter-wise -Topic-wise Multiple choice questions based on the special scheme of assessment for Board Examination for Class 12th.

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CBSE Class 12th - MATHEMATICS : Chapterwise Case Study Question & Solution

CBSE will ask two Case Study Questions in the CBSE class 12 Mathematics questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them.

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Class 12 Maths Chapter 13 Probability MCQs

Class 12 Maths Chapter 13 Probability MCQs with solutions are provided here for students. The multiple-choice questions are prepared as per the latest exam pattern. Class 12 Maths chapter-wise MCQs are given by our BYJU’S subject experts to help students score good marks in the board exams. Students can practice these questions that are formulated as per the CBSE syllabus (2022-2023) and NCERT curriculum. Let us see some of the MCQs and solve them to prepare for the chapter probability.

Probability Class 12 MCQs with Solutions

Find MCQs for Class 12 Chapter 13 Probability with solutions here.

Download PDF – Chapter 13 Probability MCQs

Q.1: P(A ∩ B) is equal to:

A. P(A) . P(B|A)

B. P(B) . P(A|B)

C. Both A and B

D. None of these

Answer: C. Both A and B

Explanation: By multiplication theorem of probability .

Q.2: If P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4, what is the value of P (A ∩ B)?

Answer: A. 0.32

Explanation: Given, P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4

By conditional probability, we have;

P (B|A) = P(A ∩ B)/P(A)

⇒ P (A ∩ B) = P(B|A). P(A) = 0.4 x 0.8 = 0.32

Q.3: If P (A) = 6/11, P (B) = 5/11 and P (A ∪ B) = 7/11, what is the value of P(B|A)?

D. None of the above

Answer: B. ⅔

Explanation: By definition of conditional probability we know;

P(B|A) = P(A ∩ B)/P(A) …(i)

P(A ∩ B) = P(A) + P(B) – P(A U B)

= 6/11 + 5/11 – 7/11

Now putting the value of P(A ∩ B) in eq.(i), we get;

P(B|A) = (4/11)/(6/11) = 4/6 = ⅔

Q.4: Find P(E|F), where E: no tail appears, F: no head appears, when two coins are tossed in the air.

Answer: A. 0

Explanation: Given,

E: no tail appears

And F: no head appears

⇒ E = {HH} and F = {TT}

⇒ E ∩ F = ϕ

As we know, two coins were tossed;

P(E ∩ F) = 0/4 = 0

Thus, by conditional probability, we know that;

P(E|F) = P(E ∩ F)/P(F)

Q.5: If P(A ∩ B) = 70% and P(B) = 85%, then P(A/B) is equal to:

Answer: B.14/17

Explanation: By conditional probability, we know;

P(A|B) = P(A ∩ B)/P(B)

= (70/100) x (100/85)

Q.6: If P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6. Find P(A ∪ B).

Answer: B. 0.86

P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6

By conditional probability, we know;

P(B|A) = P(A ∩ B)/P(A)

⇒ 0.6 × 0.4 = P (A ∩ B)

⇒ P(A ∩ B) = 0.24

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 0.4 + 0.7 – 0.24

Q.7: An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Answer: A. 3/7

Explanation: Let E and F denote the events that the first and second ball drawn is black, respectively.

We need to find P(E ∩ F) or P (EF).

P(E) is the probability of black ball first drawn.

P(E) = 10/15

Now, 9 black balls are left in the urn.

P(F|E) = 9/14

By multiplication rule;

P (E ∩ F) = P (E) P(F|E)

= 10/15 x 9/14

Q.8: If E and F are independent events, then;

A. P(E ∩ F) = P(E)/ P(F)

B. P(E ∩ F) = P(E) + P(F)

C. P(E ∩ F) = P(E) . P(F)

Answer: C. P(E ∩ F) = P(E) . P(F)

Explanation: Two events E and F are said to be independent if;

P(F|E) = P (F) given P (E) ≠ 0

P (E|F) = P (E) given P (F) ≠ 0

Now, by the multiplication rule of probability, we have

P(E ∩ F) = P(E) . P (F|E) … (1)

If E and F are independent, then;

P(E ∩ F) = P(E) . P(F)

Q.9: If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by:

A. 1+ P(A′) P (B′)

B. 1− P(A′) P (B′)

C. 1− P(A′) + P (B′)

D. 1− P(A′) – P (B′)

Answer: B. 1− P(A′) P (B′)

Explanation: P(at least one of A and B) = P(A ∪ B)

= P(A) + P(B) − P(A ∩ B)

= P(A) + P(B) − P(A) P(B)

= P(A) + P(B). P(A′)

= 1− P(A′) + P(B) P(A′)

= 1− P(A′) P (B′)

Q.10: The probability of solving the specific problems independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that exactly one of them solves the problem.

Answer: B. ½

Explanation: P(A) = ½

Since, A and B are independent events, therefore;

⇒ P (A ∩ B) = P (A). P (B)

⇒ P (A ∩ B) = ½ × 1/3 = 1/6

P (A’) = 1 – P (A) = 1 – 1/2 = 1/2

P (B’) = 1 – P (B) = 1 – 1/3 = 2/3

Now the probability that exactly one of them solved the problem is either the problem is solved by A and not B or vice versa.

P (A).P (B’) + P (A’).P (B)

= ½ (2/3) + ½ (1/3)

= 1/3 + 1/6 = 3/6

⇒ P (A).P (B’) + P (A’).P (B) = ½

Related Articles for Class 12 Maths

• Class 12 Syllabus
• Probability For Class 12
• Conditional Probability
• Events and Types of Events in Probability
• CBSE Class 12 Maths Board Exam: Important 4 marks questions
• CBSE Class 12 Maths Board Exam: Important 6 marks questions
• Important Questions for Class 12 Maths chapter 13 Probability
• Important Topics and Tips Prepare for Class 12 Maths Exam

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Case study questions on Relations & Functions, Matrices, Determinants, Applications of Derivatives, Applications of Integral, Differential equations, Vector Algebra, three dimensional Geometry and Probability etc.

CASE STUDY-1 CHAPTER -1 

Relations & functions.

A general election of Lok Sabha is a gigantic exercise. About 911 million people were eligible to vote and voter turnout was about 67%, the highest ever

Let I be the set of all citizens of India who were eligible to exercise their voting right in general election held in 2019. A relation ‘R’ is defined on I as follows:

R = {( 𝑉 1, 𝑉 2) ∶ 𝑉 1, 𝑉 2 ∈ 𝐼 and both use their voting right in general election – 2019}

Q1) Two neighbors X and Y ∈ I. X exercised his voting right while Y did not cast her vote in general election – 2019. Which of the following is true?

a) (X, Y) ∈ R

b) (Y, X) ∈ R

c) (X, X) ∉ R

d) (X, Y) ∉ R

2. Mr.’ 𝑋 ’ and his wife ‘ 𝑊 ’both exercised their voting right in general election -2019, Which of the following is true?

a) both (X, W) and (W, X) ∈ R

b)  (X, W) ∈ R but (W, X) ∉ R

c)  both (X, W) and (W, X) ∉ R

d)  (W, X) ∈ R but (X, W) ∉ R

Q3) Three friends F1, F2 and F3 exercised their voting right in general election-2019, then which of the following is true?

a) (F1, F2 ) ∈ R, (F2, F3) ∈ R and (F1,F3) ∈ R

b) (F1, F2 ) ∈ R, (F2,F3) ∈ R and (F1,F3) ∉ R

c) (F1, F2 ) ∈ R, (F2,F2) ∈ R but (F3,F3) ∉ R

d) (F1, F2 ) ∉ R, (F2,F3) ∉ R and (F1,F3) ∉ R

4. The above defined relation R is ______

a) Symmetric and transitive but not reflexive

b) Universal relation

c) Equivalence relation

d) Reflexive but not symmetric and transitive

5. Mr. Shyam exercised his voting right in General Election – 2019, then Mr. Shyam is related to which of the following?

a) All those eligible voters who cast their votes

b) Family members of Mr. Shyam

c) All citizens of India

d) Eligible voters of India

CASE STUDY-2 CHAPTER -1 

Sherlin and Danju are playing Ludo at home during Covid-19. While rolling the dice, Sherlin’s sister Raji observed and noted the possible outcomes of the throw every time belongs to set {1,2,3,4,5,6}. Let A be the set of players while B be the set of all possible outcomes

A = {S, D}, B = {1,2,3,4,5,6}

Q1. Let 𝑅 ∶ 𝐵 → 𝐵 be defined by R = {( 𝑥 , 𝑦 ): 𝑦 𝑖𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 𝑥 } is

a) Reflexive and transitive but not symmetric

b) Reflexive and symmetric and not transitive

c) Not reflexive but symmetric and transitive

d) Equivalence

2. Raji wants to know the number of functions from A to B. How many number of functions are possible?

3. Let R be a relation on B defined by R = {(1,2), (2,2), (1,3), (3,4), (3,1), (4,3), (5,5)}. Then R is

a) Symmetric

b) Reflexive

c) Transitive

d) None of these three

4. Raji wants to know the number of relations possible from A to B. How many numbers of relations are possible?

5. Let 𝑅 : 𝐵 → 𝐵 be defined by R={(1,1),(1,2), (2,2), (3,3), (4,4), (5,5),(6,6)}, then R is

b) Reflexive and Transitive

c) Transitive and symmetric

CASE STUDY-1 CHAPTER -3 

A manufacture produces three stationery products Pencil, Eraser and Sharpener which he sells in two markets. Annual sales are indicated below

If the unit Sale price of Pencil, Eraser and Sharpener are Rs. 2.50, Rs. 1.50 and Rs. 1.00 respectively, and unit cost of the above three commodities are Rs. 2.00, Rs. 1.00 and Rs. 0.50 respectively, then,

Based on the above information answer the following:

1) Total revenue of market A

a) Rs. 64,000

b) Rs. 60,400

c) Rs. 46,000

d) Rs. 40600

2) Total revenue of market B

a) Rs. 35,000

b) Rs. 53,000

c) Rs. 50,300

d) Rs. 30,500

3) Cost incurred in market A

a) Rs. 13,000

b) Rs.30,100

c) Rs. 10,300

d) Rs. 31,000

4) Profit in market A and B respectively are

a) (Rs. 15,000, Rs. 17,000)

b) (Rs. 17,000, Rs. 15,000)

c) (Rs. 51,000, Rs. 71,000)

d) ( Rs. 10,000, Rs. 20,000)

5) Gross profit in both market

a) Rs.23,000

b) Rs. 20,300

c) Rs. 32,000

d) Rs. 30,200

CASE STUDY-2 CHAPTER -3 

CASE STUDY 2: 

Amit, Biraj and Chirag were given the task of creating a square matrix of order 2. Below are the matrices created by them. A, B , C are the matrices created by Amit, Biraj and Chirag respectively. 

  If a = 4 and b = −2, based on the above information answer the following:

1. Sum of the matrices A, B and C , A+( 𝐵 + 𝐶 ) is

2. ( 𝐴 𝑇 ) 𝑇 is equal to

3. ( 𝑏𝐴 ) 𝑇  is equal to

4. AC−𝐵𝐶 is equal to

5. (𝑎 + 𝑏)𝐵 is equal to

  Case study based questions

Chapter - 6 Class XII

Application of Derivatives

Case Study - 1

Read the following passage and answer the questions given below.

The temperature of a person during an intestinal illness is given by

f(x) = - 0.1x 2  + mx + 98.6,  0≤  x  ≤   12, m being a constant, where f(x) is the temperature in  o F at x days.

(i) Is the function differentiable in the interval  (0, 12) ? Justify your answer.

(ii) If 6 is the critical point of the function, then find the value of the constant m

(iii) Find the intervals in which the function is strictly increasing/strictly decreasing.

(i) Yes it is differentiable in (0, 12) because f(x) is a polynomial function which is differentiable at everywhere.

(ii) m = 1.2

(iii) f is strictly increasing in [0, 6]

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

Case study based questions

Chapter -13 Class XII

Probability

Case Study 1

The reliability of a COVID PCR test is specified as follows:

Of people having COVID, 90% of the test detects the disease but 10% goes undetected.

Of people free of COVID, 99% of the test is judged COVID negative but 1% are diagnosed as showing COVID positive. From a large population of which only 0.1% have COVID, one person is selected at random, given the COVID PCR test, and the pathologist reports him/her as COVID positive.

Based on the above information, answer the following

1) What is the probability of the ‘person to be tested as COVID positive’ given that ‘he is actually having COVID?

2) What is the probability of the ‘person to be tested as COVID positive’ given that ‘he is actually not having COVID’?

3) What is the probability that the ‘person is actually not having COVID?

4) What is the probability that the ‘person is actually having COVID given that ‘he is tested as COVID positive’?

5) What is the probability that the ‘person selected will be diagnosed as COVID positive’?

Read the following passage and answer the questions given below:

In an Office three employees James, Sophia and Oliver process incoming copies of a certain form. James processes 50% of the forms, Sophia processes 20% and Oliver the remaining 30% of the forms. James has an error rate of 0.06 , Sophia has an error rate of 0.04 and Oliver has an error rate of 0.03 .

Based on the above information, answer the following questions.

(i) Find the probability that Sophia processed the form and committed an error.

(ii) Find the total probability of committing an error in processing the form.

(iii) The manager of the Company wants to do a quality check. During inspection, he selects a form at random from the days output of processed form. If the form selected at random has an error, find the probability that the form is not processed by James.

(iii) Let E be the event of committing an error in processing the form and let  E 1 , E 2 and E 3  

 be the events that James, Sophia and Oliver processed the form. Find the value of

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