Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 13- Probability

Last updated date: 04th Aug 2024
Total views: 7.5k
Views today: 0.07k

## NCERT Solutions for Class 12 Maths Chapter 13 Miscellaneous Exercise - Free PDF Download

Class 12 Maths Chapter 13 Miscellaneous Exercise solutions include solutions to all Miscellaneous Exercise problems. Probability Miscellaneous Class 12 Exercise is based on the ideas presented in Maths Chapter 13. This practice is crucial for competitive exams as well as the CBSE Board exams. Get the Class 12 Maths Probability Miscellaneous Exercise NCERT Solutions in PDF format and practise them offline to do well in the board exam. Students can download the revised Class 12 Maths NCERT Solutions from our page, which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 13 Miscellaneous Exercise - Free PDF Download
2. Access NCERT Class 12 Maths Chapter 13 Probability Miscellaneous Exercise
2.1Miscellaneous Exercise
3. Class 12 Maths Chapter 13: Exercises Breakdown
4. CBSE Class 12 Maths Chapter 13 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Class 12 Chapter 13 Maths Miscellaneous Exercise Solutions are aligned with the updated CBSE guidelines for Class 12, ensuring students are well-prepared for exams. Access the Class 12 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERT Class 12 Maths Chapter 13 Probability Miscellaneous Exercise

### Miscellaneous Exercise

1. If A and B are two events such that $P\left( A \right)\ne 0$. Find $P\left( B\left| A \right. \right)$ if

(i) A is a subset of B

Ans: It is given in the question that $P\left( A \right)\ne 0$ and A is a subset of B

$\therefore P\left( A\bigcap B \right)=P\left( A \right)$

$\therefore P\left( B\left| A \right. \right)=\dfrac{P\left( A\bigcap B \right)}{P\left( A \right)}$

$\Rightarrow P\left( B\left| A \right. \right)=\dfrac{P\left( A \right)}{P\left( A \right)}=1$

(ii) $A\bigcap B=\phi$

Ans: since it is given that $A\bigcap B=\phi$

Therefore $\text{ }P\left( A\bigcap B \right)=0$

$\therefore P\left( B\left| A \right. \right)=\dfrac{P\left( A\bigcap B \right)}{P\left( A \right)}=0$

2. A couple has two children,

(i) Find the probability that both children are males, if it is known that at least one of the children is male.

Ans: The sample space for a family to have two children is given by

$S=\left\{ BB,GG,BG,GB \right\}$

Let us have the following notations and their probabilities given as shown

N: Both children are males

$P\left( N \right)=\dfrac{1}{4}$

K: at least one of the children is male

$P\left( K \right)=\dfrac{3}{4}$

$\because N\bigcap K=BB$

$\therefore P\left( N\bigcap K \right)=\dfrac{1}{4}$

Now probability that both of the children are males provided at least one of the child is male is given by

$P\left( N\left| K \right. \right)=\dfrac{P\left( N\bigcap K \right)}{P\left( K \right)}$

$P\left( N\left| K \right. \right)=\dfrac{\dfrac{1}{4}}{\dfrac{3}{4}}$

Hence the probability that both of the children are males provided at least one of the children is male is $\dfrac{1}{3}$

(ii) Find the probability that both children are females if it is known that the elder child is a female.

Ans: The sample space for a family to have two children is given by

$S=\left\{ BB,GG,BG,GB \right\}$

Let us have the following notations and their probabilities given as shown

N: Both children are females

$P\left( N \right)=\dfrac{1}{4}$

K: the elder child is a female

$P\left( K \right)=\dfrac{2}{4}$

$\because N\bigcap K=GG$

$\therefore P\left( N\bigcap K \right)=\dfrac{1}{4}$

Now probability that both of the children are males provided at least one of the children is male is given by

$P\left( N\left| K \right. \right)=\dfrac{P\left( N\bigcap K \right)}{P\left( K \right)}$

$P\left( N\left| K \right. \right)=\dfrac{\dfrac{1}{4}}{\dfrac{2}{4}}$

Hence the probability that both of the children are males provided at least one of the children is male is $\dfrac{1}{2}$

3. Suppose that $5$% of men and $0.25$% of women have grey hair. A haired person is selected at random. What is the probability of this person being male? Assume that there are equal numbers of males and females.

Ans: It is given that $5$percent of males and $0.25$ per cent of females have grey hair

Therefore total people having grey hair is $\left( 5+0.25 \right)=5.25$ percent

Hence probability of male being haired is $\dfrac{5}{5.25}$

4. Suppose that $90$ % of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

Ans: Given the probability of a person being right-handed is$p=0.9$

Since a person can only be right-handed or left handed

Therefore it follows a binomial distribution with

$n=10,p=\dfrac{9}{10},q=\dfrac{1}{10}$

The probability that at least $6$ people are right handed is given by

${{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \dfrac{9}{10} \right)}}^{k}}{{\left( \dfrac{1}{10} \right)}^{10-k}}$

Therefore probability that at most $6$ people are right handed is given by

$1-{{\sum\limits_{k=0}^{k=7}{^{10}{{C}_{k}}\left( \dfrac{9}{10} \right)}}^{k}}{{\left( \dfrac{1}{10} \right)}^{10-k}}$

5. If a leap year is selected at random, what is the chance that it will contain $53$Tuesdays?

Ans: In a leap year, we have $366$ days i.e., $52$weeks and $2$ days.

In $52$ weeks, we have $52$ Tuesdays.

Therefore, the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining $2$ days will be Tuesdays.

The remaining $2$days can be any of the following

Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday and Sunday and Monday

Total number of cases = 7

⸫ Probability that a leap year will have $53$ Tuesdays $=\dfrac{2}{7}$

6. Suppose we have four boxes. A, B, C and D contain coloured marbles as given below.

 Box Red White Black A 1 6 3 B 6 2 2 C 8 1 1 D 0 6 4

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

Ans:  Let us have the following notations

R be the event of drawing a red marble

E be the event of selecting the box A

F be the event of selecting the box B

G be the event of selecting the box C

The total number of marbles is $40$

The number of red marbles is $15$

$\therefore P\left( R \right)=\dfrac{15}{40}$

Number of red marbles in box A i.e $n\left( R\bigcap E \right)=1$

Number of red marbles in box B i.e $n\left( R\bigcap F \right)=6$

Number of red marbles in box C i.e $n\left( R\bigcap G \right)=8$

Now the probability that red marble is picked from box A is given by

$P\left( R\left| E \right. \right)=\dfrac{P\left( R\bigcap E \right)}{P\left( E \right)}$

$\Rightarrow P\left( R\left| E \right. \right)=\dfrac{\dfrac{1}{40}}{\dfrac{15}{40}}$

Hence E be the event of selecting the box A is $\dfrac{1}{15}$

Similarly, the probability that red marble is picked from box B is given by

$P\left( R\left| F \right. \right)=\dfrac{\dfrac{6}{40}}{\dfrac{15}{40}}$

Hence the probability that red marble is picked from box B is $\dfrac{2}{5}$

Similarly, the probability that red marble is picked from box C is given by

$P\left( R\left| G \right. \right)=\dfrac{\dfrac{8}{40}}{\dfrac{15}{40}}$

Hence the probability that red marble is picked from box C is $\dfrac{8}{15}$

7. Assume that the chances of the patient having a heart attack are $40$%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by $30$% and a prescription of certain drugs reduces its chances by $25$%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.

Ans: Let us have the following notations

E: The events when the person took yoga and meditation courses

$\therefore P\left( E \right)=\dfrac{1}{2}$

F: The events when the person took drugs

$\therefore P\left( F \right)=\dfrac{1}{2}$

G: the person suffered a heart attack

$\therefore P\left( G \right)=0.4$

From the question also we have

$P\left( G\left| E \right. \right)=0.4\times 0.7=0.28$

$P\left( G\left| F \right. \right)=0.4\times 0.75=0.30$

Now probability that found person has a heart attack despite having yoga and meditation courses is given by

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

$P\left( E\left| G \right. \right)=\frac{0.5\times 0.28}{P0.5\times 0.28+0.5\times 0.3}$

Hence the probability that the found person has a heart attack despite having yoga and meditation courses is $\frac{14}{29}$

8. If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$)

Ans: It is clear that the total number of determinants of second order entries being $0's\text{ or 1 }\!\!'\!\!\text{ s}$ is ${{2}^{4}}$

The value of the determinant is positive for cases as shown

$\left. \left| \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right. \right|,\left. \left| \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right. \right|,\left. \left| \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right. \right|$

i.e favourable case is $3$

Hence the probability that the value of the determinant is positive is $\dfrac{3}{16}$.

9. An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P (A fails) = $0.2$

P (B fails alone) = $0.15$

P (A and B fails) = $0.15$

Evaluate the following probabilities

(i) P (A fails | B has failed)

Ans: Let us have the following notations

E: A fails

F: B fails

Given the question

$P\left( E \right)=0.2$

$P\left( E\bigcap F \right)=0.15$

$P\left( E'\bigcap F \right)=0.15$

We know that

$P\left( E'\bigcap F \right)=P\left( E \right)-P\left( E\bigcap F \right)$

$\Rightarrow P\left( E \right)=0.3$

Now the probability that A fails given B has failed is given by

$P\left( E\left| F \right. \right)=\dfrac{0.15}{0.3}$

Hence the probability that A fails given B has failed is $0.5$

(ii) P (A fails alone)

Ans: The probability that A fails alone is given by

$P\left( E\bigcap F' \right)=P\left( F \right)-P\left( E\bigcap F \right)$

$P\left( E\bigcap F' \right)=0.05$

Hence Probability that A fails alone is $0.05$

10. Bag I contains $3$ red and $4$ black balls and Bag II contains $4$ red and $5$ black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red. Find the probability that the transferred ball is black.

Ans:  Let us have the following notations that

E: the red ball is transferred

$P\left( E \right)=\dfrac{3}{7}$

F: the black ball is transferred

$P\left( F \right)=\dfrac{4}{7}$

G: the red ball is drawn

When a red ball is transferred

$P\left( G\left| E \right. \right)=\dfrac{5}{10}$

Similarly, When a black ball is transferred

$P\left( G\left| F \right. \right)=\dfrac{4}{10}$

$P\left( F\left| G \right. \right)=\dfrac{\dfrac{4}{7}\times \dfrac{4}{10}}{\dfrac{4}{7}\times \dfrac{4}{10}+\dfrac{3}{7}\times \dfrac{5}{10}}$

Hence probability that the transferred ball is black is $\dfrac{16}{31}$

Choose the correct answer in each of the following:

11. If A and B are two events such that $P\left( A \right) \ne 0$ and $P\left( {B|A} \right) = 1$, then

(A) $A \subset B$

(B) $B \subset A$

(C) $B = \phi$

(D) $A = \phi$

Ans: Given that, $P\left( A \right) \ne 0$ and $P\left( {B|A} \right) = 1$

That is $P\left( {B|A} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}$

$\Rightarrow \dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} = 1$

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

$\therefore A \subset B$

Option (A) is correct

12. If $P\left( {A|B} \right) > P\left( A \right)$, then which of the following is correct:

(A) $P\left( {B|A} \right) < P\left( B \right)$

(B) $P\left( {A \cap B} \right) < P\left( A \right).P\left( B \right)$

(C) $P\left( {B|A} \right) > P\left( B \right)$

(D) $P\left( {B|A} \right) = P\left( B \right)$

Ans: Given that, $P\left( {A|B} \right) > P\left( A \right)$

That is $\dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} > P\left( A \right)$

$P\left( {A \cap B} \right) > P\left( A \right).P\left( B \right)$

$\dfrac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}} > P\left( B \right)$

$P\left( {B|A} \right) > P\left( B \right)$

Therefore, Option (C) is correct.

13. If A and B are any two events such that P(A)+P(B) - P(A and B) = P(A), then

(A) $P(B|A) = 1$

(B) $P(A|B) = 1$

(C) $P(B|A) = 0$

(D) $P(A|B) = 0$

Ans: Given that, A and B are any two events and P(A)+P(B) - P(A and B) = P(A)

$P(A) + P(B) - P(A \cap B) = P(A)$

$P(B) - P(A \cap B) = 0$

$P(A \cap B) = P(B)$

$\therefore P(A|B) = \dfrac{{P(A \cap B)}}{{P(B)}}$

= 1

Therefore, Option (B) is correct.

## Conclusion

Class 12 Maths Chapter 13 Miscellaneous Exercise solutions are important for understanding various concepts thoroughly. It covers a wide range of issues that call for the use of several methods and formulas. It's crucial to concentrate on comprehending the fundamental ideas behind every question, as opposed to simply learning the answers by heart. You must understand the theory underlying each concepts, practise frequently, and refer to solved examples.

## Class 12 Maths Chapter 13: Exercises Breakdown

 Exercise Number of Questions Exercise 13.1 17 Questions and Solutions Exercise 13.2 18 Questions and Solutions Exercise 13.3 14 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 13- Probability

1. What kind of problems are covered in the Miscellaneous Exercise?

The miscellaneous exercise section in Chapter 13 deals with various probability concepts beyond the basics, including:

• Conditional probability (finding the probability of an event happening given another event has already occurred).

• Baye's theorem (applying Bayes' theorem to calculate conditional probabilities in certain scenarios).

• Independent vs. dependent events (determining if events are independent or influence each other's probabilities).

• Problems involving multiple trials (calculating probabilities in situations with repeated events).

• Probability distributions (exploring concepts like binomial and Poisson distributions).

2. Are there any tips for using the NCERT solutions effectively?

Here are some tips:

• Grasp the fundamentals: Ensure you have a solid understanding of basic probability concepts like sample space, events, probability of an event, and independent/dependent events.

• Identify the type of problem: Recognize whether the problem involves conditional probability, Bayes' theorem, multiple trials, or probability distributions.

• Focus on the approach: When referring to the NCERT solutions, pay close attention to the thought process behind choosing the formula and setting up the probability calculations for the specific problem.

3. Where can I find additional resources for practising miscellaneous exercise problems?

• The NCERT textbook itself might provide solutions to some problems within the miscellaneous exercise section.

• Vedantu offers comprehensive solutions and explanations for these problems. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 13 Miscellaneous Exercise Probability."

4. Sample Questions from Probability Miscellaneous Class 12?

1. Two cards are drawn without replacement from a well-shuffled deck of 52 cards. Find the probability that the first card drawn is a spade and the second card is a heart. (This question involves conditional probability)

2. A box contains 4 red balls and 6 white balls. A ball is drawn at random, its colour is noted and then replaced. This process is repeated 3 times. Find the probability that exactly 2 red balls are drawn in the 3 trials. (This question might involve using the binomial distribution)

3. In a factory, 2% of the products are defective. What is the probability that in a random sample of 100 products, none of them are defective? (This question could involve using the Poisson distribution or complementary probability)

4. A coin is tossed 4 times. Events A and B are defined as getting at least 2 heads and getting exactly 2 heads, respectively. Are events A and B independent? Explain your answer. (This question requires understanding independence of events)

5. How many questions exist in the Miscellaneous Exercise Chapter 13 Class 12?

There are 13 questions designed to integrate and apply various concepts from the chapter.

6. What is the purpose of the Miscellaneous Exercise in Chapter 13?

The purpose of the Miscellaneous Exercise is to provide a thorough review of all the concepts learned in the chapter, helping students to apply these concepts in different and more complex scenarios.

7. What are the benefits of solving the Miscellaneous Exercise for exam preparation?

Solving the Miscellaneous Exercise helps reinforce the concepts, improve problem-solving skills, and prepare students for exam questions that often require the application of multiple topics.

8. How should I approach solving the Miscellaneous Exercise questions?

Approach the questions systematically by breaking them down into smaller parts, revising related concepts, and using the NCERT Solutions for guidance and to understand the correct methods of solving complex problems.

9. Are there any specific strategies to tackle the Miscellaneous Exercise effectively?

Yes, practice regularly, understand the underlying concepts, and review the NCERT Solutions to learn the correct methods of solving complex problems.

10. How many questions from the Miscellaneous Exercise Chapter 13 Class 12 typically appear in the board exams?

While the exact number can vary, questions from the miscellaneous probability class 12  often appear in the board exams, usually around 1 to 2 questions, as they provide a comprehensive review of the chapter.