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# NCERT Solutions for Class 11 Maths Chapter 14 - Probability Miscellaneous Exercise

Last updated date: 06th Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 14 Miscellaneous Exercise - Free PDF Download

The Miscellaneous Exercise of Chapter 14 Probability in Class 11 Maths is an important part of scoring probability concepts. It provides many kinds of exercises that help students learn and apply probability rules in different situations. This practice offers an in-depth knowledge of probability distributions, individual incidents, and conditional probability through its set of questions from Class 11 Maths NCERT Solutions.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 14 Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 11 Maths Chapter 14 Probability
2.1Miscellaneous Exercise
3. Conclusion
4. CBSE Class 11 Maths Chapter 14 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Students should focus on learning the fundamental ideas and completely dealing with every problem. The key to success is knowing how to answer various kinds of questions. Students can improve their knowledge and learn practical ways to solve problems with the help of the CBSE Class 11 Maths Syllabus, making sure that they have enough resources for examinations.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 14 Probability

### Miscellaneous Exercise

1. A box contains $10$ red marbles, $20$ blue marbles and $30$ green marbles. $5$ marbles are drawn from the box, what is the probability that:

(i) All will be blue?

Ans. Total number of marbles $= 60$

The number of ways of drawing $5$ marbles from $60$ marbles is $^{60}{C_5}$ .

All the marbles drawn will be blue if we draw $5$ marbles out of the $20$ blue marbles. $5$ blue marbles can be drawn out of the $20$ blue marbles in $^{20}{C_5}$ ways.

Probability that all marbles will be blue is $\frac{{^{20}{C_5}}}{{^{60}{C_5}}}$ .

(ii) at least one will be green?

Ans. Total number of marbles $= 60$

Number of ways of drawing $5$ marbles from $60$ marbles is $^{60}{C_5}$ .

The number of ways in which the drawn marbles is not green, that is, when red or blue marbles are drawn, is $^{20 + 10}{C_5}{ = ^{30}}{C_5}$.

So, probability that no marble is green is $\frac{{^{30}{C_5}}}{{^{60}{C_5}}}$ .

Therefore, the probability that at least one marble will be green is $1 - \frac{{^{30}{C_5}}}{{^{60}{C_5}}}$.

2. $4$ cards are drawn from a well-shuffled deck of $52$ cards. What is the probability of obtaining three diamonds and one spade?

Ans. Number of ways of drawing $4$ cards from a pack of $52$ cards are $^{52}{C_4}$ .

In a deck, there are $13$ diamonds and $13$ spades.

So, the number of ways of drawing three diamonds and one spade is $^{13}{C_3}{ \times ^{13}}{C_1}$.

Hence, the probability of drawing three diamonds and one spade is $\frac{{^{13}{C_3}{ \times ^{13}}{C_1}}}{{^{52}{C_4}}}$.

3. A die has two faces each with the number ‘ $1$ ’, three faces each with number ‘ $2$ ’ and one face with number ‘ $3$ ’. If die is rolled once, determine:

(i) $P(2)$

Ans. Total number of faces of the die is $6$.

The number of faces with the number ‘ $2$ ’ is $3$.

Therefore, probability of obtaining the number $2$ is $P(2) = \frac{{n(2)}}{{n(S)}}$ .

$\Rightarrow P(2) = \frac{3}{6}$

$\Rightarrow P(2) = \frac{1}{2}$

(ii) P(1 or 3)

Ans. Total number of faces of the die is $6$.

Number of faces with the number ‘ $1$ ’ is $2$ and with the number ‘ $3$ ’ is $1$.

Therefore, probability of obtaining the number $2$ or $3$ is $P(1 \cup 3) = \frac{{n(1 \cup 3)}}{{n(S)}}$ .

$\Rightarrow P(1 \cup 3) = \frac{{1 + 2}}{6}$

$\Rightarrow P(1 \cup 3) = \frac{1}{2}$

(iii) P(not 3)

Ans. Total number of faces of the die is $6$.

Number of faces with the number ‘ $3$ ’ is $1$.

Therefore, probability of obtaining the number $3$ is $P(3) = \frac{1}{6}$ .

$\Rightarrow P(3') = 1 - P(3)$

$\Rightarrow P(3') = 1 - \frac{1}{6}$

$\Rightarrow P(3') = \frac{5}{6}$

4. In a certain lottery, $10,000$ tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy:

(i) One ticket?

Ans. Total number of tickets sold is $10,000$.

Number of prizes awarded is $10$.

If A is the event we buy $1$ ticket, the probability of winning will be $P(A) = \frac{{n(A)}}{{n(S)}}$ .

$\Rightarrow P(A) = \frac{{10}}{{10,000}}$

$\Rightarrow P(A) = \frac{1}{{1,000}}$

So, the probability of not getting a prize is $P(A') = 1 - P(A)$.

$\Rightarrow P(A') = 1 - \frac{1}{{1,000}}$

$\Rightarrow P(A') = \frac{{999}}{{1,000}}$

(ii) two tickets?

Ans. Total number of tickets sold is $10,000$.

Number of prizes awarded is $10$.

Number of tickets on which a prize is not awarded is $10,000 - 10 = 9,990$.

If B is the event we buy $2$ tickets, the probability of not winning will be $P(B) = \frac{{^{9,990}{C_2}}}{{^{10,000}{C_2}}}$ .

(iii) ten tickets?

Ans. Total number of tickets sold is $10,000$.

Number of prizes awarded is $10$.

Number of tickets on which a prize is not awarded is $10,000 - 10 = 9,990$.

If C is the event we buy $10$ tickets, the probability of not winning will be $P(C) = \frac{{^{9,990}{C_{10}}}}{{^{10,000}{C_{10}}}}$ .

5. Out of $100$ students, two sections of $40$ and $60$ are formed. If you and your friend are among the $100$ students, what is the probability that

(i) You both enter the same section?

Ans. You and your friend are among the $100$ students.

Total number of ways of selecting $2$ students out of $100$ students is ¹⁰⁰C₂.

If both of you enter the same section, you are either among the section having $40$ students or the one having $60$ students.

The number of ways in which you both can be in the same section is $^{40}{C_2}{ + ^{60}}{C_2}$.

The probability that you both can be in the same section is $\frac{{^{40}{C_2}{ + ^{60}}{C_2}}}{{^{100}{C_2}}}$

$\Rightarrow \frac{{\frac{{40!}}{{2!38!}} + \frac{{60!}}{{2!58!}}}}{{\frac{{100!}}{{2!98!}}}}$

$\Rightarrow \frac{{(40 \times 39) + (60 \times 59)}}{{(100 \times 99)}}$

$\Rightarrow \frac{{17}}{{33}}$

(ii) You both enter different sections?

Ans. The probability that you both can be in the same section is $\frac{{17}}{{33}}$.

The probability that you both are not in the same section will be $1 - \frac{{17}}{{33}}$.

$\Rightarrow \frac{{16}}{{33}}$

6. Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Ans. Let ${L_1},{L_2},{L_3}$ be the three letters and ${E_1},{E_2},{E_3}$ be their corresponding envelopes.

There are $6$ ways of inserting $3$ letters in $3$ envelopes. These are as follows:

$\begin{array}{l}{L_1}{E_1},{L_2}{E_3},{L_3}{E_2}\\{L_2}{E_2},{L_1}{E_3},{L_3}{E_1}\\{L_3}{E_3},{L_1}{E_2},{L_2}{E_1}\\{L_1}{E_1},{L_2}{E_2},{L_3}{E_3}\\\;{L_1}{E_2},{L_2}{E_3},{L_3}{E_1}\\{L_1}{E_3},{L_2}{E_1},{L_3}{E_2}\end{array}$

We observe that there are $4$ ways in which at least one letter is inserted in its proper envelope.

Hence, the required probability is $\frac{4}{6} = \frac{1}{2}$ .

7. A and B are two events such that $P(A) = 0.54$ , $P\left( B \right) = 0.69$ and $P\left( {A \cap B} \right) = 0.35$ . Find:

(i) $P\left( {A \cup B} \right)$

Ans. It is given to us that $P(A) = 0.54$ , $P(B) = 0.69$ , $P(A \cap B) = 0.35$ .

We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\Rightarrow P(A \cup B) = 0.54 + 0.69 - 0.35$

$\Rightarrow P(A \cup B) = 0.88$

(ii) $P\left( {A' \cap B'} \right)$

Ans. It is given to us that $P(A) = 0.54$ , $P(B) = 0.69$ , $P(A \cap B) = 0.35$ .

We know that $P\left( {A' \cap B'} \right) = P(A \cup B)'$

$\Rightarrow P\left( {A' \cap B'} \right) = 1 - P(A \cup B)$

$\Rightarrow P\left( {A' \cap B'} \right) = 1 - 0.88$

$\Rightarrow P\left( {A' \cap B'} \right) = 0.12$

(iii) $P\left( {A \cap B'} \right)$

Ans. It is given to us that $P(A) = 0.54$ , $P(B) = 0.69$ , $P(A \cap B) = 0.35$ .

We know that $P\left( {A \cap B'} \right) = P\left( A \right)--P\left( {A \cap B} \right)$

$\Rightarrow P\left( {A \cap B'} \right) = 0.54--0.35$

$\Rightarrow P\left( {A \cap B'} \right) = 0.19$

(iv) $P\left( {B \cap A'} \right)$

Ans. It is given to us that $P(A) = 0.54$ , $P(B) = 0.69$ , $P(A \cap B) = 0.35$ .

We know that $P\left( {B \cap A'} \right) = P\left( B \right)--P\left( {A \cap B} \right)$

$\Rightarrow P\left( {B \cap A'} \right) = 0.69--0.35$

$\Rightarrow P\left( {B \cap A'} \right) = 0.34$

8. From the employees of a company, $5$ persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

 S. No Name Sex Age in years 1 Harish M $30$ 2 Rohan M $33$ 3 Sheetal F $46$ 4 Alis F $28$ 5 Salim M $41$

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over $35$ years?

Ans. Let E be the event that the spokesperson will be a male and F be the event that the spokesperson will be over $35$ years of age.

Subsequently, $P(E) = \frac{3}{5}$ and $P(F) = \frac{2}{5}$ .

Since there is only one male who is above  $35$ years of age, $P(E \cap F) = \frac{1}{5}$ .

We know that $P(E \cup F) = P(E) + P(F) - P(E \cap F)$

$\Rightarrow P(E \cup F) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5}$

$\Rightarrow P(E \cup F) = \frac{4}{5}$

Hence, the probability that the spokesperson will either be a male or over $35$ years of age is $\frac{4}{5}$.

9. If four digit numbers greater than 5,000  are randomly formed from the digits $0$ , $1$, $3$, $5$, and $7$, what is the probability of forming a number divisible by $5$ when:

(i) the digits are repeated?

Ans. Since four-digit numbers greater than $5,000$ are formed, the thousand’s digit is either $5$ or $7$ . The remaining three digits can be filled by any of the given digits with repetition.

So, the total number of ways of forming four-digit numbers are $2 \times 5 \times 5 \times 5$, and subtract $1$ from the obtained answer, as we cannot count $5,000$ in the numbers.

$\Rightarrow 250 - 1$

$\Rightarrow 249$ ways

A number is divisible by $5$ if the digit at its unit’s place is either $0$ or $5$ .

Total number of four-digit   numbers   greater than $5,000$   that   are   divisible by $5$ are $(2 \times 5 \times 5 \times 2) - 1$ .

$\Rightarrow 100 - 1$

$\Rightarrow 99$

Therefore, the probability of forming a number divisible by $5$ when the digits are repeated is $\frac{{99}}{{249}} = \frac{{33}}{{83}}$

(ii) the repetition of digits is not allowed?

Ans. Since four-digit numbers greater than $5,000$ are formed, the thousand’s digit is either $5$ or $7$ . The remaining three digits can be filled by any of the given digits without repetition.

So, the total number of ways of forming four-digit numbers is $2 \times 4 \times 3 \times 2 = 48$.

A number is divisible by $5$ if the digit at its unit’s place is either $0$ or $5$.

When the digit at the thousands place is $5$, the unit place can be filled only with $0$, so the number of possible ways is $1 \times 3 \times 2 \times 1 = 6$.

When the digit at the thousands place is $7$ , the units place can be filled only with $0$ or $5$ , so the number of possible ways are $1 \times 3 \times 2 \times 2 = 12$ .

Total number of four-digit   numbers   greater than $5,000$   that   are   divisible by $5$ are $12 + 6 = 18$ .

Therefore, the probability of forming a number divisible by $5$ when the digits are repeated is $\frac{{18}}{{48}} = \frac{3}{8}$ .

10. The number lock of a suitcase has $4$ wheels, each labelled with ten digits i.e., from $0$ to $9$ . The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Ans. The number lock of a suitcase has $4$ wheels, each of which is labelled with ten digits, that is, from $0$ to $9$.

Number of ways of selecting $4$ different digits out of $10$ digits is $^{10}{C_4}$ .

Further, each combination of $4$ different digits can be arranged in $4$ ways.

Number of four digits with no repetitions ${ = ^{10}}{C_4} \times 4!$

$\Rightarrow \frac{{10!}}{{4!6!}} \times 4!$

$\Rightarrow 10 \times 9 \times 8 \times 7$

$\Rightarrow 5040$

There is only one number that can open the suitcase.

Thus, the required probability is $\frac{1}{{5040}}$ .

## Conclusion

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 14 on Probability provides a thorough understanding of various probability concepts. Students may develop their understanding and develop their problem-solving abilities by practising these solutions. Because it covers so many various topics and problem-solving categories, this exercise is important to understanding probability. Students who use Vedantu's solutions can take their exams with confidence, understanding that they have studied and identified the basic concepts and strategies required for success in probability.

## Class 11 Maths Chapter 14: Exercises Breakdown

 Exercise Number of Questions Exercise 14.1 7 Questions & Solutions Exercise 14.2 21 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 14 - Probability Miscellaneous Exercise

1. What is the main focus of Chapter 14 Probability in Class 11 Probability Miscellaneous Exercise?

The main focus of the Class 11 Probability Miscellaneous Exercise is to provide students with a comprehensive understanding of probability concepts. It covers various topics such as conditional probability, Bayes' theorem, and probability distributions. By learning these concepts, students can solve real-life problems involving uNCERTainty and chance. The chapter also emphasizes the theoretical and practical aspects of probability, preparing students for higher-level mathematics.

2. How can NCERT Solutions help in understanding the Class 11 Probability Miscellaneous Exercise?

NCERT Solutions provides detailed, step-by-step explanations for each problem in the Miscellaneous Exercise. They help students understand the methodology and reasoning behind each solution. This guidance ensures that students can grasp complex concepts more easily. Additionally, these solutions serve as a reliable reference for students when practising independently, enhancing their problem-solving skills.

3. Why is the Miscellaneous Exercise important in Probability Miscellaneous Class 11?

The Probability Miscellaneous Class 11 is crucial because it offers a diverse set of problems that cover all the key concepts in Chapter 14. This exercise helps students apply what they have learned in different contexts, reinforcing their understanding. By solving these problems, students can identify their strengths and areas needing improvement. It also prepares them for exams by providing practice with varied and challenging questions.

4. What types of problems are included in the Probability Miscellaneous Class 11?

The Probability Miscellaneous Class 11 includes a wide range of problems related to probability. These problems encompass conditional probability, independent and dependent events, random variables, and probability distributions. The variety ensures that students encounter different types of questions, which aids in a deeper understanding of the subject. This diversity also helps in developing versatile problem-solving skills.

5. How should students approach the problems in the Miscellaneous Exercise?

Students should start by carefully reading and understanding each problem before attempting a solution. They should identify the relevant concepts and formulas required for each question. It is important to solve the problems step-by-step, ensuring clarity at each stage. Reviewing solved examples and NCERT Solutions can provide valuable insights into the correct approaches. Regular practice and revision are key to mastering these problems.

6. What is conditional probability and why is it important according to Class 11 Maths Chapter 14 Miscellaneous Exercise?

Conditional probability of Class 11 Maths Chapter 14 Miscellaneous Exercise is the probability of an event occurring given that another event has already occurred. It is an essential concept in probability theory because many real-world problems involve dependent events. Understanding conditional probability helps in analyzing situations where the occurrence of one event affects the likelihood of another. This concept is widely used in fields such as statistics, finance, and science.

7. How do independent events differ from dependent events in Class 11 Maths Chapter 14 Miscellaneous Exercise?

Independent events are those whose occurrence does not affect the probability of another event occurring. In contrast, dependent events are those where the occurrence of one event influences the probability of the other. Understanding the difference is crucial for correctly solving probability problems. Independent events have a simple multiplication rule for their probabilities, while dependent events require conditional probability methods.

8. What is Bayes' theorem and where is it applied in Probability Class 11 NCERT Solutions Miscellaneous?

Bayes' theorem is a formula that describes how to update the probabilities of hypotheses when given evidence. It is used to find the probability of a cause given the observed effect. This theorem is widely applied in various fields, including statistics, machine learning, and medical diagnostics. Bayes' theorem helps in making informed decisions by incorporating new information.

9. What are random variables and how are they used in Probability Class 11 NCERT Solutions Miscellaneous?

A random variable is a variable that takes on different numerical values based on the outcome of a random event. Random variables are used to quantify outcomes in probability and statistics. They can be discrete or continuous, depending on whether they take on a countable or uncountable set of values. Understanding random variables is essential for studying probability distributions and expected values.

10. What are probability distributions and why are they important?

Probability distributions describe how the probabilities of different outcomes are distributed for a random variable. They are important because they provide a complete description of the randomness of a situation. Common probability distributions include the binomial, normal, and Poisson distributions. These distributions are used in various fields to model and analyze random phenomena.

11. How do NCERT Solutions assist in exam preparation for Probability Class 11 NCERT Solutions Miscellaneous?

Probability Class 11 NCERT Solutions Miscellaneous offers clear and detailed explanations for all problems, helping students understand the correct methods and approaches. They provide solved examples that are similar to exam questions, giving students a practical understanding of how to solve them. These solutions also highlight important concepts and formulas that are frequently tested in exams. Regular practice with NCERT Solutions can boost confidence and improve performance.

12. What is the benefit of practising miscellaneous exercises in Probability Class 11 NCERT Solutions Miscellaneous?

Practicing miscellaneous exercises helps students apply their knowledge to a variety of problems, enhancing their understanding and problem-solving skills. These exercises cover all key topics and provide exposure to different types of questions. By solving these problems, students can identify gaps in their understanding and improve their proficiency. This practice is essential for building confidence and ensuring success in exams.