NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

# NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

Edited By Vishal kumar | Updated on Nov 23, 2023 08:10 AM IST

## NCERT Solutions for Class 11 Maths Chapter 16 Probability Miscellaneous Exercise- DOwnload Free PDF

In this article, you will get Class 11 Maths chapter 16 miscellaneous exercise solutions. As the name suggests the miscellaneous exercise chapter 16 Class 11 consist of mixed questions from all the topics covered in this chapter. There are many questions related to finding the probability of an event, probability of the complement of an event in the NCERT book miscellaneous exercise chapter 16 Class 11. Also, you will get questions related to finding the probability of mutually exclusive events, exhaustive and equally likely outcomes in the NCERT solutions for Class 11 Maths chapter 16 miscellaneous exercise.

You can try to solve these NCERT syllabus problems which will check the conceptual understanding of this chapter. As miscellaneous exercise chapter 16 Class 11 is considered to be a bit difficult as compared to other exercises of this chapter you may find the difficulties while solving them. You can go through these Class 11 Maths chapter 16 miscellaneous solutions to get conceptual clarity.

These class 11 chapter 16 maths miscellaneous solutions are curated by subject experts at Careers360 in easy-to-understand language, explaining each step in detail. PDF versions of the class 11 maths ch 16 miscellaneous exercise solutions are also available, allowing students to access them anytime, anywhere at no cost. Also, check If you are looking for NCERT solutions for Class 6 to Class 12 at one place.

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**According to the CBSE Syllabus for the academic year 2023-24, this chapter has been renumbered as Chapter 14.

## Question:1.(i) A box contains $\small 10$ red marbles, $\small 20$ blue marbles and $\small 30$ green marbles. $\small 5$ marbles are drawn from the box, what is the probability that all will be blue?

Given,

No. of red marbles = 10

No. of blue marbles = 20

No. of green marbles = 30

Total number of marbles = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from 60 marbles = $^{60}\textrm{C}_{5}$

(i) .

Number of ways of drawing 5 blue marbles from 20 blue marbles = $^{20}\textrm{C}_{5}$

$\therefore$ Probability of drawing all blue marbles = $\frac{^{20}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}$

Given,

No. of red marbles = 10

No. of blue marbles = 20

No. of green marbles = 30

Total number of marbles = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from 60 marbles = $^{60}\textrm{C}_{5}$

(ii). We know,

The probability that at least one marble is green = 1 - Probability that no marble is green

Now, Number of ways of drawing no green marbles = Number of ways of drawing only red and blue marbles

= $^{(20+10)}\textrm{C}_{5} = ^{30}\textrm{C}_{5}$

$\therefore$ The probability that no marble is green = $\frac{^{30}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}$

$\therefore$ The probability that at least one marble is green = $1-\frac{^{30}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}$

Total number of ways of drawing 4 cards from a deck of 52 cards = $^{52}\textrm{C}_{4}$

We know that there are 13 diamonds and 13 spades in a deck.

Now, Number of ways of drawing 3 diamonds and 1 spade = $^{13}\textrm{C}_{3}.^{13}\textrm{C}_{1}$

$\therefore$ Probability of obtaining 3 diamonds and 1 spade

$= \frac{^{13}\textrm{C}_{3}.^{13}\textrm{C}_{1}}{^{52}\textrm{C}_{4}}$

$\small P(2)$

Total number of faces of a die = 6

(i) Number of faces with number '2' = 3

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

Therefore, required probability P(2) is 0.5

P(1 or 3)

Total number of faces of a die = 6

(ii) P (1 or 3) = P (not 2) = 1 − P (2)

$= 1-P(2) = 1-\frac{3}{6}$ $=\frac{1}{2}$

Therefore, required probability P(1 or 3) is 0.5

P(not 3)

Total number of faces of a die = 6

(iii) Number of faces with number '3' = 1

$\therefore$ $P(3) = \frac{1}{6}$

$\therefore$ P(not 3) = 1 - P(3)

$= 1-\frac{1}{6} = \frac{5}{6}$

Therefore, required probability P(not 3) is $\frac{5}{6}$

Given, Total number of tickets sold = 10,000

Number of prizes awarded = 10

(a) If one ticket is bought,

P(getting a prize) = $\frac{10}{10000} = \frac{1}{1000}$

$\therefore$ P(not getting a prize) = 1 - P(getting a prize)

$1- \frac{1}{1000} = \frac{999}{1000}$

Given, Total number of tickets sold = 10,000

Number of prizes awarded = 10

(b) If two tickets are bought:

Number of tickets not awarded = 10000 - 10 = 9990

$\therefore$ P(not getting a prize) = $\frac{^{9990}\textrm{C}_{2}}{^{10000}\textrm{C}_{2}}$

Given, Total number of tickets sold = 10,000

Number of prizes awarded = 10

(c) If ten tickets are bought:

Number of tickets not awarded = 10000 - 10 = 9990

$\therefore$ P(not getting a prize) = $\frac{^{9990}\textrm{C}_{10}}{^{10000}\textrm{C}_{10}}$

you both enter the same section?

Total number of students = 100

Let A and B be the two sections consisting of 40 and 60 students respectively.

Number of ways of selecting 2 students out of 100 students.= $^{100}\textrm{C}_{n}$

If both are in section A:

Number of ways of selecting 40 students out of 100 = $\dpi{100} ^{100}\textrm{C}_{40}$ (The remaining 60 will automatically be in section B!)

Remaining 38 students are to be chosen out of (100-2 =) 98 students

$\dpi{100} \therefore$ Required probability if they both are in section A = $\frac{^{98}\textrm{C}_{38}}{^{100}\textrm{C}_{40}}$

Similarly,

If both are in section B:

Number of ways of selecting 60 students out of 100 = $\dpi{100} = ^{100}\textrm{C}_{60} = ^{100}\textrm{C}_{40}$ (The remaining 40 will automatically be in section A!)

Remaining 58 students are to be chosen out of (100-2 =) 98 students

$\dpi{100} \therefore$ Required probability if they both are in section B = $\frac{^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{60}}$

Required probability that both are in same section = Probability that both are in section A + Probability that both are in section B

= $\frac{^{98}\textrm{C}_{38}}{^{100}\textrm{C}_{40}}+\frac{^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{60}}$

$\\ = \frac{^{98}\textrm{C}_{38}+^{98}\textrm{C}_{58}}{^{100}\textrm{C}_{40}} \\ \\ =\frac{\frac{98!}{38!.60!} + \frac{98!}{58!.40!}}{\frac{100!}{40!.60!}}$

$=\frac{85}{165} = \frac{17}{33}$

Hence, the required probability that both are in same section is $\dpi{100} \frac{17}{33}$

you both enter the different sections?

Total number of students = 100

Let A and B be the two sections consisting of 40 and 60 students respectively.

We found out in (a) that the probability that both students are in same section is $\dpi{100} \frac{17}{33}$

(b) Probability that both the students are in different section = $\dpi{100} 1 - \frac{17}{33}$ $\dpi{100} = \frac{16}{33}$

Given, 3 letters are put in 3 envelopes.

The number of ways of putting the 3 different letters randomly = 3!

Number of ways that at least one of the 3 letters is in the correct envelope

= No. of ways that exactly 1 letter is in correct envelope + No. of ways that 2 letters are in the correct envelope(The third is automatically placed correctly)

= No. of ways that exactly 1 letter is in correct envelope + No. of ways that all the 3 letters are in the correct envelope

= $(^{3}\textrm{C}_{1}\times1) + 1 = 4$

(Explanation for $^{3}\textrm{C}_{1}\times1$ :

No. of ways of selecting 1 envelope out of 3 = $^{3}\textrm{C}_{1}$.

If we put the correct letter in it, there is only one way the other two are put in the wrong envelope! )

Therefore, the probability that at least one letter is in its proper envelope = $\frac{4}{3!} = \frac{4}{6} = \frac{2}{3}$

Given, P(A) = 0.54, P(B) = 0.69, P(A $\dpi{100} \cap$ B) = 0.35

(i) We know, P(A $\dpi{100} \cup$ B) = P(A) + P(B) - P(A $\dpi{100} \cap$ B)

$\dpi{100} \implies$ P(A $\dpi{100} \cup$ B) = 0.54 + 0.69 - 0.35 = 0.88

$\dpi{100} \implies$ P(A $\dpi{100} \cup$ B) = 0.88

Given, $P(A) = 0.54, P(B) = 0.69, P(A \cap B) = 0.35$

And, $P(A \cup B) = 0.88$

(ii) $A' \cap B'= (A \cup B)'$ [De Morgan’s law]

So, $P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B) = 1 - 0.88 = 0.12$

$\therefore$ $P(A' \cap B') = 0.12$

Given, P(A) = 0.54, P(B) = 0.69, P(A $\dpi{100} \cap$ B) = 0.35

And, P(A $\dpi{100} \cup$ B) = 0.88

(iii) P(A $\dpi{100} \cap$ B') = P(A-B) = P(A) - P(A $\dpi{100} \cap$ B)

= 0.54 - 0.35 = 0.19

$\therefore$ P(A $\dpi{100} \cap$ B') = 0.19

Given, P(A) = 0.54, P(B) = 0.69, P(A $\dpi{100} \cap$ B) = 0.35

And, P(A $\dpi{100} \cup$ B) = 0.88

(iv) P(B $\dpi{100} \cap$ A') = P(B-A) = P(B) - P(A $\dpi{100} \cap$ B)

= 0.69 - 0.35 = 0.34

$\therefore$ P(B $\dpi{100} \cap$ A') = 0.34

 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 $\small 35$ years?

Given,

Total number of persons = 5

No. of male spokesperson = 3

No. of spokesperson who is over 35 years of age = 2

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

$\therefore$ $P(E) = \frac{3}{5}\ and\ P(F) = \frac{2}{5}$

Since only one male is over 35 years of age,

$\therefore$ $P(E\cap F) = \frac{1}{5}$

We know,

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

$\implies P(E \cup F) = \frac{3}{5} + \frac{2}{5} - \frac{1}{5} = \frac{4}{5}$

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

Since 4-digit numbers greater than 5000 are to be formed,

The $1000's$ place digit can be filled up by either 7 or 5 in $^{2}\textrm{C}_{1}$ ways

Since repetition is allowed,

Each of the remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 in $5$ ways.

$\therefore$ Total number of 4-digit numbers greater than 5000 = $^{2}\textrm{C}_{1}\times5\times5\times5 -1$

$= 250 - 1 = 249$ (5000 cannot be counted, hence one less)

We know, a number is divisible by 5 if unit’s place digit is either 0 or 5.

$\therefore$ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = $^{2}\textrm{C}_{1}\times5\times5\times^{2}\textrm{C}_{1} -1$ $= 100 - 1 = 99$

Therefore, the required probability =

$P(with\ repetition) = \frac{99}{249} = \frac{33}{83}$

Since 4-digit numbers greater than 5000 are to be formed,

The $1000's$ place digit can be filled up by either 7 or 5 in $^{2}\textrm{C}_{1}$ ways

Since repetition is not allowed,

The remaining 3 places can be filled by remaining 4 digits in $^{4}\textrm{C}_{3}\times3!$ ways.

$\therefore$ Total number of 4-digit numbers greater than 5000 = $^{2}\textrm{C}_{1}\times(^{4}\textrm{C}_{3}\times3!)= 2\times4\times6 = 48$

We know, a number is divisible by 5 if unit’s place digit is either 0 or 5.

Case 1. When digit at $1000's$ place is 5, the units place can be filled only with 0.

And the $100's$ & $10's$ places can be filled with any two of the remaining digits {1,3,7} in $^{3}{C}_{2}\times2!$

$\therefore$ Number of 4-digit numbers starting with 5 and divisible by 5 = $1\times ^{3}{C}_{2}\times2! = 1.3.2 = 6$

Case 2. When digit at $1000's$ place is 7, the units place can be filled by 0 or 5 in 2 ways.

And the $100's$ & $10's$ places can be filled with any two of the remaining 3 digits in $^{3}{C}_{2}\times2!$

$\therefore$ Number of 4-digit numbers starting with 7 and divisible by 5 = $1\times2\times( ^{3}{C}_{2}\times2! )= 1.2.3.2 = 12$

$\therefore$ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18

Therefore, the required probability =

$P(without\ repetition) = \frac{18}{48} = \frac{3}{8}$

Given, Each wheel can be labelled with 10 digits.

Number of ways of selecting 4 different digits out of the 10 digits = $^{10}\textrm{C}_{4}$

These 4 digits can arranged among themselves is $4!$ ways.

$\therefore$ Number of four digit numbers without repetitions =

$^{10}\textrm{C}_{4}\times4! = \frac{10!}{4!.6!}\times4! = 10\times9\times8\times7 = 5040$

Number of combination that can open the suitcase = 1

$\therefore$ Required probability of getting the right sequence to open the suitcase = $\frac{1}{5040}$

## More About NCERT Solutions for Class 11 Maths Chapter 16 Miscellaneous Exercise:-

In the Class 11 Maths chapter 16 miscellaneous exercise solutions there are 10 questions from topics of all the previous exercises of this chapter. There are some solved examples given before the miscellaneous exercise chapter 16 Class 11. You can start with solving these examples which will make it easy to solve exercise problems by yourself.

Also Read| Probability Class 11 Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 16 Miscellaneous Exercise:-

• Miscellaneous exercise chapter 16 Class 11 is very important for the probability concepts in the higher classes.
• The questions from Class 11 Maths chapter 16 miscellaneous exercise solutions are not much asked in the 11th final exams but it is very important for understanding the concept of probability.
• In Class 11 Maths chapter 16 miscellaneous solutions, you have learned about the classical theory of probability and axiomatic approach of probability.

## Key Features of Class 11 Chapter 16 Maths Miscellaneous Solutions

1. Comprehensive Coverage: The miscellaneous exercise class 11 chapter 16 cover a wide range of miscellaneous problems, ensuring a thorough understanding of diverse mathematical concepts.

2. Clarity in Explanation: Subject experts provide clear and concise explanations for each problem, making it easy for students to comprehend the class 11 chapter 16 maths miscellaneous solutions.

3. Step-by-Step Approach: Class 11 maths miscellaneous exercise chapter 16 solutions are presented in a step-by-step format, helping students follow the logical progression of solving mathematical problems.

4. Application of Formulas: The class 11 maths ch 16 miscellaneous exercise solutions showcase the proper application of relevant formulas, aiding students in mastering the mathematical techniques required for problem-solving.

5. Practice Questions: The miscellaneous solutions include a variety of practice questions, offering students ample opportunities to reinforce their understanding of the chapter.

6. Accessible Language: Written in accessible language, the solutions cater to a broad audience, ensuring that students can easily grasp the concepts presented.

7. Digital and Printable Formats: The class 11 chapter 16 miscellaneous exercise solutions are available in digital formats, and students also have the option to download and print them, providing flexibility in study preferences.

8. Alignment with CBSE Syllabus: The solutions are aligned with the Class 11 CBSE syllabus, ensuring relevance and suitability for exam preparation.

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### Frequently Asked Question (FAQs)

1. If a coin tossed two times then what is the probability getting both head ?

Sample space this experiment = { HH, HT, TH, TT}

Probability of both head  = 1/4

2. If a coin tossed two times then what is the probability getting atleast one head ?

Sample space this experiment = { HH, HT, TH, TT}

Probability of at least one head  = 3/4

3. What is the probability of getting a number less than or equal to 3 when a dice is rolled ?

Sample space = { 1, 2, 3, 4, 5, 6}

The probability of getting a number less than or equal to 3 = 1/2

4. What is the syllabus for CBSE Class 11 ?

Here you will get Syllabus for CBSE Class 11.

5. Can I get complete syllabus for CBSE Class 11 Maths ?
6. What is the weightage of calculus in CBSE Class 11 Maths ?

Calculus has 5 marks weightage in the CBSE Class 11 Maths.

7. What is the weightage of statistics and probability in CBSE Class 11 Maths ?

statistics and probability have 10 marks weightage in the CBSE Class 11 Maths.

8. What is the weightage of Mathematical Reasoning in CBSE Class 11 Maths ?

Mathematical Reasoning has 2 marks weightage in the CBSE Class 11 Maths.

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