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NCERT Solutions for Miscellaneous Exercise Chapter 14 Class 11 - Probability

NCERT Solutions for Miscellaneous Exercise Chapter 14 Class 11 - Probability

Edited By Komal Miglani | Updated on May 07, 2025 02:23 PM IST


Have you ever wondered how these weather apps predict the chance of rain or how games of chance like dice or cards work? Probability has an answer to it! Probability is the branch of mathematics that helps us understand and measure uncertainty in everyday life, whether in predicting the weather, outcomes in a game or chances in real-life decisions. It uses logic, data and formulas to calculate how likely an event is to occur. This exercise contains questions from a series of concepts like sample space, types of events, conditional probability etc.

This Story also Contains
  1. Class 11 Maths Chapter 14 Probability Miscellaneous Exercise Solutions - Download PDF
  2. NCERT Solutions Class 11 Maths Chapter 14: Miscellaneous Exercise
  3. Topics covered in Chapter 14 Probability Miscellaneous Exercise
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions

The NCERT Solutions for Chapter 14 Miscellaneous Exercise are designed to help you in understanding the concepts by offering detailed explanations and step-wise calculations. The NCERT solutions will help you sharpen your reasoning and analytical thinking. Follow the NCERT notes page to learn more!

Class 11 Maths Chapter 14 Probability Miscellaneous Exercise Solutions - Download PDF

Download PDF

NCERT Solutions Class 11 Maths Chapter 14: Miscellaneous Exercise

Question 1: (i) 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 all will be blue?

Answer:

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 = 60C5

(i) .

Number of ways of drawing 5 blue marbles from 20 blue marbles = 20C5

Probability of drawing all blue marbles = 20C560C5

Question 1: (ii) 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 atleast one will be green?

Answer:

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 = 60C5

(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)C5=30C5

The probability that no marble is green = 30C560C5

The probability that at least one marble is green = 130C560C5

Question 2: 4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

Answer:

Total number of ways of drawing 4 cards from a deck of 52 cards = 52C4

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

Now, Number of ways of drawing 3 diamonds and 1 spade = 13C3.13C1

Probability of obtaining 3 diamonds and 1 spade

=13C3.13C152C4

Question 3: (i) A die has two faces each with number '1', three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

P(2)

Answer:

Total number of faces of a die = 6

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

P(2)=36 =12

Therefore, required probability P(2) is 0.5

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

P(1 or 3)

Answer:

Total number of faces of a die = 6

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

=1P(2)=136 =12

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

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

P(not 3)

Answer:

Total number of faces of a die = 6

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

P(3)=16

P(not 3) = 1 - P(3)

=116=56

Therefore, required probability P(not 3) is 56

Question 4: (a) 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 one ticket

Answer:

Given, Total number of tickets sold = 10,000

Number of prizes awarded = 10

(a) If one ticket is bought,

P(getting a prize) = 1010000=11000

P(not getting a prize) = 1 - P(getting a prize)

111000=9991000

Question 4: (b) 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 two tickets

Answer:

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

P(not getting a prize) = 9990C210000C2

Question 4: (c) 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 10 tickets.

Answer:

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

P(not getting a prize) = 9990C1010000C10

Question 5: (a) 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

you both enter the same section?

Answer:

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.= 100Cn

If both are in section A:

Number of ways of selecting 40 students out of 100 = 100C40 (The remaining 60 will automatically be in section B!)

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

Required probability if they both are in section A = 98C38100C40

Similarly,

If both are in section B:

Number of ways of selecting 60 students out of 100 = =100C60=100C40 (The remaining 40 will automatically be in section A!)

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

Required probability if they both are in section B = 98C58100C60

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

= 98C38100C40+98C58100C60

=98C38+98C58100C40=98!38!.60!+98!58!.40!100!40!.60!

=85165=1733

Hence, the required probability that both are in same section is 1733

Question 5: (b) 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

you both enter the different sections?

Answer:

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 1733

(b) Probability that both the students are in different section = 11733 =1633

Question 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.

Answer:

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

= (3C1×1)+1=4

(Explanation for 3C1×1 :

No. of ways of selecting 1 envelope out of 3 = 3C1.

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 = 43!=46=23

Question 7: (i) A and B are two events such that P(A)=0.54 , P(B)=0.69 and P(AB)=0.35 Find P(AB)

Answer:

Given, P(A) = 0.54, P(B) = 0.69, P(A B) = 0.35

(i) We know, P(A B) = P(A) + P(B) - P(A B)

P(A B) = 0.54 + 0.69 - 0.35 = 0.88

P(A B) = 0.88

Question 7: (ii) A and B are two events such that P(A)=0.54, P(B)=0.69 and P(AB)=0.35. Find P(AB)

Answer:

Given, P(A)=0.54,P(B)=0.69,P(AB)=0.35

And, P(AB)=0.88

(ii) AB=(AB) [De Morgan’s law]

So, P(AB)=P(AB)=1P(AB)=10.88=0.12

P(AB)=0.12

Question 7: (iii) A and B are two events such that P(A)=0.54 , P(B)=0.69 and P(AB)=0.35. Find P(AB)

Answer:

Given, P(A) = 0.54, P(B) = 0.69, P(A B) = 0.35

And, P(A B) = 0.88

(iii) P(A B') = P(A-B) = P(A) - P(A B)

= 0.54 - 0.35 = 0.19

P(A B') = 0.19

Question 7: (iv) A and B are two events such that P(A)=0.54, P(B)=0.69 and P(AB)=0.35 . Find P(BA)

Answer:

Given, P(A) = 0.54, P(B) = 0.69, P(A B) = 0.35

And, P(A B) = 0.88

(iv) P(B A') = P(B-A) = P(B) - P(A B)

= 0.69 - 0.35 = 0.34

P(B A') = 0.34

Question 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?

Answer:

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.

P(E)=35 and P(F)=25

Since only one male is over 35 years of age,

P(EF)=15

We know,

P(EF)=P(E)+P(F)P(EF)

P(EF)=35+2515=45

Therefore, the probability that the spokesperson will either be a male or over 35 years of age is 45.

Question 9: (i) If 4-digit numbers greater than 5000 are randomly formed from the digits 0,1,3.5 and 7, what is the probability of forming a number divisible by 5 when, the digits are repeated?

Answer:

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

The 1000s place digit can be filled up by either 7 or 5 in 2C1 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.

Total number of 4-digit numbers greater than 5000 = 2C1×5×5×51

=2501=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.

Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2C1×5×5×2C11 =1001=99

Therefore, the required probability =

P(with repetition)=99249=3383

Question 9: (ii) If 4-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, the repetition of digits is not allowed?

Answer:

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

The 1000s place digit can be filled up by either 7 or 5 in 2C1 ways

Since repetition is not allowed,

The remaining 3 places can be filled by remaining 4 digits in 4C3×3! ways.

Total number of 4-digit numbers greater than 5000 = 2C1×(4C3×3!)=2×4×6=48

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

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

And the 100s & 10s places can be filled with any two of the remaining digits {1,3,7} in 3C2×2!

Number of 4-digit numbers starting with 5 and divisible by 5 = 1×3C2×2!=1.3.2=6

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

And the 100s & 10s places can be filled with any two of the remaining 3 digits in 3C2×2!

Number of 4-digit numbers starting with 7 and divisible by 5 = 1×2×(3C2×2!)=1.2.3.2=12

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

Therefore, the required probability =

P(without repetition)=1848=38

Question 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?

Answer:

Given, Each wheel can be labelled with 10 digits.

Number of ways of selecting 4 different digits out of the 10 digits = 10C4

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

Number of four digit numbers without repetitions =

10C4×4!=10!4!.6!×4!=10×9×8×7=5040

Number of combination that can open the suitcase = 1

Required probability of getting the right sequence to open the suitcase = 15040

Also read

Topics covered in Chapter 14 Probability Miscellaneous Exercise

1. Experiments and Outcomes
An experiment is considered random if it produces diverse results even when it is conducted again under the same circumstances. Each possible result is called an outcome.

2. Sample Space and Events
The sample space is the set of all possible outcomes of an experiment. An event is a subset of the sample space that we are interested in.

3. Classical Probability
If each outcome is equally likely then
P(E)= Number of favourable outcomes  Total number of outcomes 

4. Addition Theorem of Probability
For any two events A and B -
P(AB)=P(A)+P(B)P(AB)

5. Multiplication Theorem of Probability
For independent events-
P(AB)=P(A)P(B)
For conditional probability-
P(AB)=P(A)P(BA)

6. Conditional Probability
It refers to the probability of an event occurring given that another event has already occurred.

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NCERT Solutions of Class 11 Subject Wise

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Subject-Wise NCERT Exemplar Solutions

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Frequently Asked Questions (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 weightage of calculus in CBSE Class 11 Maths ?

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

5. 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.

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

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

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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