Class 9 Maths Ganita Manjari Chapter 7 End of Chapter Exercises

Probability is the measure of how likely an event is to happen, using values from 0 to 1.
Class 9 Maths Ganita Manjari Chapter 7 End of Chapter Exercises connect probability with sample spaces, relative frequency, dice, coins, cards, tree diagrams and area-based probability.

Chapter 7, The Mathematics of Maybe: Introduction to Probability Class 9, introduces uncertainty through random events such as coin tosses, die rolls, lucky draws and survey results. Class 9 Maths Ganita Manjari Chapter 7 End of Chapter Exercises revise probability scale, experimental probability, theoretical probability, sample space, events and tree diagrams. These Class 9 Maths Chapter 7 End of Chapter Exercise Solutions cover fill-in-the-blanks, equally likely outcomes, sample space questions, probability of dice Class 9 problems, card draws, tyre data, two-stage ball draws and area probability. The textbook summary states that probability is written as P(E), where 0 ≤ P(E) ≤ 1.

Key Takeaways

• Probability Scale: Probability lies between 0 and 1.
• Experimental Probability: It uses actual data from trials or observations.
• Theoretical Probability: It uses favourable outcomes divided by total possible outcomes.
• Sample Space: It lists every possible outcome of a random experiment.

Class 9 Maths Ganita Manjari Chapter 7 End of Chapter Exercises Structure 2026

Exercise Type	Topic	Question Count
Basic Concepts	Probability scale, frequency and sample space	4
Direct Probability	Coins, dice, cards, surveys and tyres	5
Higher-Order	Tree diagrams, dice pairs and area probability	7

Class 9 Maths Ganita Manjari Chapter 7 End of Chapter Exercises for Basic Probability

The first three questions revise definitions and probability language. Students identify impossible events, certain events, sample space, relative frequency and equally likely outcomes.

Q1. Fill in the blanks.

Q1(i). The probability of an impossible event is _______.

The probability of an impossible event is 0.

Answer:

0

Q1(ii). The set of all possible outcomes of a random experiment is called the __________.

The set of all possible outcomes is called the sample space.

Answer:

sample space

Q1(iii). The probability of an event that is certain to happen is _______.

The probability of a certain event is 1.

Answer:

1

Q1(iv). Tossing a fair coin has a probability of ______ for getting heads.

A fair coin has two equally likely outcomes: heads and tails.

So:

Probability of heads = 1/2

Answer:

1/2

Ganita Manjari Class 9 Chapter 7 End of Chapter Exercises: Relative Frequency

Relative frequency compares how often an event occurs with the total number of observations. It is used in experimental probability and survey-based probability.

Q2. In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ is __________.

The missing term is relative frequency, and its value is 15/50 = 3/10 = 0.3.

Given:

Students who like football = 15

Total students = 50

Relative frequency = 15/50

Relative frequency = 3/10

Relative frequency = 0.3

Answer:

The number of students who like football is 15, and the relative frequency is 15/50 = 0.3.

Q3. Which of the following experiments have equally likely outcomes? Explain.

Equally likely outcomes occur when each outcome has the same chance of happening. The textbook uses fair coins and fair dice as standard examples of equally likely outcomes.

Q3(i). A driver attempts to start a car. The car starts or does not start.

The outcomes are not equally likely.

Reason:

A working car usually has a higher chance of starting.

Answer:

The outcomes are not equally likely.

Q3(ii). Tossing a fair coin once.

The outcomes are equally likely.

Sample space:

S = {H, T}

Probability of heads = 1/2

Probability of tails = 1/2

Answer:

The outcomes are equally likely.

Q3(iii). Rolling a fair 6-sided die.

The outcomes are equally likely.

Sample space:

S = {1, 2, 3, 4, 5, 6}

Each face has probability:

1/6

Answer:

The outcomes are equally likely.

Q3(iv). Choosing a marble randomly from a bag that contains 3 red marbles and 7 blue marbles.

The colour outcomes are not equally likely.

Reason:

There are 3 red marbles and 7 blue marbles.

Probability of red = 3/10

Probability of blue = 7/10

Answer:

The outcomes red and blue are not equally likely.

Q3(v). A baby is born. It is a boy or a girl.

The textbook-level answer can treat the outcomes as roughly equally likely.

Reason:

In a basic probability model, the two outcomes boy and girl are considered comparable.

Answer:

The outcomes may be treated as equally likely in this simplified model.

Class 9 Maths Chapter 7 End of Chapter Exercise Solutions for Sample Space and Probability

Question 4 asks students to write the sample space and calculate probability. The formula used is favourable outcomes divided by total outcomes.

Q4(i). Two coins are tossed at the same time. What is the probability of getting at least one head?

The probability of getting at least one head is 3/4.

Sample space:

S = {HH, HT, TH, TT}

Total outcomes:

4

Favourable outcomes:

HH, HT, TH

Number of favourable outcomes:

3

Probability = 3/4

Answer:

The probability of getting at least one head is 3/4.

Q4(ii). Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?

The probability of drawing an even number is 1/2.

Sample space:

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Total outcomes:

10

Even numbers:

2, 4, 6, 8, 10

Number of favourable outcomes:

5

Probability = 5/10

Probability = 1/2

Answer:

The probability of drawing an even number is 1/2.

Q4(iii). A die is rolled once. What is the probability of getting a number greater than 4?

The probability of getting a number greater than 4 is 1/3.

Sample space:

S = {1, 2, 3, 4, 5, 6}

Total outcomes:

6

Numbers greater than 4:

5, 6

Number of favourable outcomes:

2

Probability = 2/6

Probability = 1/3

Answer:

The probability of getting a number greater than 4 is 1/3.

Q4(iv). A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?

The probability that the ball is not red is 1/2.

Total balls:

3 + 2 + 1 = 6

Balls that are not red:

2 blue + 1 green = 3

Probability = 3/6

Probability = 1/2

Answer:

The probability that the ball is not red is 1/2.

Q4(v). Three coins are tossed simultaneously. What is the probability of getting exactly two heads?

The probability of getting exactly two heads is 3/8.

Sample space:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total outcomes:

8

Exactly two heads:

HHT, HTH, THH

Number of favourable outcomes:

3

Probability = 3/8

Answer:

The probability of getting exactly two heads is 3/8.

The Mathematics of Maybe Introduction to Probability Class 9: Direct Probability Questions

Questions 5 to 9 use direct probability from equally likely outcomes or observed frequency tables. These questions check whether students can identify total outcomes and favourable outcomes.

Q5. A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

The probability of picking strawberry candy is 1/3.

Sample space:

S = {Strawberry, Lemon, Mint}

Total outcomes:

3

Favourable outcome:

Strawberry

Probability = 1/3

Answer:

The probability of picking strawberry candy is 1/3.

Q6. A child has 2 shirts and 3 types of pants. List all possible outfit combinations in a table format.

There are 6 possible outfit combinations.

Shirts:

Red, Blue

Pants:

Jeans, Khakis, Shorts

Shirt	Pants	Outfit
Red	Jeans	Red shirt and Jeans
Red	Khakis	Red shirt and Khakis
Red	Shorts	Red shirt and Shorts
Blue	Jeans	Blue shirt and Jeans
Blue	Khakis	Blue shirt and Khakis
Blue	Shorts	Blue shirt and Shorts

Total combinations:

2 × 3 = 6

Answer:

There are 6 possible outfits.

Q7. A tyre company records distances before replacement in 1000 cases. Find the probabilities.

The probabilities are calculated using the number of cases divided by 1000. The table gives 20, 210, 325 and 445 cases in the four distance groups.

Q7(i). Less than 4000 km

Cases less than 4000 km:

20

Total cases:

1000

Probability = 20/1000

Probability = 1/50

Probability = 0.02

Answer:

The probability is 0.02.

Q7(ii). Between 4000 and 14000 km

Cases between 4000 and 14000 km:

210 + 325 = 535

Total cases:

1000

Probability = 535/1000

Probability = 107/200

Probability = 0.535

Answer:

The probability is 0.535.

Q7(iii). More than 14000 km

Cases more than 14000 km:

445

Total cases:

1000

Probability = 445/1000

Probability = 89/200

Probability = 0.445

Answer:

The probability is 0.445.

Q8. The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.

The word PEACE has 5 letters: P, E, A, C, E.

Q8(i). What is the probability that it is a P, E or C?

The probability is 4/5.

Cards:

P, E, A, C, E

Total cards:

5

Favourable cards:

P, E, E, C

Number of favourable cards:

4

Probability = 4/5

Answer:

The probability is 4/5.

Q8(ii). What is the probability that it is not an E?

The probability is 3/5.

Total cards:

5

Cards that are not E:

P, A, C

Number of favourable cards:

3

Probability = 3/5

Answer:

The probability is 3/5.

Q9. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1 to 8, and these are equally likely outcomes. Find the probabilities.

The spinner has 8 equally likely outcomes, from 1 to 8.

Sample space:

S = {1, 2, 3, 4, 5, 6, 7, 8}

Total outcomes:

8

Q9(i). Pointing at 8

Favourable outcome:

8

Probability = 1/8

Answer:

The probability is 1/8.

Q9(ii). An odd number

Odd numbers:

1, 3, 5, 7

Number of favourable outcomes:

4

Probability = 4/8

Probability = 1/2

Answer:

The probability is 1/2.

Q9(iii). A number greater than 2

Numbers greater than 2:

3, 4, 5, 6, 7, 8

Number of favourable outcomes:

6

Probability = 6/8

Probability = 3/4

Answer:

The probability is 3/4.

Q9(iv). A number less than 9

All numbers from 1 to 8 are less than 9.

Number of favourable outcomes:

8

Probability = 8/8

Probability = 1

Answer:

The probability is 1.

Q9(v). A multiple of 3

Multiples of 3:

3, 6

Number of favourable outcomes:

2

Probability = 2/8

Probability = 1/4

Answer:

The probability is 1/4.

Class 9 Maths Probability Solutions for Tree Diagram Questions

Tree diagrams help organise two-stage experiments. In Question 10, the first ball is laid aside, so the total number of balls decreases before the second draw.

Q10. A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn.

Q10(i). What is the probability of drawing a red ball and then a blue ball?

The probability of drawing red and then blue is 5/18.

Total balls:

4 red + 5 blue = 9

First draw red:

Probability = 4/9

After drawing one red ball, remaining balls:

3 red + 5 blue = 8

Second draw blue:

Probability = 5/8

So:

P(Red then Blue) = 4/9 × 5/8

P(Red then Blue) = 20/72

P(Red then Blue) = 5/18

Answer:

The probability is 5/18.

Q10(ii). What is the probability of drawing 2 blue balls?

The probability of drawing two blue balls is 5/18.

First draw blue:

Probability = 5/9

After drawing one blue ball, remaining balls:

4 red + 4 blue = 8

Second draw blue:

Probability = 4/8

So:

P(Blue then Blue) = 5/9 × 4/8

P(Blue then Blue) = 20/72

P(Blue then Blue) = 5/18

Answer:

The probability is 5/18.

Text Tree Diagram for Q10

Start
→ Red: 4/9
→ Red: 3/8, outcome RR
→ Blue: 5/8, outcome RB

Start
→ Blue: 5/9
→ Red: 4/8, outcome BR
→ Blue: 4/8, outcome BB

Class 9 Maths Chapter 7 End of Chapter Exercise Solutions for Higher-Order Probability

The starred questions combine dice, coins, balls, cards, tree diagrams and area probability. Students must first identify the correct sample space.

Q11. I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an event that has a probability of 1.

An event with probability 0 is getting a sum of 13.

Reason:

The largest possible sum on two dice is:

6 + 6 = 12

So sum 13 cannot occur.

Event with probability 0:

Getting a sum of 13

Probability = 0

An event with probability 1 is getting a sum between 2 and 12.

Reason:

The smallest possible sum is 2, and the largest possible sum is 12.

Event with probability 1:

Getting a sum from 2 to 12

Probability = 1

Answer:

Probability 0 event: getting a sum of 13.
Probability 1 event: getting a sum from 2 to 12.

Ganita Manjari Class 9 Chapter 7 End of Chapter Exercises: Advanced Sample Space Questions

Question 12 uses standard sample spaces for two dice, coloured balls, three coins, four-digit numbers and multiple-choice tests. Each part uses favourable outcomes divided by total outcomes.

Q12(i). Two dice are rolled. What is the probability that the sum is a prime number greater than 5?

The probability is 2/9.

Total outcomes when two dice are rolled:

36

Prime numbers greater than 5 that can occur as sums:

7, 11

Ways to get sum 7:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

Number of ways:

6

Ways to get sum 11:

(5,6), (6,5)

Number of ways:

2

Total favourable outcomes:

6 + 2 = 8

Probability = 8/36

Probability = 2/9

Answer:

The probability is 2/9.

Q12(ii). A bag contains 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?

The probability is 13/18.

Total balls:

4 + 3 + 2 = 9

Total ways to draw 2 balls:

9C2 = 36

Same-colour pairs:

Red pairs = 4C2 = 6

Green pairs = 3C2 = 3

Blue pairs = 2C2 = 1

Total same-colour pairs:

6 + 3 + 1 = 10

Different-colour pairs:

36 - 10 = 26

Probability = 26/36

Probability = 13/18

Answer:

The probability is 13/18.

Q12(iii). Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?

The probability is 1/4.

Sample space:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total outcomes:

8

First coin heads and exactly two heads:

HHT, HTH

Number of favourable outcomes:

2

Probability = 2/8

Probability = 1/4

Answer:

The probability is 1/4.

Q12(iv). A four-digit number is formed using the digits 1, 2, 3, and 4 with no repetition. What is the probability that the number is even?

The probability is 1/2.

Total four-digit numbers:

4! = 24

For the number to be even, the last digit must be 2 or 4.

Choices for last digit:

2

Arrangements of remaining 3 digits:

3! = 6

Favourable outcomes:

2 × 6 = 12

Probability = 12/24

Probability = 1/2

Answer:

The probability is 1/2.

Q12(v). A student takes a multiple-choice test with 3 questions, each having 4 options. What is the probability that the student guesses and gets exactly 2 answers correct?

The probability is 9/64.

For each question:

Probability of correct answer = 1/4

Probability of wrong answer = 3/4

Exactly 2 correct out of 3:

Number of ways = 3C2 = 3

Probability = 3 × (1/4)² × (3/4)

Probability = 3 × 1/16 × 3/4

Probability = 9/64

Answer:

The probability is 9/64.

The Mathematics of Maybe Introduction to Probability Class 9: Tree Diagram Sample Spaces

Question 13 compares drawing with replacement and without replacement. Replacement allows repeated ordered pairs, while drawing without replacement uses ordered pairs with different numbers.

Q13. A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram.

Q13(i). A ball is drawn, returned, and a second ball is drawn.

With replacement, each draw has 4 possible outcomes.

Sample space:

S = {(1,1), (1,2), (1,3), (1,4),
(2,1), (2,2), (2,3), (2,4),
(3,1), (3,2), (3,3), (3,4),
(4,1), (4,2), (4,3), (4,4)}

Sample size:

n(S) = 16

Answer:

The sample space has 16 outcomes.

Q13(ii). A ball is drawn and recorded. Without replacing the first ball, a second ball is drawn and recorded.

Without replacement, the second number must be different from the first.

Sample space:

S = {(1,2), (1,3), (1,4),
(2,1), (2,3), (2,4),
(3,1), (3,2), (3,4),
(4,1), (4,2), (4,3)}

Sample size:

n(S) = 12

Answer:

The sample space has 12 outcomes.

Q13(iii). What are the sizes of these two sample spaces?

With replacement:

n(S) = 16

Without replacement:

n(S) = 12

Answer:

The sizes are 16 and 12.

Q14. List the elements of a sample space for the simultaneous tossing of a coin and drawing of a card from a set of 6 cards numbered 1 through 6.

The sample space has 12 outcomes.

Coin outcomes:

H, T

Card outcomes:

1, 2, 3, 4, 5, 6

Sample space:

S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

Sample size:

n(S) = 12

Answer:

S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

Q15. Three coins are tossed, and the number of heads is recorded. Which list is a sample space for this experiment?

The correct sample space is {0, 1, 2, 3}.

Reason:

When three coins are tossed, the number of heads can be:

0 heads

1 head

2 heads

3 heads

So:

S = {0, 1, 2, 3}

Option (i) {1, 2, 3} misses 0 heads.

Option (ii) {0, 1, 2} misses 3 heads.

Option (iii) {0, 1, 2, 3, 4} includes 4 heads, which cannot occur with three coins.

Option (iv) {0, 1, 2, 3} is complete.

Answer:

The correct list is option (iv), {0, 1, 2, 3}.

Class 9 Maths Probability Solutions for Area-Based Probability

Area-based probability is calculated by dividing the favourable area by the total area. Question 16 uses a circle inside a rectangle.

Q16. Suppose you drop a dye at random on the rectangular region shown in Fig. 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?

The probability is π/24.

From Fig. 7.8:

Rectangle length = 3 m

Rectangle breadth = 2 m

Area of rectangle:

Area = 3 × 2

Area = 6 m²

Circle diameter:

1 m

Circle radius:

1/2 m

Area of circle:

Area = πr²

Area = π × (1/2)²

Area = π/4 m²

Probability of landing inside circle:

Probability = Area of circle / Area of rectangle

Probability = (π/4) / 6

Probability = π/24

Using π ≈ 3.14:

Probability ≈ 3.14/24

Probability ≈ 0.131

Answer:

The probability is π/24, or about 0.131.

Class 9 Maths Ganita Manjari Chapter 7 Solutions: Concepts Used in End Exercises

The End of Chapter Exercises combine probability definitions, sample spaces, events, tree diagrams and area probability. These are the main concepts students need for revision.

Probability Class 9

Probability measures the likelihood of an event.

Copy-friendly result:

0 ≤ P(E) ≤ 1

Here:

0 = impossible

1 = certain

Experimental Probability Class 9

Experimental probability uses actual observations.

Formula:

Experimental probability = Number of times the event occurred / Total number of trials

Example:

Relative frequency of football = 15/50 = 0.3

Theoretical Probability Class 9

Theoretical probability uses equally likely outcomes.

Formula:

P(A) = Number of favourable outcomes / Number of possible outcomes

Example:

P(even number on a die) = 3/6 = 1/2

Sample Space Class 9

Sample space is the list of all possible outcomes.

Formula-style notation:

S = {all possible outcomes}

Example:

For one die:

S = {1, 2, 3, 4, 5, 6}

Tree Diagram Class 9

A tree diagram lists outcomes step by step.

Useful rule:

Path probability = product of branch probabilities

Example:

P(Red then Blue) = 4/9 × 5/8 = 5/18

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 7 End of Chapter Exercises

Concept	Copy-Friendly Formula	Used In
Theoretical probability	P(A) = favourable outcomes / possible outcomes	Q4, Q5, Q8, Q9, Q12
Experimental probability	Event frequency / total trials	Q2, Q7
Area probability	Favourable area / total area	Q16

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7

Section	NCERT Solutions
Class 9 Maths Ganita Manjari 2026	NCERT Class 9 Maths Ganita Manjari 2026
Chapter 7	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7
Exercise 7.1	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.1
Exercise 7.2	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.2
Exercise 7.3	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.3
Exercise 7.4	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4
End of Chapter Exercises	NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 End of Chapter Exercises