NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability (2026-27)

Think and Reflect

Question 1: Such unpredictability can be useful sometimes! For example, in a cricket match, the fact that a coin is tossed to decide which team will bat first is considered to be a fair method. Can you explain why?

Solution:

A coin toss is considered a fair method because a fair coin has two possible outcomes: Heads and Tails.

• Probability of Heads = 1/2

• Probability of Tails = 1/2

Thus, both teams have an equal 50% chance of winning the toss. The result cannot be predicted beforehand, so neither team receives an unfair advantage.

Only one outcome occurs during a toss, making the decision clear. Moreover, each coin toss is independent, which means previous results do not affect the result of the next toss.

Since the toss is conducted openly in front of both captains and officials, it is also a transparent and unbiased way of deciding which team will bat or bowl first.

Think and Reflect

Question 1: Ask your friend to predict the outcome of a ₹ 1 coin you toss. Do you see that your friend could guess heads or tails but could not know for certain? That’s randomness! All possible results are known, but each individual try is unpredictable.

Solution: When a coin is tossed, the possible outcomes are known in advance: Heads or Tails. However, it is not possible to predict with certainty which outcome will occur in a particular toss.

A friend may guess Heads or Tails, but the actual result remains uncertain until the coin lands. This unpredictability, even when all possible outcomes are known, is called randomness.

Think and Reflect

Question 1: If I have rolled a 4 on a die 8 times in succession, the probability of rolling a 4 again is still only ≈ 0.16 (assuming the die is fair). Probability does not tell you what will happen next but predicts what will happen in the long run.

Solution:

Even after rolling a 4 eight times continuously, the probability of getting another 4 on the ninth roll remains 1/6, or approximately 0.16.

For a fair six-sided die:

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

• Total number of possible outcomes = 6

• Favourable outcome = {4}

• Number of favourable outcomes = 1

Therefore:

P(rolling a 4) = 1/6 ≈ 0.16

Each roll of a die is an independent event. The die does not remember the outcomes of earlier rolls, so previous results do not change the probability of the next result.

The false belief that a repeated outcome is less likely to occur again is known as the gambler’s fallacy. Probability describes expected behaviour over many trials, not what must happen during the next trial.

Think and Reflect

Question 1: When we used the sample space {Rain, No Rain} in Example 1, we focused only on whether it will rain or not. However, if we want to include different amounts of rainfall like drizzle, light rain or heavy rain, we need to expand the sample space to {No Rain, Drizzle, Light Rain, Heavy Rain} so that it better matches the level of detail required for the question. It is important to ensure the sample space is detailed enough to suit the specific problem being studied.

Solution:

The sample space must contain all outcomes relevant to the question being studied.

For a basic question about whether it will rain, the sample space may be:

S = {Rain, No Rain}

However, when the amount or intensity of rainfall is important, the sample space should be expanded to:

S = {No Rain, Drizzle, Light Rain, Heavy Rain}

This expanded sample space provides more detailed information and allows the situation to be analyzed more accurately. Therefore, a sample space should always match the required level of detail.

Think and Reflect 

Question 1: Can you calculate the probability of getting one head and one tail?

Solution:

When a coin is tossed twice, the sample space is:

S = {HH, HT, TH, TT}

Total number of possible outcomes, n(S) = 4

The outcomes containing one Head and one Tail are:

E = {HT, TH}

Number of favourable outcomes, n(E) = 2

Therefore:

P(one Head and one Tail) =  Number of favourable outcomes /  Number of all possible outcomes = 2/4

= 1/2

= 0.5 or 50%

Exercise Set 7.1 Class 9 Maths Solutions

Question 1: Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.

(i) The next Monday will come after Sunday.

(ii) It will snow in Mumbai in July.

(iii) An elephant will walk through your classroom today.

(iv) You will greet at least one friend at school tomorrow.

Solution:

(i) The next Monday will come after Sunday.

Rank: 1

Label: Certain

Reason: The days of the week follow a fixed order, and Monday always comes after Sunday. Therefore, the event is certain.

(ii) It will snow in Mumbai in July.

Rank: 0

Label: Impossible

Reason: Mumbai has a tropical climate, and July is part of the monsoon season. The temperature does not normally fall low enough for snowfall. Therefore, this event is considered impossible.

(iii) An elephant will walk through your classroom today.

Rank: 0

Label: Impossible

Reason: An elephant cannot normally enter and walk through a classroom during a regular school day. Therefore, the event is treated as impossible under ordinary circumstances.

(iv) You will greet at least one friend at school tomorrow.

Rank: Close to 1, such as 0.8 or 0.9

Label: More likely

Reason: Students regularly meet and interact with friends at school. Therefore, greeting at least one friend is highly likely, although it is not certain.

Exercise Set 7.2 Class 9 Maths Solutions

Question 1: A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts the number of sweets of each colour:

10 red sweets | 8 green sweets | 7 yellow sweets | 5 blue sweets

(i) Calculate the probability that a randomly picked sweet from the sample is green.

(ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.

Solution:

Total number of sweets in the sample:

10 + 8 + 7 + 5 = 30

(i) Probability of selecting a green sweet

Number of green sweets = 8

P(green sweet) = 8/30

= 4/15

≈ 0.267 or 26.7%

Therefore, the probability of selecting a green sweet is approximately 0.267.

(ii) Estimated number of yellow sweets

Number of yellow sweets in the sample = 7

Estimated proportion of yellow sweets = 7/30

For 600 sweets:

Estimated number of yellow sweets = 7/30 × 600

= 7 × 20

= 140

Therefore, approximately 140 sweets are likely to be yellow.

Question 2: A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are:

14 students: Science Club | 11 students: Arts Club | 9 students: Sports Club | 6 students: Debate Club

Assume there are 800 students in the whole school.

(i) What is the probability that a randomly chosen student from the sample prefers the Arts Club?

(ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.

Solution:

Total number of students in the sample:

14 + 11 + 9 + 6 = 40

(i) Probability that a student prefers the Arts Club

Number of students preferring the Arts Club = 11

P(Arts Club) = 11/40

= 0.275 or 27.5%

(ii) Estimated number of students preferring the Sports Club

Number of students preferring the Sports Club = 9

Estimated proportion = 9/40

For 800 students:

Estimated number = 9/40 × 800

= 9 × 20

= 180

Therefore, approximately 180 students are likely to prefer the Sports Club.

Question 3: Toss a coin 20 times and record the result each time (heads or tails).

(i) How many times did you get heads?

(ii) How many times did you get tails?

(iii) Calculate the experimental probability of getting heads.

(iv) If you toss the coin once more, what is the probability of getting tails?

Solution:

This is an activity-based question, and the answers to parts (i), (ii), and (iii) will depend on the results of the 20 coin tosses.

Record every outcome in a table.

Experimental probability of getting Heads:

P(Heads) = Number of times Heads occurs/20

For the next toss of a fair coin:

P(Tails) = 1/2

The result of the earlier 20 tosses does not affect the probability of the next toss.

Question 4: Toss a paper cup into the air 100 times. After each toss, record whether the cup lands on its bottom, upside down on its top or on its side (See Fig. 7.5). Assign probabilities to the outcomes by using experimental probability.

Image 1

Solution:

This is an activity-based question.

Toss the paper cup 100 times and record whether it lands:

On its bottom

Upside down on its top

On its side

Then calculate:

• P(bottom) = Number of times it lands on the bottom/100

• P(top) = Number of times it lands on the top/100

• P(side) = Number of times it lands on the side/100

The three probabilities should add up to 1.

Question 5: What is the probability of getting an even number when rolling a fair 6-sided die?

Solution:

The sample space is:

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

Total number of possible outcomes = 6

Even-numbered outcomes are:

E = {2, 4, 6}

Number of favourable outcomes = 3

Therefore:

P(even number) = 3/6

= 1/2

= 0.5 or 50%

Question 6: Suppose you roll a 6-sided die 12 times and get a ‘3’ three times.

(i) What is the experimental probability of rolling a ‘3’?

(ii) What is the theoretical probability of rolling a ‘3’?

(iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times?

Solution:

(i) Experimental probability

Total number of rolls = 12

Number of times 3 appears = 3

Experimental probability:

P(3) = 3/12

= 1/4

= 0.25 or 25%

(ii) Theoretical probability

A fair die has six equally likely outcomes:

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

Only one outcome is favourable.

Therefore:

P(3) = 1/6

≈ 0.167 or 16.7%

(iii) Reason for the difference

Experimental probability is calculated from actual observations, while theoretical probability is based on all equally likely outcomes in an ideal situation.

Because only 12 trials were performed, random variation may cause the experimental probability to differ from 1/6.

If the die is rolled 60, 600 or 6000 times, the experimental probability is expected to move closer to the theoretical probability of 1/6. A larger number of trials generally gives a more reliable estimate.

Exercise Set 7.3 Class 9 Maths Solutions

Question 1: When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?

Solution:

The sample space is:

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

Therefore:

n(S) = 6

The total number of possible outcomes is 6.

Question 2: For the following experiments write down the sample space S.

(i) Rolling a die and tossing a coin together.

(ii) Choosing a random integer between – 5 and + 5.

(iii) A box containing 5 green and 7 red balls. One ball is drawn at random.

Solution:

(i) Rolling a die and tossing a coin together

The possible die outcomes are 1, 2, 3, 4, 5 and 6.

The possible coin outcomes are H and T.

Therefore:

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

Number of possible outcomes:

n(S) = 6 × 2 = 12

(ii) Choosing a random integer between – 5 and + 5

The integers strictly between −5 and +5 are:

S = {−4, −3, −2, −1, 0, 1, 2, 3, 4}

Therefore:

n(S) = 9

(iii) A box containing 5 green and 7 red balls. One ball is drawn at random.

At the colour level, the possible results are:

S = {Green, Red}

However, to show all individual equally likely outcomes, label the balls as:

S = {G₁, G₂, G₃, G₄, G₅, R₁, R₂, R₃, R₄, R₅, R₆, R₇}

Therefore:

n(S) = 12

There are five favourable outcomes for drawing a green ball and seven favourable outcomes for drawing a red ball.

Question 3: In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.

(i) List the sample space of all possible snack and drink combinations a person could choose at the fair.

(ii) List the event ‘Selecting Samosa as a snack.’

Solution:

(i) Sample space

Each snack can be paired with either Chai or Lassi.

S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}

Therefore:

n(S) = 3 × 2

= 6

(ii) Event of selecting Samosa

E = {(Samosa, Chai), (Samosa, Lassi)}

Therefore, the event contains two outcomes.

Exercise Set 7.4 Class 9 Maths Solutions

Question 1: There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.

(i) Draw a tree diagram showing all possible pairs of fruits.

(ii) List the sample space.

(iii) What is the probability of picking one apple and one banana?

Solution:

Let:

A = Apple

O₁ and O₂ = Two oranges

B = Banana

M = Mango

(i) Tree diagram

Image 2

From Basket A, draw the first three branches labelled A, O₁ and O₂.

From each branch, draw two further branches labelled B and M.

This gives six complete outcomes.

(ii) Sample space

S = {(A, B), (A, M), (O₁, B), (O₁, M), (O₂, B), (O₂, M)}

Therefore:

n(S) = 6

(iii) Probability of picking one apple and one banana

The only favourable outcome is:

{(A, B)}

Number of favourable outcomes = 1

Therefore:

P(Apple and Banana) = 1/6

≈ 0.167 or 16.7%

Question 2: Let us say that you have a box containing 3 red pens, 4 black pens and 2 green pens. You pick a pen (without looking) from the box and put it back. Then your friend does the same.

(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?

(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?

Solution:

There are nine pens in total:

3 red, 4 black and 2 green.

Since the first pen is replaced, both selections have the same probabilities.

• P(R) = 3/9

• P(B) = 4/9

• P(G) = 2/9

(i) Possible colour outcomes

Image 3

S = {(R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)}

The tree diagram begins with three branches: R, B and G. Each of these is followed by another set of R, B and G branches.

(ii) Probability of selecting the same colour

The matching-colour outcomes are:

(R, R), (B, B) and (G, G)

Probability of both selecting red:

P(R, R) = 3/9 × 3/9

= 9/81

Probability of both selecting black:

P(B, B) = 4/9 × 4/9

= 16/81

Probability of both selecting green:

P(G, G) = 2/9 × 2/9

= 4/81

Therefore:

P(same colour) = 9/81 + 16/81 + 4/81

= 29/81

≈ 0.358 or 35.8%

End-of-Chapter Exercises Solutions

Question 1: Fill in the blanks.

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

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

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

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

Solution:

(i) The probability of an impossible event is 0.

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

(iii) The probability of an event that is certain to happen is 1.

(iv) Tossing a fair coin has a probability of 1/2, or 0.5, of getting Heads.

Question 2: In a survey of 50 students, 15 students said they liked football. The number of students who like football is 15, and the ________ (frequency/relative frequency) is __________ (fill in the fraction or decimal).

Solution:

The number of students who like football is 15, and the relative frequency is:

15/50 = 3/10 = 0.3

Therefore, the completed statement is:

The number of students who like football is 15, and the relative frequency is 3/10 or 0.3.

Question 3: Which of the following experiments has equally likely outcomes? Explain.

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

(ii) Tossing a fair coin once.

(iii) Rolling a fair 6-sided die.

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

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

Solution: Outcomes are equally likely when each possible result has the same probability.

(i) A driver attempts to start a car.

Not equally likely.

Whether a car starts depends on factors such as fuel, battery condition and engine performance. The two outcomes do not necessarily have equal probabilities.

(ii) Tossing a fair coin once.

Equally likely.

The sample space is:

S = {H, T}

For a fair coin:

P(H) = 1/2

P(T) = 1/2

(iii) Rolling a fair 6-sided die.

Equally likely.

Each of the six faces has an equal probability of appearing.

P(each number) = 1/6

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

Not equally likely.

P(Red) = 3/10

P(Blue) = 7/10

Since the probabilities are different, the outcomes are not equally likely.

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

Equally likely for this probability model.

The two outcomes are treated as having a probability of 1/2 each.

Question 4: Write the sample space and calculate the probability based on the given information.

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

(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?

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

(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?

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

Solution:

(i) Two coins are tossed at the same time.

Sample space:

S = {HH, HT, TH, TT}

Total outcomes = 4

The event of getting at least one Head:

E = {HH, HT, TH}

Favourable outcomes = 3

Therefore:

P(at least one Head) = 3/4

= 0.75 or 75%

(ii) Ten cards numbered 1 to 10

Sample space:

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

Even-numbered cards:

E = {2, 4, 6, 8, 10}

Favourable outcomes = 5

Therefore:

P(even number) = 5/10

= 1/2

= 0.5 or 50%

(iii) A die is rolled once.

Sample space:

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

Numbers greater than 4:

E = {5, 6}

Therefore:

P(number greater than 4) = 2/6

= 1/3

≈ 0.333 or 33.3%

(iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball.

Total number of balls:

3 + 2 + 1 = 6

Balls that are not red:

2 blue balls and 1 green ball

Favourable outcomes = 3

Therefore:

P(not red) = 3/6

= 1/2

= 0.5 or 50%

(v) Three coins are tossed simultaneously.

Image 4

Sample space:

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

Total outcomes = 8

Outcomes containing exactly two Heads:

E = {HHT, HTH, THH}

Favourable outcomes = 3

Therefore:

P(exactly two Heads) = 3/8

= 0.375 or 37.5%

Question 5: A bag has 3 candies: strawberry, lemon, and mint. One is picked at random. What is the probability of picking a strawberry candy?

Solution:

Sample space:

S = {Strawberry, Lemon, Mint}

Total number of possible outcomes = 3

Favourable outcome = {Strawberry}

Therefore:

P(Strawberry) = 1/3

≈ 0.333 or 33.3%

Question 6: A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.

Solution:

Each shirt can be paired with each of the three types of pants.

Number of possible outfits:

= Number of shirt choices × Number of pants choices

= 2 × 3

= 6

Therefore, six different outfits are possible.

Question 7: A tyre company records distances before replacement in 1000 cases.

Image 5

Find the probability that a randomly chosen tyre lasts:

(i) Less than 4000 km.

(ii) Between 4000 and 14000 km.

(iii) More than 14000 km.

Solution:

Total number of cases = 1000

This question uses experimental probability because the values are based on observed data.

(i) Less than 4000 km

Number of favourable cases = 20

P(less than 4000 km) = 20/1000

= 0.02 or 2%

(ii) Between 4000 and 14000 km

Number of favourable cases:

= 210 + 325

= 535

P(between 4000 and 14000 km) = 535/1000

= 0.535 or 53.5%

(iii) More than 14000 km

Number of favourable cases = 445

P(more than 14000 km) = 445/1000

= 0.445 or 44.5%

Question 8: The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.

Image 6

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

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

Solution:

The cards contain the letters:

P, E, A, C, E

Total number of cards = 5

(i) Probability of drawing P, E or C

Favourable cards:

P, E, C, E

Number of favourable outcomes = 4

Therefore:

P(P, E or C) = 4/5

= 0.8 or 80%

(ii) Probability of not drawing E

Cards that are not E:

P, A, C

Number of favourable outcomes = 3

Therefore:

P(not E) = 3/5

= 0.6 or 60%

Question 9: A game of chance consists of spinning an arrow (see Fig. 7.7.) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at

Image 7

(i) 8?

(ii) An odd number?

(iii) A number greater than 2?

(iv) A number less than 9?

(v) A multiple of 3?

Solution:

Sample space:

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

Total outcomes = 8

(i) The arrow points at 8

Favourable outcome = {8}

P(8) = 1/8

= 0.125 or 12.5%

(ii) The arrow points at an odd number

Odd numbers:

{1, 3, 5, 7}

P(odd number) = 4/8

= 1/2

= 0.5 or 50%

(iii) The arrow points at a number greater than 2

Numbers greater than 2:

{3, 4, 5, 6, 7, 8}

P(number greater than 2) = 6/8

= 3/4

= 0.75 or 75%

(iv) The arrow points at a number less than 9

Every number in the sample space is less than 9.

P(number less than 9) = 8/8

= 1

Therefore, this is a certain event.

(v) The arrow points at a multiple of 3

Multiples of 3:

{3, 6}

P(multiple of 3) = 2/8

= 1/4

= 0.25 or 25%

Question 10: A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.

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

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

Solution:

Initially, the basket contains:

4 red balls

5 blue balls

Total balls = 9

Because the first ball is laid aside, the second ball is selected from the remaining eight balls.

Image 8

First-draw probabilities:

P(R) = 4/9

P(B) = 5/9

If the first ball is red, the basket contains 3 red and 5 blue balls:

P(R on second draw | first R) = 3/8

P(B on second draw | first R) = 5/8

If the first ball is blue, the basket contains 4 red and 4 blue balls:

P(R on second draw | first B) = 4/8

P(B on second draw | first B) = 4/8

(i) Red and then blue

P(RB) = 4/9 × 5/8

= 20/72

= 5/18

≈ 0.278

(ii) Two blue balls

P(BB) = 5/9 × 4/8

= 20/72

= 5/18

≈ 0.278

Question 11: I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.

Solution:

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

An impossible event with probability 0:

Getting a sum of 13

P(sum of 13) = 0

A certain event with probability 1:

Getting a sum less than 14

P(sum less than 14) = 1

Question 12: Write the sample space and calculate the probability based on the given information.

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

(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?

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

(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?

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

Solution:

(i) Two dice are rolled.

The sample space consists of 36 ordered pairs:

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

Prime sums greater than 5 are 7 and 11.

Outcomes with sum 7:

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

Outcomes with sum 11:

{(5,6), (6,5)}

Number of favourable outcomes = 8

Therefore:

P(prime sum greater than 5) = 8/36

= 2/9

≈ 0.222 or 22.2%

(ii) Two balls of different colours

Total balls:

4 + 3 + 2 = 9

Total ways of choosing two balls:

= 9C2

= 36

Ways to choose one red and one green:

= 4 × 3

= 12

Ways to choose one red and one blue:

= 4 × 2

= 8

Ways to choose one green and one blue:

= 3 × 2

= 6

Total favourable outcomes:

= 12 + 8 + 6

= 26

Therefore:

P(different colours) = 26/36

= 13/18

≈ 0.722 or 72.2%

(iii) First coin is Heads and exactly two Heads occur

Sample space:

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

The favourable outcomes are:

E = {HHT, HTH}

Number of favourable outcomes = 2

Therefore:

P(first coin is Heads and exactly two Heads occur) = 2/8

= 1/4

= 0.25 or 25%

(iv) Four-digit numbers using 1, 2, 3 and 4

Sample space:

S = {1234, 1243, 1324, 1342, 1423, 1432,
2134, 2143, 2314, 2341, 2413, 2431,
3124, 3142, 3214, 3241, 3412, 3421,
4123, 4132, 4213, 4231, 4312, 4321}

Total outcomes = 24

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

For each choice of final digit, the remaining three digits can be arranged in 3! = 6 ways.

Number of favourable outcomes:

= 2 × 6

= 12

Therefore:

P(even number) = 12/24

= 1/2

= 0.5 or 50%

(v) Exactly two correct answers

For each question:

P(correct) = 1/4

P(wrong) = 3/4

The possible correct-wrong patterns are:

S = {CCC, CCW, CWC, WCC, CWW, WCW, WWC, WWW}

The outcomes containing exactly two correct answers are:

{CCW, CWC, WCC}

Probability of one such pattern:

= 1/4 × 1/4 × 3/4

= 3/64

There are three such patterns.

Therefore:

P(exactly two correct) = 3 × 3/64

= 9/64

≈ 0.141 or 14.1%

Question 13: A box contains 4 balls numbered 1 to 4. Record a sample space using a tree diagram for the following experiments:

(i) A ball is drawn, and the number is recorded. Then the ball is returned, and a second ball is drawn and recorded.

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

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

Solution:

(i) With replacement

Since the first ball is replaced, all four balls are available for the second draw.

Image 9

S₁ = {(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)}

(ii) Without replacement

The first ball is not returned, so the same number cannot appear in both draws.

Image 10

S₂ = {(1,2), (1,3), (1,4),
(2,1), (2,3), (2,4),
(3,1), (3,2), (3,4),
(4,1), (4,2), (4,3)}

(iii) Sizes of the sample spaces

With replacement:

n(S₁) = 4 × 4

= 16

Without replacement:

n(S₂) = 4 × 3

= 12

Question 14: 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.

Solution:

The coin outcomes are:

{H, T}

The card outcomes are:

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

Pair each coin result with each card number.

Therefore:

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)}

Number of elements:

n(S) = 2 × 6

= 12

Question 15: Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?

(i) {1, 2, 3}

(ii) {0, 1, 2}

(iii) {0, 1, 2, 3, 4}

(iv) {0, 1, 2, 3}

Solution:

When three coins are tossed, the possible number of Heads is:

0, 1, 2 or 3

Therefore, the correct sample space is:

(iv) {0, 1, 2, 3}

Explanation of the other options:

(i) {1, 2, 3} is incomplete because 0 Heads is possible when all three coins show Tails.

(ii) {0, 1, 2} is incomplete because 3 Heads is possible when all three coins show Heads.

(iii) {0, 1, 2, 3, 4} is incorrect because getting 4 Heads is impossible when only three coins are tossed.

A valid sample space must include every possible outcome and must not include impossible outcomes.

Question 16: 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?

Image 11

Solution:

This question uses geometric probability.

Probability = Favourable area/Total area

The rectangular region has:

Length = 3 m

Breadth = 2 m

Area of rectangle:

= 3 × 2

= 6 m²

The circle has a diameter of 1 m.

Therefore:

Radius = 1/2 m

Area of circle:

= πr²

= π × (1/2)²

= π/4 m²

Therefore:

P(landing inside the circle) = (π/4)/6

= π/24

Using π ≈ 3.14:

P ≈ 3.14/24

≈ 0.131

Therefore, the probability is π/24, or approximately 0.13 or 13%.

What We Provide in NCERT Solutions for Class 9 Maths Chapter 7

The NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7, The Mathematics of Maybe: Introduction to Probability, provide clear and step-by-step answers to all Think and Reflect questions, Exercise Sets 7.1, 7.2, 7.3 and 7.4, and the End-of-Chapter Exercises. Through Vedantu, students can understand important concepts such as randomness, probability scales, experimental and theoretical probability, sample spaces, events and tree diagrams using simple explanations and solved examples. These solutions support homework, answer checking, quick revision and exam preparation, while the free PDF makes offline study easier for the 2026-27 academic session.

Access Exercise Wise NCERT Solutions for Chapter 7 Maths Class 9

CBSE Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 9 Maths