Class 9 · MathsNCERT 2026 Revised Syllabus

Chapter 7: The Mathematics of Maybe: Introduction to Probability (0904CH07)

Ganita Manjari Part I · 28 questions · 5 exercises

Chapter code: 0904ch07 · Book: Ganita Manjari Part I · Class 9 Maths

Find complete NCERT solutions for Class 9 Maths, Ganita Manjari Part I Chapter 7. Every exercise has a clear, step-by-step answer for chapter code 0904CH07.

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Exercise Set 7.1(1 questions)

1Rank 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.

School Level Solution

i) The next Monday will come after Sunday.It is certain because Monday always comes after Sunday.ii) It will snow in Mumbai in July.It is less likely because snowfall in Mumbai in July is extremely unlikely.iii) An elephant will walk through your classroom today.It is impossible in the normal classroom setting.iv) You will greet at least one friend at school tomorrow.It is more likely because students usually meet friends at school and may greet them.Thus the rankings are: i) Certain, ii) Less likely, iii) Impossible, iv) More likely.

Exercise Set 7.2(6 questions)

1A 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:(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.

School Level Solution

Colour	Number of sweets
Red	10
Green	8
Yellow	7
Blue	5

i) Probability of green = 830 = 415 Ans.ii) Probability of yellow = 730.Estimated no. of yellow sweets in 600 sweets = 730 × 600 = 140 Ans.

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

School Level Solution

Club	Students
Science Club	14
Arts Club	11
Sports Club	9
Debate Club	6

i) Probability of Arts Club = 1140 = 0.275 Ans.ii) Probability of Sports Club = 940.Estimated no. of Sports Club students = 940 × 800 = 180 Ans.

3Toss 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?

School Level Solution

The answers to (i) and (ii) depend on the actual 20 tosses.If heads occur h times, tails occur = 20 − h.Experimental probability of heads = h20.For one more toss of a fair coin, probability of tails = 12 Ans.

4Toss 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. Assign probabilities to the outcomes by using experimental probability.

School Level Solution

Fig. 7.5

Let bottom = b, top = t and side = s be the numbers of times the cup lands in each position.b + t + s = 100.Experimental probability of bottom = b100.Experimental probability of top = t100.Experimental probability of side = s100.

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

School Level Solution

Favourable outcomes = 2, 4, 6 = 3 outcomes.Total possible outcomes = 6.P(even number) = 36 = 12 Ans.

6Suppose 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?

School Level Solution

i) Experimental probability of rolling 3 = 312 = 14 Ans.

ii) Theoretical probability of rolling 3 = 16 Ans.

iii) The experimental result can differ because only 12 trials were made.As the no. of trials increases to 60, 600 or 6000, the experimental probability is expected to get closer to 16.

Exercise Set 7.3(3 questions)

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

School Level Solution

A standard 6-sided die has outcomes 1, 2, 3, 4, 5 and 6.Therefore, sample size = 6 Ans.

2For 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.

School Level Solution

i) 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) }

ii) S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}.

iii) S = {Green, Red}.

3In 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.’

School Level Solution

i) S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}.ii) Event of selecting Samosa = {(Samosa, Chai), (Samosa, Lassi)}.

Exercise Set 7.4(2 questions)

1There 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?

School Level Solution

Sample space = {AB, AM, OB, OM}.P(Apple and Banana) = 13 × 12 = 16 Ans.

2Let 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?

School Level Solution

The possible colour pairs are RR, RB, RG, BR, BB, BG, GR, GB and GG.P(RR) = 39 × 39 = 19P(BB) = 49 × 49 = 1681P(GG) = 29 × 29 = 481∴ P(same colour) = 9+16+481 = 2981 Ans.

End-of-Chapter Exercises(16 questions)

1Fill 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.

School Level Solution

(i) Probability of an impossible event = 0.(ii) The set of all possible outcomes is called the sample space.(iii) Probability of a certain event = 1.(iv) Probability of getting heads on a fair coin = 12.

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

School Level Solution

Number who like football = 15.Total students = 50.Therefore it is the relative frequency,= 1550 = 0.3.

3Which of the following experiments have 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.

School Level Solution

i) Driver startsdoes not start: not necessarily equally likely.ii) Fair coin: equally likely.iii) Fair 6-sided die: equally likely.iv) 3 red and 7 blue: not equally likely.v) Boygirl: treated as equally likely in the elementary model.

4Write 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?

School Level Solution

i) S = {HH, HT, TH, TT}.P(at least one head) = 34 = 0.75.

ii) S = {1,2,3,4,5,6,7,8,9,10}.P(even) = 510 = 12.

iii) S = {1,2,3,4,5,6}.P(>4) = 26 = 13.

iv) Total balls = 6; non-red = 3.P(not red) = 12.

v) Favourable outcomes = HHT, HTH, THH.P(exactly 2 heads) = 38 = 38.

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

School Level Solution

Favourable outcome = strawberry.Total outcomes = 3.P(strawberry) = 13 Ans.

6A 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.

School Level Solution

Shirt	Jeans	Khakis	Shorts
Red	Red + Jeans	Red + Khakis	Red + Shorts
Blue	Blue + Jeans	Blue + Khakis	Blue + Shorts

Therefore, there are 6 possible combinations.

7A tyre company records distances before replacement in 1000 cases. Find the probability that a randomly chosen tyre lasts:

School Level Solution

Distance	Number of cases
Less than 4000 km	20
4001 to 9000 km	210
9001 to 14000 km	325
More than 14000 km	445

Total cases = 1000.i) P(<4000) = 201000 = 0.02.ii) P(4000 to 14000) = 210+3251000 = 0.535.iii) P(>14000) = 4451000 = 0.445.

8The letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking.(i) What is the probability that it is a P, E or C?(ii) What is the probability that it is not an E?

School Level Solution

The word PEACE has 5 letters, with E occurring twice.i) P(P, E or C) = 1+2+15 = 45.ii) P(not E) = 35 = 35.

★9A game of chance consists of spinning an arrow 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:

School Level Solution

Fig. 7.7

There are 8 equally likely outcomes.i) P(8) = 18.ii) Odd numbers = {1,3,5,7}; P = 12.iii) Numbers greater than 2 = 6 outcomes; P = 34.iv) Numbers less than 9 = all 8 outcomes; P = 1.v) Multiples of 3 = {3,6}; P = 14.

★10A 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.

School Level Solution

P(R then B) = 49 × 58 = 518.P(B then B) = 59 × 48 = 518.

★11I 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.

School Level Solution

An impossible event: sum of the two dice is 13.A certain event: the sum is one of {2,3,…,12}.There is no single elementary outcome of probability 1 for a pair of fair dice.

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

School Level Solution

i) Prime sums greater than 5 are 7 and 11. Number of ways = 6+2 = 8.P = 836 = 29.

ii) Total ways = C(9,2) = 36.Same colour ways = C(4,2)+C(3,2)+C(2,2)=10.Different colours = 36−10 = 26.P = 1318.

iii) Required outcomes are HHT and HTH.P = 28 = 14.

iv) Even number means last digit 2 or 4.Favourable = 2×3! = 12; total = 4! = 24.P = 12.

v) P(exactly 2 correct) = C(3,2)(14)²(34) = 964.

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

School Level Solution

i) With replacement: 4×4 = 16 ordered outcomes.ii) Without replacement: 4×3 = 12 ordered outcomes.iii) Sizes = 16 and 12.

★14List 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.

School Level Solution

S = {H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}.Therefore sample space size = 12.

★15Three 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?

School Level Solution

(i) {1,2,3} is not a sample space because 0 heads is missing.(ii) {0,1,2} is not a sample space because 3 heads is missing.(iii) {0,1,2,3,4} is not a sample space because 4 heads is impossible.(iv) {0,1,2,3} contains all and only the possible numbers of heads.Therefore, the correct sample space is {0,1,2,3}.

★16Suppose 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?

School Level Solution

Fig. 7.8

Area of rectangle = 3×2 = 6 m².Diameter of circle = 1 m ⇒ radius = 0.5 m.Area of circle = π(0.5)² = π/4 m².Probability = π/46 = π24 ≈ 0.131 Ans.

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