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By Karan Singh Bisht
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Updated on 7 Jul 2026, 15:23 IST
NCERT Solutions for Class 9 is an important stage in a student’s academic journey. It builds the foundation for the Class 10 board exams and prepares students for future studies. The National Council of Educational Research and Training (NCERT) offers textbooks that are widely followed across India, and understanding these books thoroughly is essential for strong academic performance.
To support students in their Class 9 Maths preparation for the 2026-27 academic session, NCERT Solutions for Class 9 Maths act as a dependable study resource. Created by Infinity Learn experts, the NCERT Solutions for Class 9 Maths The Mathematics of Maybe: Introduction to Probability provide clear, step-by-step answers to textbook questions, helping students understand concepts easily and develop confidence in the subject.
NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7, The Mathematics of Maybe: Introduction to Probability, include detailed solutions for Exercise Sets 7.1, 7.2, 7.3, 7.4, and the End-of-Chapter Exercises as per the 2026-27 syllabus.
Chapter 7 of Ganita Manjari Grade 9 Part I is one of the most practical and useful chapters in Class 9 Mathematics for the 2026-27 academic session. Probability helps students understand uncertainty in a mathematical way and is used in many real-life situations, such as weather forecasting, cricket match predictions, insurance planning, and scientific research.
Through this chapter, students learn how to go beyond simple guesswork and calculate the likelihood of events using clear mathematical methods. The chapter builds a strong foundation in probability and helps students develop logical thinking and problem-solving skills.
Students preparing for Class 9 Maths can download the PDF of NCERT Solutions for Class 9 Maths Chapter 7: The Mathematics of Maybe – Introduction to Probability from Infinity Learn. These solutions are carefully created by experienced Infinity Learn subject experts to help students understand probability concepts in a clear and structured manner.
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Designed as per the latest 2026-27 academic session, the solutions provide step-by-step explanations for all important questions from the chapter. The PDF format makes it easy for students to access the solutions anytime, revise concepts quickly, and practice questions at their own pace.
Infinity Learn NCERT Solutions focus on accuracy, conceptual clarity, and exam-oriented preparation. With expert-prepared answers, students can strengthen their understanding of sample space, favourable outcomes, experimental probability, theoretical probability, and real-life applications of probability.
Practicing questions with the help of Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 7: The Mathematics of Maybe: Introduction to Probability NCERT Solutions helps students strengthen their understanding and build confidence in solving probability problems.
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.
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
(iii) An elephant will walk through your classroom today.
(iv) You will greet at least one friend at school tomorrow.
Solution:
(i) Certain Event — Probability = 1
Reason:
The days of the week follow a fixed sequence. Since Sunday is always followed by Monday, this event will definitely occur. Therefore, its probability is 1.
(ii) Impossible Event — Probability = 0
Reason:
Mumbai has a tropical climate and experiences the monsoon season in July, with hot and humid weather. Snowfall does not occur in Mumbai during this time. Hence, this event is impossible, and its probability is 0.
(iii) Impossible Event — Probability = 0
Reason:
Under normal circumstances, an elephant walking into a school classroom is practically impossible. Since this event cannot realistically happen, its probability is 0 or extremely close to 0.
(iv) More Likely Event — Probability close to 1, but not exactly 1
Reason:
In school, students are usually surrounded by friends, so greeting at least one friend is very likely. The chances of this event happening are high, which makes it a more likely event. However, it is not certain because rare situations, such as being absent or the school being closed, may prevent it from happening.
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 | 8 green | 7 yellow | 5 blue
(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:
(i) Number of green sweets = 8
Total number of sweets in the sample = 30
So, the probability of getting a green sweet is:
P(green) = 8/30 = 4/15 ≈ 0.267
Therefore, the probability of selecting a green sweet is 4/15, or approximately 0.267.
(ii) Number of yellow sweets = 7
Total number of sweets in the sample = 30
So, the probability of getting a yellow sweet is:
P(yellow) = 7/30
To estimate the number of yellow sweets in a pack of 600 sweets:
Estimated number = (7/30) × 600 = 7 × 20 = 140
Therefore, approximately 140 sweets are likely to be yellow.
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:
(i) Number of students who prefer the Arts Club = 11
Total number of students in the sample = 40
So, the probability that a student prefers the Arts Club is:
P(Arts Club) = 11/40 = 0.275
Therefore, the probability of selecting a student who prefers the Arts Club is 11/40, or 0.275.
(ii) Number of students who prefer the Sports Club = 9
Total number of students in the sample = 40
So, the probability that a student prefers the Sports Club is:
P(Sports Club) = 9/40
To estimate the number of students in the whole school who prefer the Sports Club:
Estimated number = (9/40) × 800 = 9 × 20 = 180
Therefore, approximately 180 students in the whole school are likely to prefer the Sports Club.
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:
(i) 11 times
(ii) 9 times
(iii) P(heads) = (Number of heads)/(Total number of tosses) = 11/20 = 0.55
(iv) This is a new independent toss. The theoretical probability of getting tails on any single fair coin toss is always: P(tails) = 1/2
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 bottom, upside down its top or on its side (See Fig. 7.5). Assign probabilities to the outcomes by using experimental probability.
Solution: This is a hands-on activity based on an experiment. After performing the experiment, the following observations were recorded:
Bottom = 35 times
Top = 15 times
Side = 50 times
Total number of trials = 100
Therefore,
P(bottom) = Number of times it landed on bottom / Total number of trials
P(bottom) = 35/100
P(top) = Number of times it landed on top / Total number of trials
P(top) = 15/100
P(side) = Number of times it landed on side / Total number of trials
P(side) = 50/100
Thus, the probabilities are:
P(bottom) = 35/100, P(top) = 15/100, and P(side) = 50/100.
5. What is the probability of getting an even number when rolling fair 6-sided die?
Solution: A fair 6-sided die has numbers: 1, 2, 3, 4, 5, 6
Even numbers are: 2, 4, 6
Number of favourable outcomes = 3
Total number of outcomes = 6
Probability of getting an even number = 3/6 = 1/2
Therefore, the probability of getting an even number is 1/2.
6. Suppose you roll a 6-sided die 12 times and get a ‘3’ 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 of getting 3: P(3) = 3/12 = 1/4
(ii) Theoretical probability of getting 3: P(3) = 1/6
(iii) The experimental probability is calculated using the actual results obtained from the experiment. Since the die was rolled only 12 times, the outcomes may vary due to chance.
The theoretical probability, on the other hand, is based on the assumption that the die is fair and that all six outcomes are equally likely.
1. When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?
Solution: When a single 6-sided die is rolled, the possible outcomes are: 1, 2, 3, 4, 5, 6
So, the total number of possible outcomes in the sample space is 6.
2. For the following experiments write down the sample space S.
(i) Rolling a die and tossing a coin together
Solution: The possible outcomes of a die are: 1, 2, 3, 4, 5, 6
The possible outcomes of a coin are: H (Heads) and T (Tails)
Each outcome is written as a pair: (die number, coin result)
Therefore, the sample space is:
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)}
Hence, n(S) = 12
(ii) Choosing a random integer between -5 and +5
Solution: The integers between -5 and +5, including both endpoints, are:
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
Therefore, the sample space is:
S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
So, n(S) = 11
(iii) A box contains 5 green balls and 7 red balls. One ball is drawn at random.
Solution: If the balls are considered identical except for their colour, then there are two possible types of outcomes: Green ball (G) and Red ball (R).
Therefore, the sample space is:
S = {Green ball, Red ball}
or
S = {G, R}
Hence, n(S) = 2
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) S = {Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}
n(S) = 6
(ii) E = {(Samosa, Chai), (Samosa, Lassi)}
The event has 2 outcomes out of 6 total outcomes.
P(Samosa) = 2/6 = 1/3
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:
(i) Tree Diagram
Basket A
/ \
Apple Orange
/ \ / \
Banana Mango Banana Mango
Possible pairs:
(Apple, Banana)
(Apple, Mango)
(Orange, Banana)
(Orange, Mango)
(ii) Sample Space
S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}
(iii) Probability of picking one apple and one banana
Number of favourable outcomes = 1 [(Apple, Banana)]
Total number of outcomes = 4
Probability = 1/4
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 you 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 are your friend pick pens of the same colour?
Solution:
(i) Possible Outcomes of the Pen Colours
Solution:
The possible pen colours are:
Red (R), Black (B), and Green (G)
Since each person can pick any one of the three colours, the sample space is:
S = {(R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)}
Tree Diagram:
First Pick R B G / | \ / | \ / | \ R B G R B G R B G(ii) Probability That Both Pick Pens of the Same Colour
Solution: From the sample space, the outcomes in which both people pick pens of the same colour are:
(R, R), (B, B), and (G, G)
Number of favourable outcomes = 3
Total number of outcomes = 9
Therefore,
Probability = 3/9 = 1/3
Hence, the probability that both people pick pens of the same colour is 1/3.
1. Fill in the blanks.
(i) The probability of an important 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 important 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 for getting heads.
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.
3. Which of the following experiments have equally likely outcome? 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.
Solution:
(i) Not equally likely
Solution:
A car starting depends on several factors, such as the condition of the car, fuel level, battery, and engine performance. Therefore, the two outcomes — the car starts and the car does not start — are not equally likely under normal conditions.
(ii) Equally likely
Reason:
When a fair coin is tossed, the two possible outcomes are Heads and Tails. Each outcome has an equal probability of 1/2. Therefore, the outcomes are equally likely.
(iii) Equally likely
Reason:
When a fair six-sided die is rolled, each face numbered 1 to 6 has an equal chance of appearing. The probability of each outcome is 1/6. Hence, the outcomes are equally likely.
(iv) Not equally likely
Reason:
The probability of selecting a red ball is 3/10, while the probability of selecting a blue ball is 7/10. Since the probabilities are different, the outcomes are not equally likely.
(v) Equally likely
Reason:
The chances of a baby being a boy or a girl are generally considered approximately equal. Therefore, these two outcomes can be treated as equally likely.
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?
Solution:
The sample space is:
S = {HH, HT, TH, TT}
So, n(S) = 4
The event of getting at least one head is:
E = {HH, HT, TH}
So, n(E) = 3
Therefore,
P(at least one head) = n(E)/n(S) = 3/4
Hence, the probability of getting at least one head is 3/4.
(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?
Solution: The sample space is:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
So, n(S) = 10
The even-numbered cards are:
E = {2, 4, 6, 8, 10}
So, n(E) = 5
Therefore,
P(even number) = n(E)/n(S) = 5/10 = 1/2
Hence, the probability of drawing an even-numbered card is 1/2.
(iii) A die is rolled once. What is the probability of getting a number greater than 4?
Solution: The sample space is:
S = {1, 2, 3, 4, 5, 6}
So, n(S) = 6
The numbers greater than 4 are:
E = {5, 6}
So, n(E) = 2
Therefore,
P(number greater than 4) = n(E)/n(S) = 2/6 = 1/3
Hence, the probability of getting a number greater than 4 is 1/3.
(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?
Solution: Total number of balls:
3 + 2 + 1 = 6
Number of balls that are not red:
Blue balls + Green ball = 2 + 1 = 3
Therefore,
P(not red) = 3/6 = 1/2
Hence, the probability of picking a ball that is not red is 1/2.
(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
Solution: The sample space is:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
So, n(S) = 8
The outcomes with exactly two heads are:
E = {HHT, HTH, THH}
So, n(E) = 3
Therefore,
P(exactly two heads) = n(E)/n(S) = 3/8
Hence, the probability of getting exactly two heads is 3/8.
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: There are 3 candies in the bag: strawberry, lemon, and mint.
Number of strawberry candies = 1
Total number of candies = 3
Therefore,
P(strawberry) = 1/3
Hence, the probability of picking a strawberry candy is 1/3.
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: A child has 2 shirts and 3 types of pants.
Total possible outfits = 2 × 3 = 6
| S. No. | Shirt | Pants |
| 1 | Red shirt | Jeans |
| 2 | Red shirt | Khakis |
| 3 | Red shirt | Shorts |
| 4 | Blue shirt | Jeans |
| 5 | Blue shirt | Khakis |
| 6 | Blue shirt | Shorts |
Therefore, there are 6 possible outfit combinations.
7. A tyre company records distances before replacement in 1000 cases.
Distance (km) | Less than 4000 | 4001 to 9000 | 9001 to 14000 | More than 14000
Number of cases | 20 | 210 | 325 | 445
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:
(i) Probability that a tyre lasts less than 4000 km
Total number of tyres = 1000
Number of tyres that lasted less than 4000 km = 20
Therefore,
P(less than 4000 km) = 20/1000 = 1/50
Hence, the probability that a tyre lasts less than 4000 km is 1/50.
(ii) Probability that a tyre lasts between 4001 km and 14,000 km
Total number of tyres = 1000
Number of tyres that lasted between 4001 km and 14,000 km:
210 + 325 = 535
Therefore,
P(between 4001 km and 14,000 km) = 535/1000 = 107/200
Hence, the probability that a tyre lasts between 4001 km and 14,000 km is 107/200.
(iii) Probability that a tyre lasts more than 14,000 km
Total number of tyres = 1000
Number of tyres that lasted more than 14,000 km = 445
Therefore,
P(more than 14,000 km) = 445/1000 = 89/200
Hence, the probability that a tyre lasts more than 14,000 km is 89/200.
8. The letter 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?
Solution:
(i) Probability of getting P, E, or C
Total number of cards in the word PEACE = 5
The cards with the letters P, E, or C are:
P = 1, E = 2, C = 1
So, total favourable cards = 1 + 2 + 1 = 4
Therefore,
P(P, E, or C) = 4/5
Hence, the probability of getting P, E, or C is 4/5.
(ii) Probability of not getting E
Number of cards with the letter E = 2
Number of cards that are not E:
5 − 2 = 3
The cards that are not E are P, A, and C.
Therefore,
P(not E) = 3/5
Hence, the probability of not getting E is 3/5.
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
(i) 8?
(ii) An odd number?
(iii) A number greater than 2?
(iv) A number less than 9?
(v) A multiple of 3?
Solution: The spinner has numbers:
1, 2, 3, 4, 5, 6, 7, 8
Total number of possible outcomes = 8
(i) Probability that it will point at 8
Favourable outcome = 8
P(8) = 1/8
(ii) Probability that it will point at an odd number
Odd numbers are: 1, 3, 5, 7
Number of favourable outcomes = 4
P(odd number) = 4/8 = 1/2
(iii) Probability that it will point at a number greater than 2
Numbers greater than 2 are: 3, 4, 5, 6, 7, 8
Number of favourable outcomes = 6
P(number greater than 2) = 6/8 = 3/4
(iv) Probability that it will point at a number less than 9
All numbers on the spinner are less than 9: 1, 2, 3, 4, 5, 6, 7, 8
Number of favourable outcomes = 8
P(number less than 9) = 8/8 = 1
(v) Probability that it will point at a multiple of 3
Multiples of 3 are: 3, 6
Number of favourable outcomes = 2
P(multiple of 3) = 2/8 = 1/4
10. A basket contains 4 red ball 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: A basket contains 4 red balls and 5 blue balls.
Total number of balls = 4 + 5 = 9
Since the first ball is laid aside, the second ball is drawn without replacement.
Tree Diagram
First Draw Start / \ Red (4/9) Blue (5/9) / \ / \ Red (3/8) Blue (5/8) Red (4/8) Blue (4/8)Possible Outcomes and Probabilities
| Outcome | Probability |
| Red, Red | 4/9 × 3/8 = 12/72 = 1/6 |
| Red, Blue | 4/9 × 5/8 = 20/72 = 5/18 |
| Blue, Red | 5/9 × 4/8 = 20/72 = 5/18 |
| Blue, Blue | 5/9 × 4/8 = 20/72 = 5/18 |
(i) Probability of drawing a red ball and then a blue ball
P(Red, Blue) = 4/9 × 5/8 = 20/72 = 5/18
Hence, the probability is 5/18.
(ii) Probability of drawing 2 blue balls
P(Blue, Blue) = 5/9 × 4/8 = 20/72 = 5/18
Hence, the probability is 5/18.
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: When a pair of 6-sided dice is thrown, the smallest possible sum is:
1 + 1 = 2
and the largest possible sum is:
6 + 6 = 12
An event with probability 0 is an impossible event.
Example:
Getting a sum of 13
This is impossible because the maximum sum is 12.
So, P(sum of 13) = 0
An event with probability 1 is a certain event.
Example:
Getting a sum less than 13
This is certain because every possible sum from two dice is between 2 and 12.
So, P(sum less than 13) = 1.
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?
Solution:
When two dice are rolled, the total number of possible outcomes is:
6 × 6 = 36
Prime numbers greater than 5 and less than or equal to 12 are:
7 and 11
Outcomes with sum 7 are:
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
Number of outcomes = 6
Outcomes with sum 11 are:
(5, 6), (6, 5)
Number of outcomes = 2
Total favourable outcomes:
6 + 2 = 8
Therefore,
P(sum is a prime number greater than 5) = 8/36 = 2/9
Hence, the probability is 2/9.
(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?
Solution:
Total number of balls:
4 + 3 + 2 = 9
Since two balls are drawn without replacement, the total number of ordered outcomes is:
9 × 8 = 72
Now, we find the favourable outcomes where both balls are of different colours.
Red and Green:
Red then Green = 4 × 3 = 12
Green then Red = 3 × 4 = 12
Total = 24
Red and Blue:
Red then Blue = 4 × 2 = 8
Blue then Red = 2 × 4 = 8
Total = 16
Green and Blue:
Green then Blue = 3 × 2 = 6
Blue then Green = 2 × 3 = 6
Total = 12
Total favourable outcomes:
24 + 16 + 12 = 52
Therefore,
P(different colours) = 52/72 = 13/18
Hence, the probability that both balls are of different colours is 13/18.
(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
Solution: The sample space is:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
So, n(S) = 8
The outcomes in which the first coin shows heads and exactly two heads occur in total are:
{HHT, HTH}
So, n(E) = 2
Therefore,
P(first coin is heads and exactly two heads occur) = 2/8 = 1/4
Hence, the probability is 1/4.
(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?
Solution:
Total arrangements of the four digits are:
4 × 3 × 2 × 1 = 24
A number is even if its last digit is even.
The even digits among 1, 2, 3, and 4 are 2 and 4.
Case 1: Last digit is 2
The remaining digits are 1, 3, and 4.
Number of arrangements:
3 × 2 × 1 = 6
Case 2: Last digit is 4
The remaining digits are 1, 2, and 3.
Number of arrangements:
3 × 2 × 1 = 6
Total favourable outcomes:
6 + 6 = 12
Therefore,
P(even number) = 12/24 = 1/2
Hence, the probability that the number is even is 1/2.
(v) A student takes a multiple-choice test with 3 questions, each having 4 options A, B, C, and D. Only one option is correct for each question. What is the probability that the student guesses and gets exactly 2 answers correct?
Solution: Each question has 4 options, out of which only 1 option is correct.
So,
P(correct answer) = 1/4
and
P(wrong answer) = 3/4
The student must get exactly 2 answers correct out of 3 questions.
This can happen in 3 ways:
CCW, CWC, WCC
Probability of one such case:
(1/4) × (1/4) × (3/4) = 3/64
Since there are 3 such cases,
P(exactly 2 correct answers) = 3 × 3/64 = 9/64
Hence, the probability that the student gets exactly 2 answers correct is 9/64.
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) Sample space of all ordered pairs
The sample space contains all possible ordered pairs:
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)}
Therefore,
n(S) = 16
(ii) Sample space of all ordered pairs where both numbers are different
Here, pairs with the same numbers are not included.
So, the sample space is:
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)}
Therefore,
n(S) = 12
(iii) Comparison
With replacement:
n(S) = 16
Without replacement:
n(S) = 12
Hence, the sample space has 16 outcomes with replacement and 12 outcomes without replacement.
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: When a coin is tossed, the possible outcomes are:
H (Heads), T (Tails)
The cards are numbered:
1, 2, 3, 4, 5, 6
Each outcome is written as a pair: (coin outcome, card number)
Therefore, the sample space is:
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)}
Hence, the total number of outcomes is:
n(S) = 12.
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 outcomes are:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
Now, the number of heads in these outcomes can be:
0 heads → TTT
1 head → HTT, THT, TTH
2 heads → HHT, HTH, THH
3 heads → HHH
So, the sample space for recording the number of heads is:
{0, 1, 2, 3}
Therefore, the correct option is:
(iv) {0, 1, 2, 3}
Why the other options are not sample spaces:
(i) {1, 2, 3}
This does not include 0 heads, which is possible when all three coins show tails.
(ii) {0, 1, 2}
This does not include 3 heads, which is possible when all three coins show heads.
(iii) {0, 1, 2, 3, 4}
This includes 4 heads, which is impossible because only three coins are tossed.
16. Suppose you drop a dye at random on the rectangular region show in fig 7.8. What is the probability that it will land inside the circle with a diameter of 1 m?
Solution: Area of rectangle = 3 × 2 = 6 m².
The circle has diameter 1 m.
So, radius:
r = 1/2 m
Area of circle:
πr² = π × (1/2)² = π/4 m²
Now,
Probability = Area of circle / Area of rectangle
P(landing inside the circle) = (π/4) / 6 = π/24
Hence, the probability that it will land inside the circle is π/24.
The NCERT Solutions for Class 9 Maths Chapter 7, The Mathematics of Maybe: Introduction to Probability, are designed to help students understand probability concepts in a simple and structured way. Created by Infinity Learn experts, these solutions are aligned with the latest CBSE 2026-27 syllabus and provide clear explanations for all exercise questions. Below are the key features:
| Chapter No. | Important Links for class 9 Maths Chapter-wise Solutions PDF |
| Chapter 1 | Orienting Yourself: The Use of Coordinates |
| Chapter 2 | Introduction to Linear Polynomials |
| Chapter 3 | The World of Numbers |
| Chapter 4 | Exploring Algebraic Identities |
| Chapter 5 | I’m Up and Down, and Round and Round |
| Chapter 6 | Measuring Space: Perimeter and Area |
Important Study Materials for Class 9 Maths
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Students can download the Class 9 maths chapter 7 ncert solutions pdf from Infinity Learn. The solutions are created by subject experts and provide step-by-step answers to help students understand probability concepts clearly.
Yes, the Class 9 Maths Chapter 7 PDF available on Infinity Learn is prepared as per the latest 2026-27 academic session. It helps students practise questions from Chapter 7, The Mathematics of Maybe: Introduction to Probability, in a simple and structured way.
Students can visit Infinity Learn to access the Class 9 maths chapter 7 ncert solutions pdf download. The PDF is designed to support quick revision, homework help, and exam preparation.
The Mathematics of Maybe Introduction to Probability PDF includes detailed solutions to questions based on probability, sample space, outcomes, experimental probability, theoretical probability, and real-life probability examples. Infinity Learn provides these solutions in easy-to-understand language.
Infinity Learn Class 9 Maths Chapter 7 PDF is created by experienced experts and follows the latest NCERT syllabus. It helps students build conceptual clarity and improve confidence in solving probability-based questions.
Yes, the Class 9 maths chapter 7 NCERT solutions pdf from Infinity Learn is useful for exam preparation. It provides clear explanations, accurate answers, and step-by-step methods that help students revise Chapter 7 effectively.