NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability

Class 9 Ganita Manjari Chapter 7 Quick Links:

• Exercise Set 7.1
• Exercise Set 7.2
• Exercise Set 7.3
• Exercise Set 7.4
• End-of-Chapter Exercises

To study with mobile or tab, download Class 9 Maths Ganita Manjari Offline App here.

The chapter 9 of 9th Maths introduces key ideas such as random experiments, the probability scale (0 to 1), experimental probability, theoretical probability, sample spaces, events and tree diagrams, each explained with relatable Indian contexts like Snakes and Ladders, lucky draws, fruit preferences in a class and cricket tosses. Whether you are a student looking for clear notes, a teacher planning your lessons or a parent trying to help your child, this guide covers everything you need from Chapter 7 of the new NCERT Ganita Manjari textbook.

Class 9 Maths Ganita Manjari Chapter 7 Exercise Set 7.1 Solutions

(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.
Answer:
(i) Certain (Probability = 1)
𝗥𝗲𝗮𝘀𝗼𝗻:
Days of the week always follow a fixed order. Sunday is always followed by Monday. This will definitely happen, so its probability is 1.

(ii) Impossible (Probability = 0)
𝗥𝗲𝗮𝘀𝗼𝗻:
Mumbai has a tropical climate. In July it experiences the monsoon season with hot and humid weather. Snowfall is not possible there. So this event cannot happen at all, its probability is 0.

(iii) Impossible (Probability = 0)
𝗥𝗲𝗮𝘀𝗼𝗻:
Under normal circumstances, it is not possible for an elephant to walk into a school classroom. This event practically cannot happen, so its probability is 0 (or extremely close to 0).

(iv) More likely (Probability is close to 1, but not exactly 1)
𝗥𝗲𝗮𝘀𝗼𝗻:
At school there are many friends around you. It is very likely that you will greet at least one friend. The chances are very high, so this event is “more likely.” It is not “certain” because there could be rare situations (like you are absent or school is closed), but normally it will happen.

Class 9 Maths Ganita Manjari Chapter 7 Exercise Set 7.2 Solutions

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.
Answer:
(i) Number of green sweets = 8
Total sweets in sample = 30
P(green) = 8/30 = 4/15 ≈ 0.267

(ii) Number of yellow sweets = 7
Total sweets in sample = 30
P(yellow) = 7/30
Estimated number of yellow sweets in 600 = (7/30) × 600 = 7 × 20 = 140
Approximately 140 sweets are likely to be yellow.

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.
Answer:
(i) Number of students who prefer Arts Club = 11
Total students in sample = 40
P(Arts Club) = 11/40 = 0.275

(ii) Number of students who Sports Club = 9
Total students in sample = 40
P(Sports Club) = 9/40
Estimated number in whole school = (9/40) × 800 = 9 × 20 = 180
Approximately 180 students in the whole school prefer the Sports Club.

(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?
Answer:
(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

Answer:
This is a hands-on activity. I get the following observation from the experiment:
bottom = 35, top = 15, side = 50

Therefore:
P(bottom) = Number of times it landed on bottom / 100
⇒ P(bottom) = 35/100

P(top) = Number of times it landed on top / 100
⇒ P(top) = 15/100

P(side) = Number of times it landed on side / 100
⇒ P(side) = 50/100

Answer:
Total possible outcomes S = {1, 2, 3, 4, 5, 6},
Even numbers = {2, 4, 6}
Therefore,the number of favourable outcomes = 3
So, P(even number) = 3/6 = 1/2.

(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?
Answer:
(i) P(3) experimentally = 3/12 = 1/4

(ii) P(3) theoretically = 1/6

(iii) Experimental probability is based on actual results from a limited number of trials. With only 12 rolls, the results can vary a lot by chance.
The theoretical probability assumes a perfectly fair die with all outcomes equally likely.

As the number of trials increases:
• At 60 rolls: Experimental probability will come closer to 1/6 but may still vary.
• At 600 rolls: It will be very close to 1/6.
• At 6000 rolls: It will be almost exactly equal to 1/6.
This is the Law of Large Numbers – as the number of trials increases, the experimental probability approaches the theoretical probability.

Class 9 Maths Ganita Manjari Chapter 7 Exercise Set 7.3 Solutions

Answer:
When a 6-sided die is rolled, the possible outcomes are: 1, 2, 3, 4, 5, 6
Sample Space S = {1, 2, 3, 4, 5, 6}
Total number of possible outcomes = 6, i.e., n(S) = 6

(i) Rolling a die and tossing a coin together.
Answer:
Die outcomes: 1, 2, 3, 4, 5, 6
Coin outcomes: H (Heads), T (Tails)
Each outcome is a pair: (die number, coin result)
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)}
Hence, n(S) = 12

(ii) Choosing a random integer between -5 and +5
Answer:
Integers between -5 and +5 (including both endpoints):
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
Therefore, S = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}
So, n(S) = 11

(iii) A box containing 5 green and 7 red balls. One ball is drawn at random.
Answer:
Case 1: If balls are identical.
There are two type of balls: Green (G) and Red (R)
S = {Green ball, Red ball}  or S = {G, R}
n(s) = 2

Case 2: If balls are not identical.
If balls are individually distinguishable, S would have 12 elements: G₁, G₂, G₃, G₄, G₅, R₁, R₂, R₃, R₄, R₅, R₆, R₇, but for probability purposes the colour-based sample space is standard at this level.

(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.”
Answer:
(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

Class 9 Maths Ganita Manjari Chapter 7 Exercise Set 7.4 Solutions

(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?
Answer:
(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

(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?
Answer:
(i) Possible outcomes of the pen colours
The possible colours are:
Red (R)
Black (B)
Green (G)

So, 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
From the sample space, the outcomes in which both pick pens of the same colour are:
(R, R), (B, B), (G, G)

Number of favourable outcomes = 3
Total number of outcomes = 9

Therefore, Probability = 3/9 = 1/3
Hence, the probability that both pick pens of the same colour is 1/3.

Class 9 Maths Ganita Manjari Chapter 7 End-of-Chapter Exercises Solutions

(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.
Answer:
(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.

Answer:
The number of students who like football is 15, and the 𝗿𝗲𝗹𝗮𝘁𝗶𝘃𝗲 𝗳𝗿𝗲𝗾𝘂𝗲𝗻𝗰𝘆 is 𝟭𝟱/𝟱𝟬 = 𝟯/𝟭𝟬 = 𝟬.𝟯.

(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.
Answer:
(i) Not equally likely.
The car starting depends on the condition of the car, fuel, battery etc. The two outcomes (starts / does not start) are not equally likely under normal conditions.

(ii) Yes, equally likely.
Heads and Tails each have probability = 1/2. A fair coin gives equal chance to both outcomes.

(iii) Yes, equally likely.
Each of the 6 faces (1 to 6) has an equal probability = 1/6.

(iv) Not equally likely.
P(red) = 3/10 and P(blue) = 7/10. The two colours have different probabilities.

(v) Yes, equally likely.
Biologically, the probability of a boy or girl is approximately 1/2 each, so these can be considered roughly equally likely.

(i) Two coins are tossed at the same time. What is the probability of getting at least one head?
Answer:
S = {HH, HT, TH, TT}
⇒ n(S) = 4
Event E = at least one head = {HH, HT, TH}
⇒ n(E) = 3
P(at least one head) = n(E)/n(S) = 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?
Answer:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
⇒ n(S) = 10
Even numbers = {2, 4, 6, 8, 10},
⇒ n(E) = 5
Hence, P(even number) = n(E)/n(S) = 5/10 = 1/2.

(iii) A die is rolled once, What is the probability of getting a number greater than 4?
Answer:
S = {1, 2, 3, 4, 5, 6}
⇒ n(S) = 6
Numbers greater than 4 = {5, 6}
⇒ n(E) = 2
P(number greater than 4) = n(E)/n(S) = 2/6 = 1/3.

(iv) A bag contain 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?
Answer:
Total balls = 3 + 2 + 1 = 6
Not red = blue + green = 2 + 1 = 3
P(not red) = 3/6 = 1/2.

(v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
Answer:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒ n(S) = 8
Exactly two heads = {HHT, HTH, THH}
⇒ n(E) = 3
P(exactly two heads) = n(E)/n(S) = 3/8.

Answer:
S = {strawberry, lemon, mint}
⇒ n(S) = 3
n(E) = 1 [strawberry]
P(strawberry) = 1/3.

Answer:
The possible combinations of outfits consisting of one shirt and one pair of pants:

Therefore the total possible combinations = 6 outfits.

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.
Answer:
(i) Total number of tyres = 1000
Number of tyre for less than 4000 km = 20
Therefore, P(Less than 4000 km) = 20/1000 = 1/50

(ii) Total number of tyres = 1000
Cases between 4001 to 14000 = 210 + 325 = 535
Therefore, P(Between 4000 and 14000 km) = 535/1000

(iii) Total number of tyres = 1000
Cases More than 14000 km = 445
Therefore, P(More than 14000 km) = 445/1000

(i) What is the probability that it is a P, E or C?
(ii) What is the probability that it is not an E?
Answer:
(i) Total number of cards in P E A C E = 5
Cards with P, E or C:
P = 1, E = 2, C = 1
⇒ total = 4 cards
Therefore, P(P, E or C) = 4/5

(ii) Number of E cards = 2
Cards that are not E = 5 – 2 = 3 (P, A, C)
Therefore, P(not E) = 3/5.

(i) 8?
(ii) An odd number?
(iii) A number greater than 2?
(iv) A number less than 9?
(v) A multiple of 3?
Answer:
(i) P (Pointing at 8)
Favourable outcomes = {8}
⇒ n(E) = 1
Therefore, P(8) = 1/8

(ii) P (odd number)?
Odd numbers = {1, 3, 5, 7}
⇒ n(E) = 4
Therefore, P (odd) = 4/8 = 1/2

(iii) P(number greater than 2)
Numbers greater than 2 = {3, 4, 5, 6, 7, 8}
⇒ n(E) = 6
Therefore, P(number greater than 2) = 6/8 = 3/4

(iv) P(number less than 9)
All numbers 1 to 8 are less than 9.
⇒ n(E) = 8
Therefore, P(number less than 9) = 8/8 = 1

(v) P(multiple of 3)
Multiple of 3 from 1 to 8 = {3, 6}
⇒ n(E) = 2
Therefore, P(multiple of 3) = 2/8 = 1/4.

Answer:
Total balls = 4 red + 5 blue = 9 balls
Since one ball is laid aside, the second ball is drawn without replacement.
Possible outcomes: {(R, R), (R, B), (B, R), (B, B)}
Tree Diagram
First Draw
/ \
R B
4/9 5/9
/ \ / \
R B R B
3/8 5/8 4/8 4/8

(i) Probability of drawing a red ball and then a blue ball
Probability of first drawing a red ball = 4/9
After drawing one red ball, remaining balls: 3 red and 5 blue
Total remaining balls = 8
Probability of drawing a blue ball next = 5/8
Therefore, P(R, B) = (4/9) × (5/8)
= 20/72
= 5/18
Hence, the probability of drawing a red ball and then a blue ball is 5/18

(ii) Probability of drawing 2 blue balls
Probability of first drawing a blue ball = 5/9
After drawing one blue ball, remaining balls: 4 red and 4 blue
Total remaining balls = 8
Probability of drawing another blue ball = 4/8
Therefore, P(B, B) = (5/9) × (4/8)
= 20/72
= 5/18
Hence, the probability of drawing 2 blue balls is 5/18.

Answer:
Event with probability 0 (Impossible):
Getting a sum of 1. (Minimum sum when throwing two dice is 1 + 1 = 2, so sum = 1 is impossible)

Outcome with probability 1 (Certain):
Getting a sum that is less than 13. (Maximum sum is 6 + 6 = 12, which is always less than 13, so this is certain.)

(i) Two dice are rolled. What is the probability that the sum is a prime number greater than 5?
Answer:
When two dice are rolled, the sample space 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)}
Total number of outcomes = 36.

Prime numbers greater than 5 up to 12: 7, 11
Outcomes with sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) ⇒ 6 outcomes
Outcomes with sum = 11: (5,6), (6,5) ⇒ 2 outcomes
Total favourable = 6 + 2 = 8
P(sum is a prime number greater than 5) = 8/36 = 2/9

(ii) A bag contain 4 red, 3 green, and 2 blue balls. Two balls are drawn without replacement. What is the probability that both are of different colours?
Answer:
Total balls = 4 red + 3 green + 2 blue = 9 balls
Two balls are drawn without replacement.
We need probability that both balls are of different colours.
Total possible outcomes:
First ball can be chosen in 9 ways.
Second ball can be chosen in 8 ways.
Total outcomes = 9 × 8 = 72

Favourable outcomes (different colours)
Red and Green:
First Red then Green = 4 × 3 = 12
First 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.

(iii) Three coins are tossed. What is the probability that the first coin shows heads and exactly two heads occur in total?
Answer:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
⇒ n(S) = 8
First coin shows H and exactly 2 heads total: {HHT, HTH}
⇒ n(E) = 2
Therefore, Probability = 2/8 = 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?
Answer:
Total arrangements of 4 different digits:
= 4 × 3 × 2 × 1
= 24

Condition for even number:
A number is even if its last digit is even.
Even digits here: 2 and 4
So, last digit can be 2 or 4.

Case 1: Last digit = 2
Remaining digits: 1, 3, 4
Number of ways to arrange them = 3 × 2 × 1 = 6

Case 2: Last digit = 4
Remaining digits: 1, 2, 3
Number of ways = 3 × 2 × 1 = 6

Total favourable outcomes = 6 + 6 = 12
Therefore, P(even number) = 12/24 = 1/2.

(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 answer correct?
Answer:
Each question has 4 options and only 1 is correct.
So for each question:
Probability of correct answer = 1/4
Probability of wrong answer = 3/4
Total questions = 3
We need probability of exactly 2 correct answers.

Possible cases for exactly 2 correct
There are 3 ways this can happen: {(C, C, W), (C, W, C), (W, C, C)}

Probability of each case
Each case has:
2 correct answers ⇒ (1/4) × (1/4)
1 wrong answer ⇒ (3/4)
So probability of one case = (1/4) × (1/4) × (3/4) = 3/64

Since there are 3 such cases, so
P(exactly 2 correct) = 3 × (3/64) = 9/64.

(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?
Answer:
(i) Sample Space (all 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)}
n(S) = 16

(ii) Sample Space (all ordered pairs where both numbers are different):
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)}
n(S) = 12

(iii) With replacement: n(S) = 16 Without replacement: n(S) = 12.

Answer:
Coin outcomes: H, T
Card outcomes: 1, 2, 3, 4, 5, 6
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)}
n(S) = 12.

(i) {1, 2, 3}
(ii) {0, 1, 2}
(iii) {0, 1, 2, 3, 4}
(iv) {0, 1, 2, 3}
Answer:
When 3 coins are tossed, possible number of heads = 0, 1, 2 or 3.
(i) {1, 2, 3} — Not a valid sample space. It does not include 0 heads (all tails = TTT).
(ii) {0, 1, 2} — Not a valid sample space. It does not include 3 heads (HHH).
(iii) {0, 1, 2, 3, 4} — Not a valid sample space. The value 4 is not possible since only 3 coins are tossed. Maximum heads = 3.
(iv) {0, 1, 2, 3} — Yes, this is the correct sample space. It includes all possible values: 0 heads (TTT), 1 head, 2 heads, and 3 heads (HHH).

Answer:
Area of rectangle = 3 × 2 = 6 m²
Diameter of circle = 1 m, so radius = 0.5 m
Area of circle = π × r²
= π × (0.5)²
= π × 0.25 = π/4

P(lands inside circle) = Area of circle / Area of rectangle
= (π/4)/6
= π/24

Probability is the branch of mathematics that measures how likely an event is to occur. Unlike most mathematical topics that deal with exact answers, probability lives in the space of uncertainty — it tells you not what will happen, but how confident you can be that something will happen. From weather forecasts to cricket match predictions to school lucky draws, we encounter uncertain situations every day. Probability gives us a precise, numerical way to talk about them. The chapter begins by distinguishing between subjective probability (based on personal opinion, like saying “it looks cloudy, so it might rain”) and objective probability (based on evidence or mathematical reasoning). The goal of Chapter 7 is to move students from the former to the latter — teaching them to measure likelihood in a way that is consistent, verifiable and mathematical.

• Probability is always expressed as a number between 0 and 1 or equivalently as a percentage between 0% and 100%.
• A random event is one where all possible outcomes are known, but which one will occur cannot be predicted in advance.
• Randomness does not mean chaos – across many trials, random events follow predictable patterns, which is what makes probability useful.
• The probability scale works like a number line: the closer a value is to 1, the more likely the event; the closer to 0, the less likely.

Chapter 7 presents two scientific methods for measuring probability objectively. The first is experimental probability, which is calculated from real data — either by running an experiment multiple times or by analysing historical records. The second is theoretical probability, which uses logic and mathematical reasoning to determine likelihood, assuming all outcomes are equally likely. These two approaches are not in competition; they complement each other. Experimental probability reflects what actually happens in the real world, including all its imperfections, while theoretical probability describes what should happen in an ideal, perfectly fair situation.
Ganita Manjari chapter 7 also introduces a third approach — statistical sampling — where data collected from a small, representative group (a sample) is used to estimate probability for a much larger population. This is how businesses forecast demand, how researchers measure public opinion and how schools might estimate preferences without surveying every single student.

• The Law of Large Numbers states that as the number of trials increases, experimental probability gets closer and closer to theoretical probability.
• Gambler’s Fallacy is the mistaken belief that past random outcomes affect future ones – a coin that has landed Heads six times still has exactly a 1/2 chance of Heads on the next toss.
• A coin, die or selection process is called fair or unbiased when all outcomes are equally likely and no outcome is favoured.
• Statistical estimates become more reliable when the sample is both large and representative of the full population.

Every probability calculation starts with a clearly defined sample space — the complete list of all possible outcomes of a random experiment, written as a set S = { }. The number of elements in this set is called the sample size, denoted n(S). An event is then any specific outcome or combination of outcomes that you are interested in — formally, it is a subset of the sample space.
The probability of an event is calculated as P(E) = n(E) ÷ n(S), where n(E) is the number of outcomes that satisfy the event’s condition. The most common mistake students make here is writing an incomplete sample space — for example, listing only {HH, TT} when tossing two coins, missing the outcomes {HT, TH}. A correct sample space must include every possible outcome without repetition, and it must match the level of detail the question requires.

• An event E is always a subset of the sample space S, and its probability is P(E) = n(E) ÷ n(S).
• Always write out the full sample space before attempting any probability calculation — it prevents missing outcomes and earns partial marks in exams.
• The sample space should be tailored to the question: {Rain, No Rain} works for a simple yes/no, but {No Rain, Drizzle, Light Rain, Heavy Rain} is needed when the amount of rainfall matters.
• For any event, 0 ≤ P(E) ≤ 1 always holds — a probability can never be negative or greater than 1.

When an experiment involves two or more sequential steps, a tree diagram is the most reliable tool for mapping every possible outcome without missing any. Starting from a single point, branches are drawn for each outcome of the first step and then further branches spread from each of those for the second step – and so on. Every complete path from the starting point to a tip of the tree represents one outcome in the sample space.
Beyond simply listing outcomes, tree diagrams also make probability calculation visual and traceable, especially for multi-step problems involving selections with or without replacement. Chapter 7 uses tree diagrams to analyse experiments like tossing a coin twice, picking from multiple baskets and drawing coloured balls in sequence – all of which are multi-step experiments where a systematic approach is essential. Mastering tree diagrams in Class 9 also lays the groundwork for more advanced probability work in later years.

• Always show your working clearly: write n(E) = ?, n(S) = ?, then P(E) = ?/? – do not jump directly to the final answer.
• For tree diagrams, label every branch and read each complete path carefully to build the sample space.
• The difference between experimental and theoretical probability and why they may not match in small samples, is a common exam question worth preparing a written explanation for.
• Chapter 7 requires conceptual understanding, not just formula recall — students who can explain why probability works the way it does consistently score higher than those who only memorise the formulas.