NCERT Solutions for Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability Exercise 7.3 | 2026-27

Think and Reflect

In Example 1, we used the sample space {Rain, No Rain} and focused only on whether it will rain. However, if we want to include different amounts of rainfall, such as drizzle, light rain, or heavy rain, we need to expand the sample space to {No Rain, Drizzle, Light Rain, Heavy Rain} to match better the level of detail the question requires. Ensure the sample space is sufficiently detailed for the specific problem being studied.

Solution: 

Expanding the sample space from {Rain, No Rain} to {No Rain, Drizzle, Light Rain, Heavy Rain} better matches the level of detail needed for the situation being studied. The sample space should include all possible outcomes relevant to the situation. This helps make more accurate predictions or analyses by accounting for rainfall variations. 

Think and Reflect

Question 1.

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

Solution:

Given: Two fair coins are tossed.

To find: Probability of getting one head and one tail.

When two coins are tossed, the possible outcomes are:

S = {HH, HT, TH, TT} where, H = Head and T = Tail

Here, total outcomes = 4

We need one head and one tail, so favourable outcomes are: HT, TH

Number of favourable outcomes = 2

Now, P(one head and one tad) = $\dfrac{\text{Favourable outcomes}}{\text{Total outcomes}}$

= $\dfrac{2}{4}$

= $\dfrac{1}{2}$

When two coins are tossed, getting one head and one tail can happen in two different ways (HT, TH), so the probability is $\dfrac{1}{2}$.

Exercise Set 7.3 

Question 1.

When a single 6-sided die is rolled, how many possible outcomes are in the sample space?

Solution: 

When a 6-sided die is rolled once, the possible outcomes are {1, 2, 3, 4, 5, 6}.
As the die has 6 distinct faces, each face represents one possible outcome.

∴ Total number of possible outcomes = 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

A 6-sided die has 6 possible outcomes: 1, 2, 3, 4, 5, and 6. A coin has 2 possible outcomes: Heads (H) and Tails (T). 

Therefore, the total number of outcomes is:

6 x 2 = 12

Each outcome consists of one die result and one coin result.

Sample Space (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 all possible outcomes = 12

(ii) Choosing an integer between -5 and +5

Since the integer must lie strictly between -5 and +5, the endpoints -5 and +5 are not included. The possible integers are:

Sample Space (S):
{-4,-3,-2,-1,0,1,2,3,4}

Number of all possible outcomes = 9

(iii) Drawing a ball from a box

The box contains 5 green balls and 7 red balls. When one ball is drawn at random, the possible colours are Green and Red.

Sample Space (S):
{Green, Red}

There are 12 balls in total:

5+7=12

Thus, we can draw 12 individual balls. However, when the sample space is described by colour, it contains only two outcomes: Green and Red.

• Number of green balls = 5

• Number of red balls = 7

If each ball is treated as a separate outcome, the sample space can be written as:

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

Here, each ball represents one possible outcome.

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

When you roll a die and toss a coin, the die has 6 possible outcomes (1, 2, 3, 4, 5, and 6), and the coin has 2 possible outcomes (Heads (H) and Tails (T). Hence, the total number of possible outcomes is:

6 x 2 = 12

Each outcome combines one die result with one coin-toss result.

Sample Space (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 all possible outcomes = 12

(ii) Choosing an integer between -5 and +5

When an integer is selected at random between -5 and +5, it does not include -5 or +5. Therefore, the sample space contains all integers that lie strictly between these two endpoints.

Sample Space (S):
{-4,-3,-2,-1,0,1,2,3,4}

Number of all possible outcomes = 9

(iii) A box contains 5 green and 7 red balls.

One ball is drawn at random.

Sample Space (S): {Green, Red}

Number of all possible outcomes: 12 (Total balls: 5 + 7)

Although there are only two types of outcomes (Green or Red), the weight, or likelihood, of each outcome is determined by the total number of items.

Number of favourable outcomes for Green: 5

Number of favourable outcomes for Red: 7

Although there are 12 balls in total, the sample space includes only the types of outcomes, not their quantities.

Also, if individual balls are considered, then

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

Question 3.
In a village fair, 3 popular snacks are 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) To find the sample space, we pair each snack with every possible drink.
There are 3 snacks (Samosa, Pakora, Bhaji) and 2 drinks (Chai, Lassi).
Sample Space (S): {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}
Each outcome represents one complete meal choice.
Number of all possible outcomes: 6

(ii) An event is a specific subset of the sample space that satisfies a given condition.
In this case, we only look for pairs where the snack is a Samosa.
Event (E): Selecting Samosa as a snack ={(Samosa, Chai), (Samosa, Lassi)}
Number of favourable outcomes: 2

Benefits of Using Class 9 Maths Chapter 7 The Mathematics of Maybe: Introduction to Probability Exercise 7.3 Solutions

Using Class 9 Maths Chapter 7 Exercise 7.3 Solutions helps students practise NCERT questions while clearly understanding the steps involved. These solutions are useful for:

• Understanding the method: Follow step-by-step solutions to learn how to solve each question in Class 9 Maths Ex 7.3.

• Checking your answers: Compare your work with the correct solution and identify calculation or concept-based mistakes.

• Improving exam preparation: Practise Exercise 7.3 questions regularly to build accuracy and confidence before exams.

• Quick revision: Use NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.3 to revise the exercise and key concepts faster.

• Independent practice: Try the questions yourself first, then use the solutions to clear doubts and strengthen your problem-solving skills.

Class 9 Maths Ex 7.3 solutions help you understand the correct steps and check your answers with ease. Practise the exercise regularly to strengthen your concepts and improve accuracy. Use the solutions for quick revision and better exam preparation.

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