Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions

Tree diagram is a visual method used to list all possible outcomes of a multi-step probability experiment.
Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 connects tree diagrams with fruit-pair outcomes, sample spaces and same-colour probability.

Chapter 7, The Mathematics of Maybe: Introduction to Probability Class 9, introduces tree diagrams after sample spaces and events. Exercise 7.4 asks students to use tree diagrams for two-step experiments: picking fruits from two baskets and picking pens with replacement. Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions cover all possible outcomes, sample space listing, probability of apple-banana, and probability that two people pick pens of the same colour. The textbook explains that a tree diagram helps list outcomes of a multi-step experiment, where each path from start to end represents one complete outcome.

Key Takeaways

• Tree Diagram: Each complete path gives one possible outcome.
• Sample Space: All outcomes from a tree diagram form the sample space.
• Fruit Basket Probability: Basket A and Basket B choices are paired together.
• Replacement: When a pen is put back, the same colour probabilities apply again.

Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions for Tree Diagrams

Exercise 7.4 contains two questions based on multi-step experiments. The first question uses fruit baskets, and the second uses pen colours with replacement.

Ganita Manjari Class 9 Chapter 7 Exercise 7.4: Fruit Basket Probability

Basket A has three fruits in total, while Basket B has two fruits in total. One fruit is picked from each basket, so every final outcome is written as a pair.

Q1. 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.

Q1(i). Draw a tree diagram showing all possible pairs of fruits.

The tree diagram has 6 final paths because Basket A has 3 fruit choices and Basket B has 2 fruit choices.

Basket A:

Apple, Orange 1, Orange 2

Basket B:

Banana, Mango

Text tree diagram:

Start
→ Apple
→ Banana: (Apple, Banana)
→ Mango: (Apple, Mango)

Start
→ Orange 1
→ Banana: (Orange 1, Banana)
→ Mango: (Orange 1, Mango)

Start
→ Orange 2
→ Banana: (Orange 2, Banana)
→ Mango: (Orange 2, Mango)

Answer:

The tree diagram gives 6 possible fruit pairs.

Q1(ii). List the sample space.

The sample space is the set of all fruit pairs from the tree diagram.

Sample space:

S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)}

Sample size:

n(S) = 6

Answer:

S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)}

Q1(iii). What is the probability of picking one apple and one banana?

The probability of picking one apple and one banana is 1/6.

Favourable outcome:

(Apple, Banana)

Number of favourable outcomes:

1

Total possible outcomes:

6

Use:

Probability = Number of favourable outcomes / Number of possible outcomes

Substitute values:

Probability of apple and banana = 1/6

Answer:

The probability of picking one apple and one banana is 1/6.

Class 9 Maths Chapter 7 Exercise 7.4 Solutions for Pen Colour Outcomes

Question 2 is a two-step experiment with replacement. Since the first pen is put back, the second draw has the same colour options and the same probabilities as the first draw.

Q2. 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.

Q2(i). What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?

The possible colour outcomes are 9 ordered pairs.

Total pens:

3 + 4 + 2 = 9

Colour probabilities for each draw:

P(Red) = 3/9 = 1/3

P(Black) = 4/9

P(Green) = 2/9

Since the pen is replaced, the second draw has the same probabilities.

Text tree diagram:

Start
→ Red (1/3)
→ Red (1/3): (Red, Red)
→ Black (4/9): (Red, Black)
→ Green (2/9): (Red, Green)

Start
→ Black (4/9)
→ Red (1/3): (Black, Red)
→ Black (4/9): (Black, Black)
→ Green (2/9): (Black, Green)

Start
→ Green (2/9)
→ Red (1/3): (Green, Red)
→ Black (4/9): (Green, Black)
→ Green (2/9): (Green, Green)

Sample space:

S = {(Red, Red), (Red, Black), (Red, Green), (Black, Red), (Black, Black), (Black, Green), (Green, Red), (Green, Black), (Green, Green)}

Sample size:

n(S) = 9 colour-pair outcomes

Answer:

The possible colour outcomes are all ordered pairs made from Red, Black and Green.

Q2(ii). Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?

The probability that both pick pens of the same colour is 29/81.

Same colour outcomes:

(Red, Red), (Black, Black), (Green, Green)

Find each probability using the tree diagram.

For two red pens:

P(Red, Red) = 3/9 × 3/9

P(Red, Red) = 9/81

For two black pens:

P(Black, Black) = 4/9 × 4/9

P(Black, Black) = 16/81

For two green pens:

P(Green, Green) = 2/9 × 2/9

P(Green, Green) = 4/81

Now add the three same-colour probabilities:

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

P(same colour) = 29/81

Answer:

The probability that both pick pens of the same colour is 29/81.

The Mathematics of Maybe Introduction to Probability Class 9: Concepts Used in Exercise 7.4

Exercise 7.4 uses tree diagrams to organise multi-step experiments. This helps students see every possible outcome before calculating probabilities.

Tree Diagram Class 9

A tree diagram shows outcomes step by step.

Copy-friendly idea:

One complete branch path = one final outcome

Example:

First toss H, second toss T gives outcome HT.

Sample Space Class 9

The sample space is the set of all final outcomes from the tree diagram.

Copy-friendly result:

S = {all final outcomes}

For two fruits:

S = {(Apple, Banana), (Apple, Mango), (Orange 1, Banana), (Orange 1, Mango), (Orange 2, Banana), (Orange 2, Mango)}

Probability Using Tree Diagram

Probability using a tree diagram is found by multiplying along branches and adding matching final outcomes.

Copy-friendly rules:

Path probability = probability on first branch × probability on second branch

Event probability = sum of probabilities of favourable paths

Fruit Basket Probability

For one fruit from Basket A and one fruit from Basket B:

Total outcomes = 3 × 2

Total outcomes = 6

Favourable apple-banana outcome = 1

So:

P(Apple, Banana) = 1/6

Pen Colour Probability

The pen question uses replacement, so both draws have the same probabilities.

Copy-friendly values:

P(Red) = 3/9

P(Black) = 4/9

P(Green) = 2/9

Same colour probability:

P(same colour) = P(Red, Red) + P(Black, Black) + P(Green, Green)

P(same colour) = 3/9 × 3/9 + 4/9 × 4/9 + 2/9 × 2/9

P(same colour) = 29/81

Class 9 Maths Probability Solutions: How to Draw Tree Diagrams

A tree diagram becomes clear when each step of the experiment is shown separately. Each final branch should show one complete outcome.

Step 1: Write the First-Stage Outcomes

For Basket A:

Apple, Orange 1, Orange 2

For the pen box:

Red, Black, Green

Step 2: Add Second-Stage Outcomes

For each first-stage branch, add every possible second-stage result.

For Basket B:

Banana, Mango

For the second pen draw:

Red, Black, Green

Step 3: List Final Outcomes

Write final results as ordered pairs.

Example:

(Red, Black)

This means:

First person picked Red, second person picked Black.

Step 4: Calculate Probability

Use:

Probability = Number of favourable outcomes / Number of possible outcomes

For unequal branch probabilities, multiply along the branches.

Example:

P(Black, Black) = 4/9 × 4/9

P(Black, Black) = 16/81

Quick Concept Table for Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 7