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Class 9 Ganita ManjariClass 9 Maths Solutions | 2026–27
Chapter 7: The Mathematics of Maybe: Introduction to ProbabilityPage 169All 2 Questions on This Page

Class 9 Maths Chapter 7 Exercise Set 7.4 Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 7 Exercise Set 7.4: Tree Diagrams (Page 169). Covers drawing tree diagrams for multi-step experiments, finding sample spaces, and calculating path probabilities.

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Key Concepts for Exercise Set 7.4

Jump to Question 1
1

Multi-Step Experiment

A multi-step experiment involves two or more trials performed in sequence.

2

Tree Diagram

A tree diagram is a branching diagram used to show all possible outcomes of a multi-step experiment.

3

Complete Path

Each complete path from the beginning to the end of a tree diagram represents one possible outcome of the combined experiment.

4

Path Probability

For independent steps, the probability of a complete path is found by multiplying the probabilities along that path.

P(A and B) = P(A) × P(B)

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) Draw a tree diagram showing all possible pairs of fruits:

Basket A contains:

1 Apple and 2 Oranges

Therefore,
P(Apple) = 13
P(Orange) = 23

Basket B contains:

1 Banana and 1 Mango

Therefore,
P(Banana) = 12
P(Mango) = 12

The tree diagram shows:

Basket A → Apple or Orange

Then from each branch:

Basket B → Banana or Mango

The four possible fruit-type pairs are:

(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)

(ii) List the sample space:
Therefore, the sample space is:

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

These represent the possible pairs of fruit types.

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

The favourable outcome is:

(Apple, Banana)

Using the branch probabilities,

P(Apple and Banana) = P(Apple) × P(Banana)
= 13 × 12
= 16
Therefore,
16

is the probability of picking one apple and one banana.

Answer:(i) See tree diagram above | (ii) S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)} | (iii) P(Apple and Banana) = 16

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 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?

Solution
(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
Given:
Number of red pens = 3
Number of black pens = 4
Number of green pens = 2

Total number of pens:

3 + 4 + 2 = 9

Since the pen is put back after the first pick, the number and composition of pens remain unchanged for the second pick.

Therefore,
P(R) = 39 = 13
P(B) = 49
P(G) = 29

The possible colour outcomes for the pair are:

RR, RB, RG, BR, BB, BG, GR, GB, GG

Thus, there are 9 possible colour pairs.
(ii) Probability that both pick pens of the same colour:

The favourable outcomes are:

RR, BB, GG
Therefore,

P(same colour) = P(RR) + P(BB) + P(GG)

Using the branch probabilities,

P(RR) = 13 × 13
= 19 = 981
P(BB) = 49 × 49
= 1681
P(GG) = 29 × 29
= 481
Therefore,

P(same colour) = 1/9 + 16/81 + 4/81

= 981 + 1681 + 481
= 2981
Hence,
19 + 1681 + 481 = 2981
Therefore,

the probability that both you and your friend pick pens of the same colour is:

2981
Answer:(i) 9 colour pairs (RR, RB, RG, BR, BB, BG, GR, GB, GG) with tree diagram above | (ii) P(same colour) = 2981

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