Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions The Mathematics of Maybe: Introduction to Probability

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 7 Exercise 7.4 provide detailed, step-by-step solutions to the questions from The Mathematics of Maybe: Introduction to Probability. This exercise helps students strengthen their understanding of probability and apply different concepts to solve problems based on possible outcomes and events.

The solutions explain each question in a simple and structured manner, making it easier for students to understand the concepts, check their answers and improve their problem-solving skills. A printable PDF is also available for quick revision and exam preparation.

Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions The Mathematics of Maybe: Introduction to Probability

Class 9 Maths Ganita Manjari Chapter 7 Exercise 7.4 Solutions The Mathematics of Maybe: Introduction to Probability

NCERT Solutions for Class 9 Maths Chapter 7 Exercise Set 7.4

Exercise Set 7.4 focuses on tree diagrams for multi-step experiments.

Question 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.

Answer:
Basket A outcomes:

Apple, Orange, Orange

Basket B outcomes:

Banana, Mango

Tree structure:

Apple → Banana, Mango
Orange → Banana, Mango
Orange → Banana, Mango

Since the two oranges are identical in fruit type, the fruit-type pairs are:

Apple-Banana, Apple-Mango, Orange-Banana, Orange-Mango

Final answer:
The tree has branches from Basket A to Apple or Orange, and from each branch to Banana or Mango.

(ii) List the sample space.

Answer:
If only fruit types are recorded:

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

If individual oranges are treated separately, there are 6 equally likely outcomes.

Final answer:
S = {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}

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

Answer:
Probability of apple from Basket A:

= 1/3

Probability of banana from Basket B:

= 1/2

So:

P(apple and banana) = 1/3 × 1/2
= 1/6

Final answer:
1/6

Question 2. A box contains 3 red pens, 4 black pens and 2 green pens. You pick a pen and put it back. Then your friend does the same.

(i) What are the possible outcomes of the pen colours? Draw a tree diagram.

Answer:
Possible colours:

Red, Black, Green

Since the pen is replaced, both picks have the same possible colour outcomes.

Sample space:

S = {(R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)}

Tree structure:

Red → Red, Black, Green
Black → Red, Black, Green
Green → Red, Black, Green

Final answer:
S = {(R, R), (R, B), (R, G), (B, R), (B, B), (B, G), (G, R), (G, B), (G, G)}

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

Answer:
Total pens = 3 + 4 + 2 = 9

P(Red) = 3/9 = 1/3
P(Black) = 4/9
P(Green) = 2/9

Since the pen is replaced, the events are independent.

Probability of same colour:

= P(R, R) + P(B, B) + P(G, G)

= (3/9 × 3/9) + (4/9 × 4/9) + (2/9 × 2/9)

= 9/81 + 16/81 + 4/81
= 29/81

Final answer:
29/81

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