Class 9 Mathematics Ganita Manjari Part 1 NCERT Solutions
Chapter-7: The Mathematics of Maybe: Introduction to Probability

Exercise Set 7.4

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.
(ii) List the sample space.
(iii) What is the probability of picking one apple and one banana?

Solution:
Basket A contains: Apple, Orange1, Orange2
Basket B contains: Banana, Mango

(i) Tree Diagram:

(ii) Sample Space

Since the two oranges are identical, the sample space (by fruit type) is:
                   S={(Apple, Banana),(Apple, Mango),(Orange, Banana),(Orange, Mango)}

There are 6 equally likely outcomes if each fruit is considered individually:
      S = {(Apple, Banana),(Apple, Mango),(Orange1​,Banana),(Orange1​,Mango),(Orange2​,Banana),(Orange2​,Mango)}

(iii) Probability of picking one apple and one banana
From Basket A:
                  P(Apple) = \frac{1}{3}
From Basket B:
                  P(Banana) = \frac{1}{2}
Therefore,
                P(Apple and Banana) = \frac{1}{3}×\frac{1}{2}=\frac{1}{6}

2. 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:
The box contains:
                     Red pens (R) = 3
                     Black pens (B) = 4
                     Green pens (G) = 2
Total number of pens = 3+4+2= 9

(i) Tree Diagram

(ii) Probability that both pick pens of the same colour

Both pick the same colour in the following outcomes:
                E = {RR, BB, GG}
Now,
           P(RR) = \frac{3}{9}×\frac{3}{9}=\frac{9}{81}

           P(BB) = \frac{4}{9}×\frac{4}{9}=\frac{16}{81}

           P(GG) = \frac{2}{9}×\frac{2}{9}=\frac{4}{81}

Therefore,
           P(Same colour) = P(RR) + P(BB) + P(GG)

                                        = \frac{9}{81}+\frac{16}{81}+\frac{4}{81}

                                        = \frac{29}{81}