NCERT Solutions for Class 9 Maths Chapter 7 — Exercise Set 7.4#
A tree diagram lists all outcomes of a multi-step experiment: one branch for each outcome of the first step, then branches from each of those for the second step. Every path from start to end is one outcome.
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.
We label the oranges Orange 1 and Orange 2 so that all fruits in basket A are equally likely to be picked.
(ii) List the sample space.
where A = apple, O = orange, Ba = banana, M = mango. So $n(S) = 3 \times 2 = 6$.
(iii) What is the probability of picking one apple and one banana?
Only one of the 6 equally likely outcomes is (Apple, Banana):
$$P(\text{apple and banana}) = \dfrac{1}{6}$$∴ The probability is $\dfrac{1}{6}$.
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?
There are 9 pens, so on each pick:
$$P(\text{R}) = \dfrac{3}{9}, \quad P(\text{B}) = \dfrac{4}{9}, \quad P(\text{G}) = \dfrac{2}{9}$$The pen is put back, so these stay the same for your friend.
The 9 possible colour outcomes are RR, RB, RG, BR, BB, BG, GR, GB, GG.
(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
Along each path, the probability is the product of the probabilities on its branches. The same-colour paths are RR, BB and GG:
$$P(\text{RR}) = \dfrac{3}{9} \times \dfrac{3}{9} = \dfrac{9}{81}$$$$P(\text{BB}) = \dfrac{4}{9} \times \dfrac{4}{9} = \dfrac{16}{81}$$
$$P(\text{GG}) = \dfrac{2}{9} \times \dfrac{2}{9} = \dfrac{4}{81}$$
Adding:
$$P(\text{same colour}) = \dfrac{9 + 16 + 4}{81}$$$$P(\text{same colour}) = \dfrac{29}{81}$$
∴ The probability is $\dfrac{29}{81}$ ≈ 0.36.
(Counting pens directly gives the same: there are $9 \times 9 = 81$ equally likely ordered pairs of pens, and $3 \times 3 + 4 \times 4 + 2 \times 2 = 29$ of them have matching colours.)