Mathematics solution NCERT
Class 9 - Chapter 7: The mathematics of maybe: Introduction to Probability
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.
(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?
Given:
- Basket A: 1 Apple (A), 2 Oranges (O)
- Basket B: 1 Banana (B), 1 Mango (M)
(i) Tree Diagram
Basket A
│
├── Apple
│ ├── Banana → (Apple, Banana)
│ └── Mango → (Apple, Mango)
│
├── Orange
│ ├── Banana → (Orange, Banana)
│ └── Mango → (Orange, Mango)
│
└── Orange
├── Banana → (Orange, Banana)
└── Mango → (Orange, Mango)
(ii) Sample Space
S = { (Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango), (Orange, Banana), (Orange, Mango) }
\(Total possible outcomes = 6\)
(iii) Probability of Picking One Apple and One Banana
Favourable outcome:
(Apple, Banana)
\(Number of favourable outcomes = 1\)
\(Total outcomes = 6\)
\(P(Apple and Banana) = \frac{1}{6}\)
Answer: 1/6
Q2: 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?
Given:
- 3 Red pens
- 4 Black pens
- 2 Green pens
\(Total pens = 9\)
Pen is replaced after the first pick.
(i) Possible Outcomes
Let:
- R = Red
- B = Black
- G = Green
Sample Space:
S = { (R,R), (R,B), (R,G), (B,R), (B,B), (B,G), (G,R), (G,B), (G,G) }
Tree Diagram
You
│
├── Red
│ ├── Red → (R,R)
│ ├── Black → (R,B)
│ └── Green → (R,G)
│
├── Black
│ ├── Red → (B,R)
│ ├── Black → (B,B)
│ └── Green → (B,G)
│
└── Green
├── Red → (G,R)
├── Black → (G,B)
└── Green → (G,G)
(ii) Probability That Both Pick Pens of the Same Colour
Since replacement is done:
\(P(Red) = \frac{3}{9} = \frac{1}{3}\)
\(P(Black) = \frac{4}{9}\)
\(P(Green) = \frac{2}{9}\)
Probability both pick Red:
\((\frac{1}{3})(\frac{1}{3}) = \frac{1}{9}\)
Probability both pick Black:
\((\frac{4}{9})(\frac{4}{9}) = \frac{16}{81}\)
Probability both pick Green:
\((\frac{2}{9})(\frac{2}{9}) = \frac{4}{81}\)
Total probability:
\(\frac{1}{9} + \frac{16}{81} + \frac{4}{81}\)
\(= \frac{9}{81} + \frac{16}{81} + \frac{4}{81}\)
\(= \frac{29}{81}\)
Answer:
\(P(Same Colour) = \frac{29}{81}\)
≈ 0.358
\(≈ 35.8%\)
Summary Table
| Question | Answer |
|---|---|
| Q1(ii) | 6 outcomes |
| Q1(iii) | 1/6 |
| Q2(i) | 9 possible colour pairs |
| Q2(ii) | 29/81 ≈ 35.8% |