Class 9 - Chapter 7: The mathematics of maybe: Introduction to Probability

(i) The probability of an impossible event is ______.

Answer: 0

(ii) The set of all possible outcomes of a random experiment is called the ______.

Answer: Sample Space

(iii) The probability of an event that is certain to happen is ______.

Answer: 1

(iv) Tossing a fair coin has a probability of ______ for getting heads.

Answer: 1/2

Given:

Total students surveyed = 50

Students who like football = 15

The number of students who like football is 15, and the relative frequency is:

Relative Frequency = 15 / 50

= 3 / 10

= 0.3

Answer:

Relative Frequency = 15/50 = 3/10 = 0.3

Experiment	Equally Likely?	Reason
A driver attempts to start a car.	No	The car is usually more likely to start than not start.
Tossing a fair coin once.	Yes	Head and Tail each have probability 1/2.
Rolling a fair 6-sided die.	Yes	Each number from 1 to 6 has probability 1/6.
Choosing a marble from a bag with 3 red and 7 blue marbles.	No	Blue is more likely than red.
A baby is born. It is a boy or a girl.	Approximately Yes	Both outcomes are nearly equally likely.

(i) Two coins are tossed. Probability of getting at least one head.

Sample Space:

S = {HH, HT, TH, TT}

Favourable outcomes:

{HH, HT, TH}

P(At least one Head) = 3/4

Answer: 3/4

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(ii) A card numbered 1 to 10 is drawn. Probability of getting an even number.

Sample Space:

{1,2,3,4,5,6,7,8,9,10}

Even numbers:

{2,4,6,8,10}

P(Even Number) = 5/10 = 1/2

Answer: 1/2

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(iii) A die is rolled. Probability of getting a number greater than 4.

Sample Space:

{1,2,3,4,5,6}

Numbers greater than 4:

{5,6}

P(Number > 4) = 2/6 = 1/3

Answer: 1/3

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(iv) A bag contains 3 red, 2 blue and 1 green ball. Probability that it is not red.

Total balls = 6

Not red balls = 2 + 1 = 3

P(Not Red) = 3/6 = 1/2

Answer: 1/2

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(v) Three coins are tossed. Probability of getting exactly two heads.

Sample Space:

{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Exactly two heads:

{HHT, HTH, THH}

P(Exactly Two Heads) = 3/8

Answer: 3/8

Given:

• Strawberry
• Lemon
• Mint

Total candies = 3

Favourable outcomes = 1

P(Strawberry) = 1/3

Answer: 1/3

Given:

• Shirts: Red, Blue
• Pants: Jeans, Khakis, Shorts

Total possible outfits = 2 × 3 = 6

Shirt	Pants
Red	Jeans
Red	Khakis
Red	Shorts
Blue	Jeans
Blue	Khakis
Blue	Shorts

Total cases = 1000

Distance (km)	Cases
Less than 4000	20
4001 to 9000	210
9001 to 14000	325
More than 14000	445

Letters available:

P, E, A, C, E

Total cards = 5

(i) Probability of drawing P, E or C

Favourable cards:

P, E, E, C

Number of favourable cards = 4

P(P or E or C) = 4/5

Answer: 4/5

---

(ii) Probability that it is not an E

Not E cards:

P, A, C

Number of favourable cards = 3

P(Not E) = 3/5

Answer: 3/5

Sample Space:

S = {1,2,3,4,5,6,7,8}

Total outcomes = 8

(i) Probability of getting 8

P(8) = 1/8

Answer: 1/8

---

(ii) Probability of getting an odd number

Odd numbers = {1,3,5,7}

P(Odd) = 4/8 = 1/2

Answer: 1/2

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(iii) Probability of getting a number greater than 2

{3,4,5,6,7,8}

P = 6/8 = 3/4

Answer: 3/4

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(iv) Probability of getting a number less than 9

All outcomes satisfy this condition.

P = 8/8 = 1

Answer: 1

---

(v) Probability of getting a multiple of 3

Multiples of 3 = {3,6}

P = 2/8 = 1/4

Answer: 1/4

Given:

4 Red balls and 5 Blue balls

Total balls = 9

Tree Diagram

Start
│
├── Red (4/9)
│     ├── Red (3/8) → RR
│     └── Blue (5/8) → RB
│
└── Blue (5/9)
      ├── Red (4/8) → BR
      └── Blue (4/8) → BB

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(i) Probability of Red then Blue

P(RB) = (4/9) × (5/8)

= 20/72

= 5/18

Answer: 5/18

---

(ii) Probability of drawing 2 Blue balls

P(BB) = (5/9) × (4/8)

= 20/72

= 5/18

Answer: 5/18

Event with probability 0:

Getting a sum of 13 when two 6-sided dice are thrown.

P(Sum = 13) = 0

Reason: The maximum possible sum is 12.

Outcome with probability 1:

Getting a sum between 2 and 12.

P(2 ≤ Sum ≤ 12) = 1

Reason: Every outcome gives a sum between 2 and 12.

(i) Two dice are rolled. Probability that the sum is a prime number greater than 5.

Total outcomes = 36

Prime numbers greater than 5:

7, 11

Sum = 7:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

Number of outcomes = 6

Sum = 11:

(5,6), (6,5)

Number of outcomes = 2

Total favourable outcomes = 8

P = 8/36 = 2/9

Answer: 2/9

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(ii) A bag contains 4 red, 3 green and 2 blue balls. Two balls are drawn without replacement. Probability that both are of different colours.

Total balls = 9

Total ways to select 2 balls:

⁹C₂ = 36

Same colour pairs:

Red = ⁴C₂ = 6

Green = ³C₂ = 3

Blue = ²C₂ = 1

Total same colour pairs = 10

Different colour pairs = 36 − 10 = 26

P(Different Colours) = 26/36 = 13/18

Answer: 13/18

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(iii) Three coins are tossed. Probability that the first coin shows heads and exactly two heads occur in total.

Sample Space:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Favourable outcomes:

HHT, HTH

Number of favourable outcomes = 2

P = 2/8 = 1/4

Answer: 1/4

---

(iv) A four-digit number is formed using 1, 2, 3 and 4 without repetition. Probability that the number is even.

Total four-digit numbers:

4! = 24

Even numbers end in 2 or 4.

Ending with 2:

3! = 6

Ending with 4:

3! = 6

Total favourable numbers:

6 + 6 = 12

P(Even) = 12/24 = 1/2

Answer: 1/2

---

(v) Multiple-choice Test

Each question has:

P(Correct) = 1/4

P(Wrong) = 3/4

Exactly 2 correct answers out of 3 questions:

Possible arrangements:

CCW, CWC, WCC

P = 3 × (1/4)² × (3/4)

= 9/64

Answer: 9/64

(i) With Replacement

Sample Space:

{
(1,1),(1,2),(1,3),(1,4),
(2,1),(2,2),(2,3),(2,4),
(3,1),(3,2),(3,3),(3,4),
(4,1),(4,2),(4,3),(4,4)
}

Size = 16

---

(ii) Without Replacement

{
(1,2),(1,3),(1,4),
(2,1),(2,3),(2,4),
(3,1),(3,2),(3,4),
(4,1),(4,2),(4,3)
}

Size = 12

---

(iii) Sizes of Sample Spaces

Experiment	Size
With Replacement	16
Without Replacement	12

Coin outcomes:

H, T

Cards:

1, 2, 3, 4, 5, 6

Sample Space:

{
(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),
(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)
}

Total outcomes = 12

The number of heads can be:

0, 1, 2, or 3

Therefore the correct sample space is:

{0, 1, 2, 3}

Answer: Option (iv)

Option	Valid?	Reason
{1,2,3}	No	Missing 0 heads
{0,1,2}	No	Missing 3 heads
{0,1,2,3,4}	No	4 heads impossible
{0,1,2,3}	Yes	All possible outcomes included

Given:

• Rectangle = 3 m × 2 m
• Diameter of circle = 1 m
• Radius = 0.5 m

Step 1: Area of Rectangle

Area = Length × Breadth

= 3 × 2

= 6 m²

Step 2: Area of Circle

Area = πr²

= π(0.5)²

= π/4

≈ 0.785 m²

Step 3: Probability

P(Lands inside circle) = Area of Circle / Area of Rectangle

= (π/4) / 6

= π/24

≈ 0.131

Answer:

P = π/24 ≈ 0.131

≈ 13.1%