Study SimplifyStudy Simplify

Class 12 Statistics Notes · GSEB

Mathematical Definition of Probability

2.1.3 Mathematical Definition of Probability — Understand equiprobable events and favourable outcomes, apply P(A) = m/n on a live probability scale, and learn the important results and limitations. GSEB Class 12 Statistics notes with eight solved illustrations.

Last updated: 25 Aug 2026

Practice Sums →

Notes

Equiprobable Events and Favourable Outcomes

Equiprobable Events
If there is no apparent reason to believe that out of one or more events of a random experiment, any one event is more or less likely to occur than the other events, then the events are called equiprobable events.

Example 1 — the manufacturer's machines

A manufacturer has two machines M₁ and M₂ that produce the same number of items during a day; lots are made by properly mixing the day's production of both machines. An item randomly selected from such a lot was made on machine M₁ or machine M₂ — elementary events which are equiprobable.

Example 2 — why sector size matters (⭐): wheels A and B, both marked 1, 2, 3, are rotated by hand and the number against the pointer is noted.

123Equiprobable ✓
123Not equiprobable ✗

All three numbers on wheel A are equiprobable (equal sectors); the numbers 1, 2 and 3 on wheel B are not equiprobable (unequal sectors). ⭐

Favourable Outcomes
If some outcomes out of all the elementary outcomes in the sample space of a random experiment indicate the occurrence of a certain event A, then these outcomes are called the favourable outcomes of the event A.

Example: a card is drawn from a pack of 52 cards. If event A denotes that the card drawn is a face card, then the set of favourable outcomes is A = {S_K, D_K, C_K, H_K, S_Q, D_Q, C_Q, H_Q, S_J, D_J, C_J, H_J} — 12 outcomes are favourable for event A. ⭐

The Mathematical (Classical) Definition

Suppose there are total n outcomes in the finite sample space of a random experiment which are mutually exclusive, exhaustive and equiprobable. If m outcomes among them are favourable for an event A, then the probability of the event A is m/n. The probability of event A is denoted by P(A).

Mathematical (classical) definition

P(A)=Favourable outcomes of event ATotal mutually exclusive, exhaustive, equiprobable outcomes=mnP(A) = \frac{\text{Favourable outcomes of event } A}{\text{Total mutually exclusive, exhaustive, equiprobable outcomes}} = \frac{m}{n}

Both the numbers m (≥ 0) and n (> 0) are integers and m ≤ n. ⭐

n cannot be zero and cannot be infinity.

The mathematical definition of probability is also called the classical definition. ⭐

Assumptions of the mathematical definition⭐ Section C Q5

1

The number of outcomes in the sample space of the random experiment is finite.

2

The number of outcomes in the sample space of the random experiment is known.

3

The outcomes in the sample space of the random experiment are equi-probable.

The formula only applies when all three assumptions hold — the statistical definition (topic 6) exists because they often do not.

The 0–1 probability scale

P(A) = m/n
0 · P(φ) = 0P(A) = 3/81 · P(U) = 1

Favourable outcomes m = 3

Total outcomes n = 8

37.5%

P(A) = 3/8

= 3/8

= 0.375

= 37.5%

Important Results (accepted without proof)

The 0–1 probability scale

P(A) = m/n
0 · P(φ) = 0P(A) = 3/81 · P(U) = 1

Complement segment: P(A′) = 1 − P(A) = 5/8 — the rose-striped part.

Favourable outcomes m = 3

Total outcomes n = 8

37.5%

P(A) = 3/8

= 3/8

= 0.375

= 37.5%

These results are accepted without proof and are the “formula sheet” for every numeric question in Exercise 1.2 and the final Exercise (Sections B and C).

Range of probability ⭐

0≤P(A)≤10 \le P(A) \le 1

Impossible event ⭐

P(ϕ)=0P(\phi) = 0

Certain event ⭐

P(U)=1P(U) = 1

Complementary event ⭐

P(A′)=1−P(A)P(A') = 1 - P(A)

Subset result

A⊂B⇒P(A)≤P(B), P(B−A)=P(B)−P(A)A \subset B \Rightarrow P(A) \le P(B),\ P(B - A) = P(B) - P(A)

Intersection is smaller

P(A∩B)≤P(A), P(A∩B)≤P(B)P(A \cap B) \le P(A),\ P(A \cap B) \le P(B)

Union is bigger

P(A)≤P(A∪B), P(B)≤P(A∪B)P(A) \le P(A \cup B),\ P(B) \le P(A \cup B)

Neither A nor B ⭐

P(A′∩B′)=P(A∪B)′=1−P(A∪B)P(A' \cap B') = P(A \cup B)' = 1 - P(A \cup B)

Not both ⭐

P(A′∪B′)=P(A∩B)′=1−P(A∩B)P(A' \cup B') = P(A \cap B)' = 1 - P(A \cap B)

Only A happens ⭐

P(A−B)=P(A∩B′)=P(A)−P(A∩B)P(A - B) = P(A \cap B') = P(A) - P(A \cap B)

Only B happens ⭐

P(B−A)=P(A′∩B)=P(B)−P(A∩B)P(B - A) = P(A' \cap B) = P(B) - P(A \cap B)

The master chain ⭐

0≤P(A∩B)≤P(A)≤P(A∪B)≤P(A)+P(B)0 \le P(A \cap B) \le P(A) \le P(A \cup B) \le P(A) + P(B)

Solved Illustrations 11–18

Solved Example

Problem

Illustration 11 (⭐ board pattern): two balanced coins are tossed simultaneously. Find the probability that (1) one head and one tail appear (2) at least one head appears.

Solution

P(A) = 1/2; P(B) = 3/4.

Step 1 — A₁: sum is 7

1/6

U = {(i, j); i, j = 1, 2, 3, 4, 5, 6}, n = 36.

A₁ = sum is 7 → m = 6: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

P(A₁) = 6/36 = 1/6 ⭐

Solved Example

Problem

Illustration 13 (⭐ board — the word RUTVA): 5 letters R, U, T, V, A are arranged in all possible ways. Find the probability that R appears in the first place.

Solution

Required probability = 1/5.

Solved Example

Problem

Illustration 14: 4 male and 2 female employees are sent one by one for training. Find the probability that the two female employees go successively.

Solution

Required probability = 1/3.

Solved Example

Problem

Illustration 15 (⭐ board — 53 Thursdays in a leap year): find the probability that a leap year has 53 Thursdays.

Solution

Required probability = 2/7.

Solved Example

Problem

Illustration 16: a bank cash department has 2 officers, 3 clerks and 2 peons. A committee of 2 is formed by the method of lottery. Find the probability that (1) both are peons (2) both are clerks (3) one is an officer and one is a clerk.

Solution

(1) 1/21 (2) 1/7 (3) 2/7.

Solved Example

Problem

Illustration 17: a box contains 20 items of which 10% are defective (2 defective, 18 non-defective). Three items are selected at random. Find the probability that (1) two are defective (2) two are non-defective (3) all three are non-defective.

Solution

(1) 3/190 (2) 51/190 (3) 68/95.

Solved Example

Problem

Illustration 18 (⭐ board — Kathan's chits): out of 10 chits, 3 are prize-eligible and 7 are not. Kathan selects two chits at random. Find the probability that Kathan gets the prize.

Solution

Required probability = 8/15 (complement shortcut).

Limitations of the Mathematical Definition

The classical definition fails whenever any of its three assumptions breaks — and in one deeper way too. ⭐ Section C Q6
1

The probability of an event cannot be found by this definition if there are infinite outcomes in the sample space of a random experiment.

2

The probability of an event cannot be found by this definition if the total number of outcomes in the sample space of a random experiment are not known.

3

The probability of an event cannot be found by this definition if the elementary outcomes in the sample space of a random experiment are not equi-probable.

4

The word "equi-probable" is mentioned in the mathematical definition. Equi-probable events are events with the same probability — so the word probability is used inside the definition of probability (the definition is circular).

Practice Checkpoint

Your turn. Work each question in your notebook — type only the final answer here.

Type 3 Q2

3 marks

A balanced coin is tossed three times. Find the probability of: (1) getting all three heads (2) not getting a single head (3) getting at least one head.

Try the experiment:

Type 3 Q4

1 mark

A number is randomly selected from the first 100 natural numbers. Find the probability that the number is divisible by 7.

Try the experiment:

From 1 to 100 (100 numbers)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100

Type 3 Q24

2 marks

Find the probability that there will be 5 Mondays in the month of February of a leap year.

Finish the rest of this question type → Practice Page· then return here for the next topic

Key Takeaways

Key Takeaways

  • Equiprobable events have no apparent reason to be more or less likely; favourable outcomes are the outcomes that make an event occur.
  • P(A) = m/n where n = total mutually exclusive, exhaustive, equiprobable outcomes and m = favourable outcomes; n is never 0 or infinite; the definition is also called the classical definition.
  • The classical definition needs a finite, known, equiprobable sample space — its three assumptions.
  • 0 ≤ P(A) ≤ 1, P(φ) = 0, P(U) = 1, P(A′) = 1 − P(A).
  • P(A − B) = P(A) − P(A ∩ B) and the chain 0 ≤ P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B).
  • Complements are the examiner's favourite shortcut: P(A) = 1 − P(A′) (Kathan's chits).