1. Chapter at a Glance
In Class 10 you calculated probabilities by counting equally likely outcomes. Now we put probability on a firm foundation using the language of sets. Outcomes form a sample space; events are subsets; "A or B" is a union and "A and B" an intersection. The modern axiomatic approach, given by the Russian mathematician A. N. Kolmogorov in 1933, defines probability through three simple rules from which everything else follows — including the addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B). This chapter is the base for conditional probability and distributions in Class 12.
- Random experiments, outcomes and sample spaces.
- Events: simple, compound, sure, impossible; complement.
- Algebra of events: A or B, A and B, A but not B.
- Mutually exclusive and exhaustive events.
- Axiomatic definition of probability; equally likely outcomes.
- Addition rule and probability of "not A".
2. Sample Space
A random experiment has more than one possible outcome, and the outcome cannot be predicted in advance. The set of all possible outcomes is the sample space S; each element is a sample point.
| Experiment | Sample space | n(S) |
|---|---|---|
| Toss a coin twice | {HH, HT, TH, TT} | 4 |
| Roll a die | {1, 2, 3, 4, 5, 6} | 6 |
| Toss a coin, then roll a die if it shows H | {H1, …, H6, T} | 7 |
| Draw 2 balls from 3 red (R₁, R₂, R₃) — unordered | {R₁R₂, R₁R₃, R₂R₃} | 3 |
Note the third row: sample points need not be equally likely.
3. Events
An event is any subset of S. An event E occurs if the outcome belongs to E.
| Type | Meaning |
|---|---|
| Impossible event | ∅ |
| Sure event | S |
| Simple event | Exactly one sample point |
| Compound event | More than one sample point |
| Complement A′ (not A) | S − A |
| In words | Set notation |
|---|---|
| A or B (at least one) | A ∪ B |
| A and B (both) | A ∩ B |
| A but not B | A − B = A ∩ B′ |
| Not A | A′ |
| Neither A nor B | A′ ∩ B′ = (A ∪ B)′ |
- A and B are mutually exclusive if A ∩ B = ∅ (cannot happen together).
- E₁, E₂, …, Eₙ are exhaustive if E₁ ∪ E₂ ∪ … ∪ Eₙ = S (one of them must happen).
- If both, they form a partition of S.
A die is rolled. A: "even number", B: "odd number", C: "number ≤ 3". Which pairs are mutually exclusive? Are A and B exhaustive?
Solution: A = {2, 4, 6}, B = {1, 3, 5}, C = {1, 2, 3}. A ∩ B = ∅ → A, B mutually exclusive; A ∩ C = {2}, B ∩ C = {1, 3} — not exclusive. A ∪ B = S → A and B are exhaustive.
4. Axiomatic Probability
A probability P assigns to each event E a number P(E) such that:
- P(E) ≥ 0 for every event E.
- P(S) = 1.
- If E and F are mutually exclusive, P(E ∪ F) = P(E) + P(F).
Consequences: P(∅) = 0; 0 ≤ P(E) ≤ 1; for sample points ωᵢ, ΣP(ωᵢ) = 1 and P(E) = sum of P(ωᵢ) over ωᵢ ∈ E.
For a finite sample space with equally likely outcomes this gives the classical formula P(E) = n(E)/n(S).
A coin is weighted so that H is twice as likely as T. Find P(H) and P(T).
Solution: P(H) = 2P(T); P(H) + P(T) = 1 → 3P(T) = 1 → P(T) = 1/3, P(H) = 2/3.
5. Addition Rule and Complement
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
- If A, B mutually exclusive: P(A ∪ B) = P(A) + P(B).
- P(A′) = 1 − P(A)
- P(A − B) = P(A) − P(A ∩ B).
- P(neither A nor B) = 1 − P(A ∪ B).
A card is drawn from a deck of 52. Find P(a king or a heart).
Solution: P(K) = 4/52, P(H) = 13/52, P(K ∩ H) = 1/52. P = (4 + 13 − 1)/52 = 4/13.
If P(A) = 0.42, P(B) = 0.48 and P(A ∩ B) = 0.16, find P(not A), P(not B), P(A or B) and P(neither).
Solution: P(A′) = 0.58; P(B′) = 0.52; P(A ∪ B) = 0.42 + 0.48 − 0.16 = 0.74; neither = 0.26.
From a group of 4 men and 3 women, 2 people are chosen at random. Find P(both women) and P(at least one man).
Solution: n(S) = ⁷C₂ = 21. Both women ³C₂ = 3 → 1/7. At least one man = 1 − 1/7 = 6/7.
6. Solved Problems (Board Pattern)
In a lottery, a person chooses 6 different numbers from 1 to 20. If these match the 6 numbers drawn, they win. Find the probability of winning.
Solution: n(S) = ²⁰C₆ = 38760. Only one winning set → 1/38760.
Can P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.6 be valid?
Solution: No — P(A ∩ B) cannot exceed P(A) = 0.5, because A ∩ B ⊆ A.
A committee of 4 is chosen at random from 6 boys and 4 girls. Find P(exactly 2 girls).
Solution: ⁴C₂ × ⁶C₂ / ¹⁰C₄ = 6 × 15/210 = 3/7.
7. Master Formula Table
| Rule | Formula |
|---|---|
| Classical | P(E) = n(E)/n(S) |
| Axioms | P(E) ≥ 0; P(S) = 1; additivity for disjoint events |
| Complement | P(A′) = 1 − P(A) |
| Addition rule | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) |
| A but not B | P(A) − P(A ∩ B) |
| Exactly one of A, B | P(A) + P(B) − 2P(A ∩ B) |
| Three events | ΣP(A) − ΣP(A∩B) + P(A∩B∩C) |
8. Master Summary Table
| Concept | Key idea | Exam focus |
|---|---|---|
| Sample space | All outcomes | Write S for compound experiments |
| Events | Subsets of S | Describe A ∪ B, A ∩ B, A′ |
| Mutually exclusive | A ∩ B = ∅ | Identify pairs |
| Exhaustive | Union = S | Partition of S |
| Axioms | Kolmogorov | Check valid assignments |
| Addition rule | Subtract overlap | "or" questions with cards, dice |
Chart — probabilities from Worked Example 4:
9. Mind Map
- Outcomes
- Sample points
- ∪, ∩, ′
- Exclusive, exhaustive
- P ≥ 0, P(S) = 1
- Additivity
- Addition rule
- Complement
10. Tips and Tricks
- Translate words into set notation before calculating.
- Use combinations (ⁿCᵣ) to count when selections are unordered.
- "At least one" → 1 − P(none).
- Adding P(A) + P(B) for overlapping events without subtracting P(A ∩ B).
- Confusing "mutually exclusive" with "exhaustive".
- Assuming all sample points are equally likely when they are not.
11. Important One-liners
- An event is a subset of the sample space.
- Mutually exclusive events cannot occur together.
- P(A ∩ B) ≤ P(A) and P(A ∩ B) ≤ P(B).
- P(A ∪ B) ≤ P(A) + P(B).
- The axiomatic approach was given by Kolmogorov (1933).
12. Exam-pattern Questions with Detailed Answers
The number of sample points when 3 coins are tossed is:
- 3
- 6
- 8
- 9
Show answer
2³ = 8.
Answer: (C)
If P(A) = 2/11, find P(not A).
Show answer
1 − 2/11 = 9/11.
Answer: 9/11
A die is rolled. Are the events "getting a prime" and "getting an even number" mutually exclusive?
Show answer
{2, 3, 5} ∩ {2, 4, 6} = {2} ≠ ∅ → not mutually exclusive.
Answer: No
Write the sample space: a coin is tossed; if it shows T, a die is thrown; if H, the coin is tossed again.
Show answer
{HH, HT, T1, T2, T3, T4, T5, T6}.
Answer: 8 sample points as listed
If P(A) = 3/5 and P(B) = 1/5, find P(A or B) when A and B are mutually exclusive.
Show answer
3/5 + 1/5 = 4/5.
Answer: 4/5
Two dice are thrown. Find P(the sum is 9 or a doublet occurs).
Show answer
Sum 9: 4 outcomes; doublets: 6; overlap: none (9 is odd). P = 10/36 = 5/18.
Answer: 5/18
In a class, 40% study Mathematics, 30% Biology and 10% both. A student is chosen at random. Find P(studies Mathematics or Biology).
Show answer
0.4 + 0.3 − 0.1 = 0.6.
Answer: 0.6
A letter is chosen at random from the word ASSASSINATION. Find P(it is a vowel).
Show answer
13 letters; vowels A, A, I, A, I, O = 6. P = 6/13.
Answer: 6/13
Check whether the assignment P(ω₁) = 0.1, P(ω₂) = 0.01, P(ω₃) = 0.05, P(ω₄) = 0.03, P(ω₅) = 0.01, P(ω₆) = 0.2, P(ω₇) = 0.6 is valid for S = {ω₁, …, ω₇}.
Show answer
All values ≥ 0 and their sum = 1.00. Valid.
Answer: Valid
Four cards are drawn at random from a deck of 52. Find the probability of getting exactly 3 diamonds and 1 spade.
Show answer
¹³C₃ × ¹³C₁ / ⁵²C₄ = 286 × 13/270725 = 3718/270725 ≈ 0.0137.
Answer: 3718/270725
If A and B are events with P(A) = 1/4, P(B) = 1/2 and P(A ∪ B) = 1/2, find P(A ∩ B), P(A ∩ B′) and P(A′ ∩ B′).
Show answer
P(A ∩ B) = 1/4 + 1/2 − 1/2 = 1/4. P(A ∩ B′) = 1/4 − 1/4 = 0. P(A′ ∩ B′) = 1 − 1/2 = 1/2.
Answer: 1/4, 0, 1/2
A bag has 9 discs: 4 red, 3 blue, 2 yellow. One disc is drawn. Find P(it is red or yellow) and P(it is not blue).
Show answer
Red or yellow = 6/9 = 2/3. Not blue = 1 − 3/9 = 2/3 (the same event!).
Answer: 2/3 and 2/3
13. Quick Revision Summary
- Sample space S = all outcomes; events = subsets of S.
- or = ∪, and = ∩, not = complement; mutually exclusive ⇔ A ∩ B = ∅.
- Axioms: P(E) ≥ 0, P(S) = 1, additivity for disjoint events.
- Equally likely: P(E) = n(E)/n(S); count with ⁿCᵣ.
- P(A ∪ B) = P(A) + P(B) − P(A ∩ B); P(A′) = 1 − P(A).