Quick Summary
Probability quantifies how likely an event is to occur. Class 11 introduces the axiomatic approach, the classical model, and key theorems — addition, multiplication, and Bayes'. Probability is the foundation of statistics, and appears in every entrance exam.
Basic Terminology
| Term | Definition |
|---|---|
| Random experiment | An experiment whose outcome cannot be predicted with certainty |
| Sample space (S) | Set of all possible outcomes |
| Event (A) | Subset of the sample space |
| Sure event | S itself; P(S) = 1 |
| Impossible event | ∅; P(∅) = 0 |
| Complementary event | A' = S − A; P(A') = 1 − P(A) |
| Mutually exclusive | A ∩ B = ∅ |
| Exhaustive events | A₁ ∪ A₂ ∪ … ∪ Aₙ = S |
Classical Probability
P(A) = Number of favourable outcomes / Total number of outcomes = n(A)/n(S)
Conditions: outcomes must be equally likely and sample space must be finite.
Range: 0 ≤ P(A) ≤ 1
Addition Theorem
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
For three events: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A∩B) − P(B∩C) − P(A∩C) + P(A∩B∩C)
For mutually exclusive events: P(A ∪ B) = P(A) + P(B)
Multiplication Theorem
| **P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B)** |
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For independent events: P(A ∩ B) = P(A) × P(B)
Conditional Probability
| **P(A | B) = P(A ∩ B) / P(B)**, where P(B) > 0 |
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| Independence check: A and B are independent ↔ P(A | B) = P(A) ↔ P(A∩B) = P(A)·P(B) |
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Total Probability Theorem
If B₁, B₂, …, Bₙ are mutually exclusive and exhaustive events:
| **P(A) = Σ P(A | Bᵢ) × P(Bᵢ)** |
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Bayes' Theorem
| **P(Bᵢ | A) = P(A | Bᵢ)·P(Bᵢ) / Σ P(A | Bⱼ)·P(Bⱼ)** |
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Used to update probabilities when new information (the occurrence of A) is available.
Worked Examples
Example 1: A card is drawn from a standard 52-card deck. Find P(King or Heart). P(King) = 4/52, P(Heart) = 13/52, P(King ∩ Heart) = 1/52. P(King ∪ Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13
Example 2: Two dice are rolled. Find P(sum = 7). Favourable: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes. Total: 36. P = 6/36 = 1/6
Example 3: In a bag: 3 red, 4 blue balls. Two are drawn without replacement. Find P(both red).
| P = P(R₁) × P(R₂ | R₁) = (3/7) × (2/6) = 6/42 = 1/7 |
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Example 4 (Bayes): A factory has machines A, B, C producing 50%, 30%, 20% of items. Defect rates are 2%, 3%, 5%. An item is defective. Find P(from machine A). P(D) = 0.5×0.02 + 0.3×0.03 + 0.2×0.05 = 0.01 + 0.009 + 0.01 = 0.029
| P(A | D) = (0.5×0.02)/0.029 = 0.01/0.029 ≈ 0.345 |
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Important Questions
One Mark
- Define a sample space. Give one example.
- State the addition theorem for mutually exclusive events.
- What is the probability of an impossible event?
- If P(A) = 0.4 and P(B) = 0.3 and A, B are independent, find P(A ∩ B).
Short Answer (3 marks)
- Two cards are drawn from a deck. Find P(both aces) without replacement.
- If P(A) = 1/3, P(B) = 1/4, P(A ∪ B) = 1/2, find P(A ∩ B) and P(A|B).
- If A and B are independent, P(A) = 0.5, P(B) = 0.4, find P(A ∪ B).
Long Answer (5 marks)
- A bag has 4 white and 6 black balls. Two balls are drawn one by one without replacement. Find the probability distribution and expected value.
- State and prove the Bayes' theorem. Apply it to a problem involving two boxes of balls.
Common Mistakes
- Adding probabilities for non-mutually-exclusive events without subtracting the intersection.
- Conditional probability: computing P(A|B) = P(A)/P(B) instead of P(A∩B)/P(B).
- Independence vs mutually exclusive: independent events CAN occur together; mutually exclusive events CANNOT. They are different concepts.
- Bayes: not correctly identifying the prior probabilities P(Bᵢ) from the problem.
Board Exam Tips
- Addition theorem problems (with Venn diagram data) appear every year as 3-mark questions.
- Bayes' theorem is a standard 5-mark question — present it in a table format showing P(Bᵢ), P(A|Bᵢ), and P(A∩Bᵢ) columns.
- "Without replacement" problems require using conditional probability at each draw.
- JEE: Conditional probability, independence, and binomial distribution (from Class 12) build on these foundations.