Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
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.