Probability — Mathematics Class 12 Notes (CBSE & HBSE)
Free NCERT Mathematics notes for Probability (Class 12) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 3 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.
Board exam focus — Probability (CBSE & HBSE)
Probability is the final chapter, carrying 8-10 marks in CBSE. Conditional probability, Bayes theorem, and binomial distribution are most frequently tested. Every year CBSE includes Bayes theorem or a binomial distribution problem.
Conditional Probability and Multiplication Theorem
Conditional Probability
P(A|B) = P(A∩B)/P(B) (provided P(B) > 0)
Reading: Probability of A given B has occurred.
Properties of Conditional Probability:
- 0 ≤ P(A|B) ≤ 1
- P(S|B) = 1 (certain event)
- P(A'|B) = 1 − P(A|B)
- P(A∪C|B) = P(A|B) + P(C|B) − P(A∩C|B)
Multiplication Theorem
P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B)
Extended: P(A∩B∩C) = P(A)·P(B|A)·P(C|A∩B)
Independent Events
A and B are independent iff P(A∩B) = P(A)·P(B)
Equivalently: P(A|B) = P(A) and P(B|A) = P(B)
Key distinction:
- Mutually exclusive: A∩B = ∅, so P(A∩B) = 0
- Independent: P(A∩B) = P(A)·P(B) ≠ 0 (generally)
- Mutually exclusive events with non-zero probabilities are NEVER independent!
Important Results
- If A and B are independent: A and B', A' and B, A' and B' are also independent pairs.
- P(A∪B) = P(A)+P(B)−P(A∩B) = P(A)+P(B)−P(A)·P(B) (if independent)
Worked Example: A coin is tossed twice. A = {first head}, B = {second head}. P(A) = 1/2, P(B) = 1/2, P(A∩B) = P(HH) = 1/4 = (1/2)(1/2) ✓ Independent.
CBSE Tip: Always distinguish between independent events and mutually exclusive events. Independent: P(A∩B)=P(A)P(B). Mutually exclusive: P(A∩B)=0. Two events with positive probability CANNOT be both mutually exclusive AND independent.
Total Probability and Bayes Theorem
Law of Total Probability
Let E₁, E₂, …, Eₙ be a partition of sample space S (mutually exclusive and exhaustive events). For any event A: P(A) = Σᵢ P(Eᵢ)·P(A|Eᵢ) = P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + … + P(Eₙ)P(A|Eₙ)
Bayes Theorem
P(Eᵢ|A) = P(Eᵢ)·P(A|Eᵢ) / Σⱼ P(Eⱼ)·P(A|Eⱼ)
In words: Given that A has occurred, find the probability that it was caused by Eᵢ.
Tree Diagram Method — Recommended Approach
- Draw tree with E₁, E₂, … as first branches with probabilities P(Eᵢ)
- Second branches: P(A|Eᵢ) and P(A'|Eᵢ)
- P(Eᵢ∩A) = P(Eᵢ)·P(A|Eᵢ) — product along the branch
- P(A) = sum of all branches leading to A
- P(Eᵢ|A) = P(Eᵢ∩A)/P(A) — Bayes
Classic Bayes Problem Structure:
Given: 3 bags. Bag I has 2R,3W; Bag II has 3R,2W; Bag III has 4R,1W balls. A bag is chosen randomly, then a ball is drawn. It is red. Find probability it came from Bag II.
- P(Bag I) = P(Bag II) = P(Bag III) = 1/3
- P(R|I) = 2/5, P(R|II) = 3/5, P(R|III) = 4/5
- P(R) = (1/3)(2/5)+(1/3)(3/5)+(1/3)(4/5) = (1/15)(2+3+4) = 9/15 = 3/5
- P(II|R) = [(1/3)(3/5)]/(3/5) = (1/5)/(3/5) = 1/3
CBSE Tip: Bayes theorem is guaranteed in CBSE. The tree diagram approach is the most reliable method. Organise in a table: List Eᵢ, P(Eᵢ), P(A|Eᵢ), P(Eᵢ)·P(A|Eᵢ). The last column sums to P(A); divide each entry by P(A) for Bayes probabilities.
Random Variables, Probability Distributions and Binomial Distribution
Random Variable and Probability Distribution
A random variable X is a real-valued function on a sample space.
Probability Distribution: A table listing X values with their probabilities P(X=xᵢ) such that ΣP(X=xᵢ) = 1.
| X | x₁ | x₂ | … | xₙ |
|---|---|---|---|---|
| P(X) | p₁ | p₂ | … | pₙ |
Mean (Expected Value): E(X) = μ = Σ xᵢpᵢ
Variance: Var(X) = σ² = Σ xᵢ²pᵢ − μ² = E(X²) − [E(X)]²
Standard Deviation: σ = √(Var(X))
Bernoulli Trials
A Bernoulli trial is an experiment with exactly two outcomes: Success (probability p) and Failure (probability q = 1−p).
Conditions for Bernoulli trials:
- Finite number of trials n
- Each trial is independent
- Each trial has exactly two outcomes: success/failure
- Probability p of success is same in each trial
Binomial Distribution
If X = number of successes in n independent Bernoulli trials with success probability p: X ~ B(n, p)
P(X = r) = ⁿCᵣ · pʳ · qⁿ⁻ʳ, where q = 1−p, r = 0,1,2,…,n
Mean of X: μ = np Variance of X: σ² = npq Standard deviation: σ = √(npq)
Note: For Binomial distribution, Variance = npq ≤ np = Mean (since q ≤ 1)
Finding most likely value: The mode is the value of r for which P(X=r) is maximum.
- If (n+1)p is an integer m: two modes (m−1 and m)
- Otherwise: mode = ⌊(n+1)p⌋
Worked Example
A die is thrown 6 times. Find P(exactly 2 sixes).
- n=6, p=1/6, q=5/6, r=2
- P(X=2) = ⁶C₂·(1/6)²·(5/6)⁴ = 15·(1/36)·(625/1296) = 9375/46656
CBSE Tip: For Binomial distribution questions, clearly identify n, p, q, r at the start. The formula P(X=r) = ⁿCᵣ pʳ qⁿ⁻ʳ must be written out. Mean = np and Variance = npq are directly tested as 1-mark questions.
Frequently asked questions
Are these Probability notes free?
Yes — the Probability notes for Mathematics (Class 12) on Siksha Sarovar are completely free to read, with no account required.
Do these notes follow CBSE and HBSE?
Yes. The Probability notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.
What does the Probability chapter cover?
Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Probability.