R
Rishtaara
Mathematics · Class 12

Probability

Learn Probability with notes, examples, and practice questions.

5 sections15–25 min read8 MCQs · 4 examples

Chapter Overview

P(E) = favourable outcomes / total outcomes — only when all outcomes are equally likely (classical definition).

Probability quantifies uncertainty — from 0 (impossible) to 1 (certain). Class 12 builds on Class 11 with conditional probability, Bayes' theorem, and random variables.

Conditional Probability

Definition

P(A|B) = P(A∩B) / P(B), P(B) > 0

Multiplication

P(A∩B) = P(B)·P(A|B) = P(A)·P(B|A)

Independent

P(A∩B) = P(A)·P(B)

  • A and B independent: P(A|B) = P(A)
  • Mutually exclusive: P(A∩B) = 0, not independent unless one has P=0

Bayes' Theorem

Bayes

P(Aᵢ|B) = P(Aᵢ)·P(B|Aᵢ) / Σⱼ P(Aⱼ)·P(B|Aⱼ)

Total probability

P(B) = Σ P(Aᵢ)·P(B|Aᵢ) for partition A₁…Aₙ

Bayes reverses conditioning — find cause probability given effect. Used in medical tests, spam filters, and diagnostic problems.

Random Variables & Distribution

PMF

P(X = xᵢ) = pᵢ, Σpᵢ = 1

Mean E(X)

= Σ xᵢpᵢ

Variance Var(X)

= E(X²) − [E(X)]²

Bernoulli

P(1)=p, P(0)=1−p; E=np, Var=npq for n trials

Binomial B(n,p)

P(X=r) = ⁿCᵣ pʳ qⁿ⁻ʳ

Solving Strategy

  • Define sample space clearly
  • Draw tree diagram for sequential events
  • Identify if with/without replacement
  • Check independence before multiplying probabilities
  • For Bayes: identify partitions (mutually exclusive, exhaustive)

Solved Examples

Step-by-step solutions — read each step before checking the final answer.

Example 1: Conditional

P(A)=0.5, P(B)=0.4, P(A∩B)=0.2. Find P(A|B).

  1. 1P(A|B) = P(A∩B)/P(B) = 0.2/0.4 = 0.5

Answer

0.5

Example 2: Bayes

Disease 1% population. Test 99% accurate. Positive test — find P(disease).

  1. 1P(D)=0.01, P(+|D)=0.99, P(+|D')≈0.01
  2. 2P(+) = 0.01·0.99 + 0.99·0.01 = 0.0198
  3. 3P(D|+) = 0.0099/0.0198 ≈ 0.5

Answer

≈ 50% (screening paradox — rare disease)

Example 3: Binomial

10 fair coin tosses. P(exactly 3 heads)?

  1. 1n=10, p=1/2, r=3
  2. 2P = ¹⁰C₃ (1/2)³ (1/2)⁷ = 120/1024 = 15/128

Answer

15/128

Example 4: Mean

X takes values 1,2,3 with P 0.2, 0.5, 0.3. Find E(X).

  1. 1E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2+1+0.9

Answer

2.1

Key Points to Remember

  • Conditional probability narrows sample space to B
  • Bayes needs partition of sample space
  • Binomial: fixed n, independent trials, two outcomes
  • Var(X) = E(X²) − [E(X)]²

Exam Tips

  • State formula before substituting numbers
  • Tree diagrams earn partial credit
  • Distinguish 'with replacement' vs 'without'
  • Bernoulli trials: same p each trial, independent

Formula Cheat Sheet

Quick reference — NCERT board exam formulas

Probability

  • P(E) = favourable outcomes / total outcomes
  • P(A∪B) = P(A) + P(B) − P(A∩B)
  • P(A|B) = P(A∩B) / P(B)
  • Independent: P(A∩B) = P(A)P(B)
  • Bayes:P(Aᵢ|B) = P(Aᵢ)P(B|Aᵢ) / Σ P(Aⱼ)P(B|Aⱼ)
  • E(X) = Σ xᵢpᵢ ; Var(X) = E(X²) − [E(X)]²
  • Binomial: P(X=r) = ⁿCᵣ pʳ qⁿ⁻ʳ

Practice MCQs — Probability

Test your understanding with topic-wise multiple choice questions. Explanations appear after each answer.

Question 1 of 8Score: 0/0

P(A|B) equals: