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Probability — Mathematics Class 11 Notes (CBSE & HBSE)

Free NCERT Mathematics notes for Probability (Class 11) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 4 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.

Board exam focus — Probability (CBSE & HBSE)

Axiomatic foundation of probability — sample spaces, events, P(A ∪ B), complementary events. CBSE asks 5–7 marks; HBSE 4–6 marks. Builds on the language of sets from Chapter 1.

Random Experiments, Sample Space, Events

Random Experiment

An experiment is random if:

  1. It has more than one possible outcome.
  2. The exact outcome cannot be predicted in advance.
  3. It can (theoretically) be repeated under identical conditions.

Examples: tossing a coin, rolling a die, drawing a card from a shuffled deck.

Outcome and Sample Space

  • Outcome: a specific result of the experiment.
  • Sample space (S): the set of all possible outcomes.

Example: rolling a die ⇒ S = {1, 2, 3, 4, 5, 6}.

Sample Space — Examples

ExperimentSample space
Toss a coin{H, T}
Toss two coins{HH, HT, TH, TT}
Roll a die{1, 2, 3, 4, 5, 6}
Roll two dice36 ordered pairs (i, j)
Draw a card52 cards
Family of 3 children (gender)8 outcomes (BBB, BBG, …)

Event

An event is a subset of the sample space — a collection of outcomes for which the event is said to occur.

Types:

TypeDefinition
Simple / Elementarya singleton subset (one outcome)
Compoundcontains more than one outcome
Sure eventS itself (always occurs)
Impossible event∅ (never occurs)

Operations on Events

Since events are sets:

  • A ∪ B = A or B (or both) occurs.
  • A ∩ B = both A and B occur.
  • A′ (complement) = A does NOT occur.
  • A − B = A occurs but B does not.

Mutually Exclusive Events

A ∩ B = ∅: A and B cannot occur together.

Exhaustive Events

A₁, A₂, …, Aₙ are exhaustive if A₁ ∪ A₂ ∪ … ∪ Aₙ = S.

Mutually Exclusive AND Exhaustive

These partition S — every outcome belongs to exactly one Aᵢ.

CBSE Tip: Always write the sample space explicitly at the start of a probability problem — it earns the first mark and prevents off-by-one errors.

Axiomatic Definition of Probability

Axiomatic Approach (Kolmogorov)

A probability function P assigns to each event A ⊆ S a real number P(A) satisfying:

  1. Non-negativity: P(A) ≥ 0 for every event A.
  2. Normalisation: P(S) = 1.
  3. Additivity: If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B).

Consequences

  • P(∅) = 0.
  • 0 ≤ P(A) ≤ 1.
  • P(A′) = 1 − P(A).
  • For any events A, B: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) (Inclusion–Exclusion).

Classical (Equally Likely) Probability

If S has n equally likely outcomes and event A has k of them: P(A) = k/n = n(A)/n(S).

Used for coin, die, card problems where outcomes are symmetric.

Sample Probability Tables

Single die: P(any face) = 1/6. Two dice (sum = 7): favourable outcomes = 6; total = 36; P = 6/36 = 1/6. Card (king): P = 4/52 = 1/13.

Useful Formulae for Cards

  • 52 cards: 26 red (hearts + diamonds), 26 black (spades + clubs).
  • 13 face values × 4 suits.
  • 12 face cards (J, Q, K of each suit).
  • 4 aces, 4 kings, 4 queens, 4 jacks.

Two-Dice Sum Probabilities

SumOutcomesP
2(1,1)1/36
76 pairs6/36 = 1/6
12(6,6)1/36

Empirical / Relative Frequency Probability

For observed data, P(A) ≈ (frequency of A) / (total trials). Stabilises around the true probability as trials grow.

HBSE Tip: When writing a probability fraction, simplify to lowest terms (6/36 → 1/6) — examiners deduct for unreduced fractions.

Equally Likely Outcomes & Compound Events

When Are Outcomes Equally Likely?

In classical probability, outcomes are equally likely by symmetry: a fair coin, an unbiased die, a well-shuffled deck.

When outcomes are not equally likely (e.g. weighted die), classical formula fails — use axiomatic / empirical methods.

Probability of Compound Events

P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

If A and B are mutually exclusive (no overlap): P(A ∪ B) = P(A) + P(B).

Complementary Events

P(A) + P(A′) = 1.

Often easier to compute P(A′) than P(A) directly:

  • "At least one" → use complement ("none").

Mutually Exclusive and Exhaustive Events

If events A₁, A₂, …, Aₙ are mutually exclusive AND exhaustive, then P(A₁) + P(A₂) + … + P(Aₙ) = 1.

Independent Events (Preview — Class 12)

A and B are independent iff P(A ∩ B) = P(A) · P(B). Not formally treated in Class 11.

Standard Card Problems

  • P(ace) = 4/52 = 1/13.
  • P(king or queen) = 8/52 = 2/13.
  • P(spade) = 13/52 = 1/4.
  • P(red king) = 2/52 = 1/26.
  • P(face card) = 12/52 = 3/13.

Common Trick — Words "and" / "or"

  • AND suggests intersection ∩ (multiply if independent).
  • OR suggests union ∪ (add and subtract intersection).

Card Word Problems Template

  1. Identify total outcomes (typically 52).
  2. Count favourable outcomes.
  3. Use union/complement as needed.
CBSE Tip: Memorise card-deck composition — it eliminates 1–2 lines of explanation in board problems.

Real-Life Probability Problems

Problems Involving People / Groups

  • Selecting a committee with constraints (uses Permutations & Combinations from Ch 6).
  • Probability of drawing specific names from a hat.
  • Birthday-paradox-style problems (informal in Class 11).

Strategy for Word Problems

  1. Read carefully and identify the sample space.
  2. Compute n(S) using counting principles (multiplication, permutations, combinations).
  3. Identify the favourable event A and compute n(A).
  4. Use P(A) = n(A)/n(S); apply union/complement if needed.
  5. Simplify to lowest terms.

Sample 'At Least' Trick

P(at least one defective) = 1 − P(none defective). This often saves casework.

Sample Trick — Drawing Without Replacement

  • Total ways = ⁿCₖ (selecting k from n).
  • Favourable = ways to satisfy the condition (use ⁿCₖ for combinations).

Common Distributions in Daily Life

  • Weather forecasts give probability of rain (relative frequency).
  • Quality control uses probability to estimate defective rates.
  • Insurance uses long-run probabilities to set premiums.

Sample Word Problem

A bag contains 5 red and 3 blue balls. Two balls are drawn at random without replacement. Find P(both same colour).

Total selections = ⁸C₂ = 28. Both red = ⁵C₂ = 10. Both blue = ³C₂ = 3. Favourable = 10 + 3 = 13. P = 13/28.

Sample 'Exactly One' Question

When choosing 2 balls from a bag of 3 red and 4 white, find P(exactly one red):

  • Favourable = ³C₁ × ⁴C₁ = 12.
  • Total = ⁷C₂ = 21.
  • P = 12/21 = 4/7.

Probability Through Venn Diagram

For a two-event problem, draw the Venn diagram; fill in P(A ∩ B), then derive P(only A), P(only B), P(neither).

HBSE Tip: For 'at least' problems, always check whether the complement is simpler — it often saves a lot of casework.

Frequently asked questions

Are these Probability notes free?

Yes — the Probability notes for Mathematics (Class 11) 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.