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CBSE Class 11 Mathematics★ 6-8 marks · CBSE Class 11 Mathematics Chapter 14

Probability

What is the chance of rain tomorrow? Of rolling a sum of 7 with two dice? PROBABILITY quantifies uncertainty. This chapter covers the classical definition, sample space, events, and the addition rule. CBSE Class 11 Mathematics Chapter 14.

⏱ 22 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Define random experiments, sample space, and events and list sample spaces for coin, die, and card experiments
2Classify events as simple, compound, mutually exclusive, exhaustive, or impossible
3Apply the classical probability formula P(E) = n(E)/n(S) for equally likely outcomes
4Use the complement rule P(E') = 1−P(E) to calculate probabilities more efficiently
5Apply the addition rule P(A∪B) = P(A)+P(B)−P(A∩B) for any two events

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Why this chapter matters

Probability is the language of uncertainty — used in every science, engineering, and economic model. This foundational chapter (classical definition, addition rule) is the prerequisite for conditional probability and Bayes' theorem in Class 12, which carries 8-10 marks.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Sets

Sample space is a set; events are subsets; union, intersection, complement of events use set operations

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Permutations and Combinations

Counting n(E) and n(S) in complex experiments requires combinations and the multiplication principle

Probability

"Probability is the mathematics of uncertainty — the tool we use when we don't know what WILL happen, but we can quantify what MIGHT happen."

1. Chapter Overview

PROBABILITY is the branch of mathematics that deals with CHANCE and UNCERTAINTY. This chapter covers: random experiments, sample space, events (simple, compound, mutually exclusive, exhaustive), the classical definition of probability (P(E) = n(E)/n(S)), and the ADDITION RULE for probability.

2. Basic Concepts

Random Experiment

An experiment whose OUTCOME CANNOT BE PREDICTED with certainty
BUT: the set of ALL POSSIBLE OUTCOMES is known
Examples: Tossing a coin. Rolling a die. Drawing a card from a deck.

Sample Space (S)

The SET OF ALL POSSIBLE OUTCOMES of a random experiment
Coin toss: S = {H, T}
Die roll: S = {1, 2, 3, 4, 5, 6}
Two coins: S = {HH, HT, TH, TT}

Event (E)

A SUBSET of the sample space
E = 'getting an even number on a die' = {2, 4, 6}

3. Types of Events

Type	Definition	Example
Simple (Elementary)	Exactly ONE outcome	Getting a '6' on a die
Compound	More than one outcome	Getting an even number {2, 4, 6}
Impossible	Can NEVER occur. E = ∅.	Getting a 7 on a standard die. P(∅) = 0.
Sure (Certain)	MUST occur. E = S.	Getting a number ≤ 6 on a die. P(S) = 1.
Mutually Exclusive	Two events CANNOT occur together. E₁ ∩ E₂ = ∅.	Getting both a Head AND Tail on one coin toss.
Exhaustive	Union of events = S (all possibilities covered)	E₁ = {odd}, E₂ = {even}. Together cover the whole die.
Complement	E' = everything in S that is NOT in E	E = {even}. E' = {odd}.

4. Classical (A Priori) Definition of Probability

If a random experiment has n EQUALLY LIKELY outcomes, and m of them are FAVOURABLE to event E:

$P (E) = \frac{Number of outcomes favourable to E}{Total number of outcomes} = \frac{n ( E )}{n ( S )}$

Key Properties

0 ≤ P(E) ≤ 1 (probability is always between 0 and 1)
P(Impossible event) = 0
P(Sure event) = 1
P(E) + P(E') = 1 (probability of an event + its complement = 1)
P(E') = 1 — P(E) (useful shortcut: sometimes it's easier to find the complement)

5. Algebra of Events

A ∪ B: A OR B (or both) occur
A ∩ B: BOTH A AND B occur
A' (complement) : A does NOT occur

Addition Rule (for Any Two Events)

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$

For Mutually Exclusive Events

If A and B are mutually exclusive (A ∩ B = ∅): P(A ∩ B) = 0 $P (A \cup B) = P (A) + P (B)$

6. Key Examples to Know

Two Dice

Total outcomes: 6 × 6 = 36
P(sum = 7) = 6/36 = 1/6 (most probable sum)
P(sum = 12) = 1/36 (only one way: 6+6)

Cards (Standard 52-Card Deck)

4 suits (♠ ♥ ♦ ♣). 13 cards per suit: Ace, 2-10, Jack, Queen, King.
P(King) = 4/52 = 1/13
P(Heart) = 13/52 = 1/4
P(King of Hearts) = 1/52

7. Exam Focus

Sample space — list all outcomes for dice/coins/cards
Classical probability — n(E)/n(S)
Mutually exclusive events — definition, addition rule
Complement rule — P(E') = 1 — P(E)
Addition rule with intersection — P(A ∪ B) = P(A) + P(B) — P(A ∩ B)

8. Key Formulas

P(E) = n(E) / n(S)
0 ≤ P(E) ≤ 1
P(E') = 1 — P(E)
P(A ∪ B) = P(A) + P(B) — P(A ∩ B)
For mutually exclusive: P(A ∪ B) = P(A) + P(B)

9. Conclusion

Probability is the language in which we speak about UNCERTAINTY:

SAMPLE SPACE: List everything that CAN happen.
EVENT: A subset — what you're INTERESTED in.
PROBABILITY: The ratio. Favourable ÷ Total. Between 0 and 1.
RULES: Addition rule. Complement rule. The basic algebra of uncertainty.

'The theory of probability is at bottom nothing but common sense reduced to calculation.' — Laplace.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Classical Probability

P(E) = n(E) / n(S)

Valid only when all outcomes are equally likely; n(E) = number of favourable outcomes, n(S) = total outcomes

Complement Rule

P(E') = 1 − P(E)

P(E) + P(E') = 1 always; use complement when 'at least one' problems are complex

Addition Rule (General)

P(A∪B) = P(A) + P(B) − P(A∩B)

Subtract P(A∩B) to avoid double-counting outcomes in both A and B

Addition Rule (Mutually Exclusive)

If A∩B = ∅, then P(A∪B) = P(A) + P(B)

Mutually exclusive events cannot occur together, so P(A∩B) = 0

Probability Bounds

0 ≤ P(E) ≤ 1 for any event E; P(∅) = 0; P(S) = 1

Probability of impossible event = 0; probability of sure (certain) event = 1

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Writing P(A∪B) = P(A)+P(B) without subtracting P(A∩B)

✓ P(A∪B) = P(A)+P(B)−P(A∩B) always. Omitting P(A∩B) is valid ONLY when A and B are mutually exclusive. Always check whether the events can overlap.

WATCH OUT

✗ Confusing mutually exclusive with independent events

✓ Mutually exclusive: A∩B=∅, so they CANNOT occur together. Independent: occurrence of A does not affect probability of B (requires conditional probability — Class 12 topic). These are different concepts.

WATCH OUT

✗ Listing duplicate outcomes in the sample space (e.g., listing {HT, TH} as one outcome for two coins)

✓ The order of outcomes matters. For two coin tosses: {HH, HT, TH, TT} — four outcomes, not three. HT (first head, second tail) and TH are DIFFERENT outcomes.

WATCH OUT

✗ Computing P(E) > 1 or P(E) < 0

✓ If you get probability > 1 or < 0, recheck your counting. The number of favourable outcomes n(E) can NEVER exceed the total outcomes n(S). Probability is always in [0,1].

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 14.1

Exercise 14.1

Random experiments, writing sample spaces for coins, dice, and cards

Questions

Ex 14.2

Exercise 14.2

Types of events — identifying mutually exclusive, exhaustive, impossible events

Questions

Ex 14.3

Exercise 14.3

Classical probability, complement rule, addition theorem with various experiments

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Classical Probability

A die is thrown once. Find the probability of getting a number greater than 4.

▶ Show solution

Sample space S = {1,2,3,4,5,6}, n(S) = 6. Event E = {5,6}, n(E) = 2. P(E) = 2/6 = 1/3.

Q2MEDIUM· Addition Rule

A card is drawn at random from a standard deck of 52 cards. Find the probability that it is a King OR a Heart.

▶ Show solution

P(King) = 4/52 = 1/13. P(Heart) = 13/52 = 1/4. P(King and Heart) = P(King of Hearts) = 1/52. P(King OR Heart) = P(King)+P(Heart)−P(King∩Heart) = 4/52+13/52−1/52 = 16/52 = 4/13.

Q3HARD· Complement Rule

Two dice are thrown simultaneously. Find the probability that the sum is at least 9.

▶ Show solution

n(S) = 6×6 = 36. Instead of listing all favourable outcomes for sum≥9, we can list for sum<9 and use complement. Favourable for sum ≥ 9: sum=9: {(3,6),(4,5),(5,4),(6,3)}=4; sum=10: {(4,6),(5,5),(6,4)}=3; sum=11: {(5,6),(6,5)}=2; sum=12: {(6,6)}=1. Total favourable = 4+3+2+1 = 10. P(sum≥9) = 10/36 = 5/18.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Random experiment: outcome unpredictable, but all possible outcomes are known
•Sample space S: set of ALL possible outcomes; events are subsets of S
•Classical probability: P(E) = n(E)/n(S), valid for equally likely outcomes only
•0 ≤ P(E) ≤ 1; P(∅)=0 (impossible); P(S)=1 (certain)
•Complement: P(E') = 1−P(E); useful shortcut for 'at least one' problems
•Mutually exclusive events: E₁∩E₂=∅; P(E₁∪E₂) = P(E₁)+P(E₂)
•Addition rule (general): P(A∪B) = P(A)+P(B)−P(A∩B)
•For two dice: 36 equally likely outcomes; most probable sum = 7 (6 ways)

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks

Question type	Marks each	Typical count	What it tests
Short Answer	2	1	Basic probability using n(E)/n(S), identifying event types
Long Answer	4-6	1	Addition theorem, complement rule with two events, card or dice problems

Prep strategy

Memorise the standard sample space sizes: 1 coin=2, 2 coins=4, 3 coins=8, 1 die=6, 2 dice=36, 52-card deck=52
For 'at least one' probability questions, always use the complement: P(at least one) = 1 − P(none)
Before applying the addition rule, always check if the events are mutually exclusive — if yes, skip computing P(A∩B)

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Insurance and Risk Assessment

Actuaries use probability to calculate premiums — the probability of a claim (car accident, illness) determines the insurance price. Addition theorem is used to compute the probability of at least one of multiple risk events.

Weather Forecasting

'40% chance of rain' is a direct application of probability. Forecasters compute P(rain) from historical data and atmospheric models, giving probabilistic rather than certain predictions.

Medical Diagnosis

Sensitivity and specificity of medical tests are probabilities. The addition rule helps compute the probability of a positive result from either of two correlated symptoms.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Always write out the sample space (or its size) before computing any probability — this anchors the denominator
For addition theorem questions, identify P(A), P(B), and P(A∩B) explicitly before plugging into the formula
In card problems, recall the 52-card deck structure: 4 suits × 13 cards; 4 Kings, 13 Hearts, only 1 King of Hearts
For 'at least' questions (at least one head, sum at least 8), use the complement to avoid enumerating many cases

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Geometric probability: when outcomes are measured by length, area, or angle rather than counts — P = favourable measure / total measure
Bayes' Theorem (Class 12 preview): P(A|B) = P(B|A)·P(A)/P(B) — updating probability given new evidence; one of the most important results in probability

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardHigh

JEE MainVery High

JEE AdvancedHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Mutually exclusive: events CANNOT occur together (E₁∩E₂=∅). Exhaustive: together they COVER the whole sample space (E₁∪E₂=S). Events can be mutually exclusive without being exhaustive, or exhaustive without being mutually exclusive.

Only if at least one event has probability zero. For events with positive probability, mutually exclusive and independent are incompatible: if A∩B=∅, then P(A∩B)=0 ≠ P(A)·P(B) (unless P(A)=0 or P(B)=0).

Outcomes in A∩B are counted once in P(A) and once in P(B), so they're double-counted in P(A)+P(B). Subtracting P(A∩B) corrects this: each outcome in A∪B is counted exactly once.

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