Chapter 14 of 14

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CBSE Class 11 Mathematics★ Unit V · part of the 12-mark Statistics and Probability block shared with Statistics — CBSE states no chapter-wise split

Probability

This chapter doesn't start from scratch — it opens by treating "random experiment" and "sample space" as things you already know, and goes straight to events, the algebra of events, and an axiom-based definition of probability. That opening choice isn't an accident: those foundational ideas are formative-only this year, not new teaching. CBSE Class 11 Mathematics Chapter 14 — the second half of the 12-mark Statistics and Probability unit.

⏱ 9 min readMedium difficulty✓ Free · NCERT-aligned

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

1Classify events as impossible, sure, simple, or compound, and identify whether two or more events are mutually exclusive and/or exhaustive
2Translate event descriptions ('A or B', 'A and B', 'not A', 'A but not B') into the corresponding set operations on the sample space
3Check whether a proposed probability assignment to a sample space's outcomes satisfies the axioms of probability
4Apply the classical formula P(A)=n(A)/n(S) for equally likely outcomes, using counting techniques from Permutations and Combinations where needed
5Apply the addition rule P(A union B)=P(A)+P(B)-P(A intersection B) and its mutually-exclusive special case
6Apply the complement rule P(not A)=1-P(A), especially for 'at least one' problems

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Why this chapter matters

This chapter builds probability from a small set of axioms rather than just a counting formula, connecting the classical n(A)/n(S) definition from earlier classes to a more general framework. It's the direct foundation for Class 12's conditional probability and Bayes' theorem, and closes out the last chapter of Class 11 Mathematics.

Before you start — revise these

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

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Sets (Class 11, Chapter 1): union, intersection, difference, and complement, since every event operation in this chapter is exactly one of these applied to subsets of the sample space

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Permutations and Combinations (Class 11, Chapter 6): counting n(A) and n(S) in card, dice, and selection problems often needs combinations or the multiplication principle

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Random experiments and sample space from earlier classes, since this chapter recalls but does not re-teach these ideas from scratch

Probability

1. Check this before you revise anything

"Random experiment" and "sample space" are not re-taught here — and that's intentional, not an oversight. The book's very first sentence is "We have studied about random experiment and sample space associated with an experiment" — it recalls these ideas in one line and moves straight to events.

CBSE's own formative-only block confirms why: "Random experiments; outcomes, sample space (set representation)" is listed as formative for this chapter, not summative. If you need the full definitions, they're genuinely prior-class material, not something this chapter derives from first principles.

The old stub's exercise map didn't match the book at all. It described Exercise 14.1 as being about sample spaces and Exercise 14.2 as being about event types — the reverse of the truth — and invented an entire third exercise (14.3, 21 questions) to hold content that's actually already inside the real Exercise 14.2.

The current book has exactly two exercises plus a Miscellaneous Exercise: 14.1 is about the algebra and classification of events (mutually exclusive, exhaustive), and 14.2 is where the numbers are — axiomatic and classical probability, the addition rule, the complement rule, 21 questions' worth.

2. What this chapter covers

Textbook section	Topic
14.1	Events: occurrence, types (impossible, sure, simple, compound), algebra of events, mutually exclusive and exhaustive events
14.2	Axiomatic approach to probability; probability of equally likely outcomes; probability of 'A or B' and 'not A'

3. Events and their algebra

Any subset $E$ of a sample space $S$ is an event; it has occurred if the actual outcome $ω$ satisfies $ω \in E$ . Events are classified by what they contain: the impossible event is $\emptyset$ ; the sure event is $S$ itself; a simple (elementary) event has exactly one sample point; a compound event has more than one.

Since events are just subsets of $S$ , every set operation carries over directly, each with a probabilistic reading:

Set notation	Event reading
$A^{'}$ (i.e. $S - A$ )	'not $A$ '
$A \cup B$	' $A$ or $B$ ' (either, or both)
$A \cap B$	' $A$ and $B$ ' (both)
$A - B$	' $A$ but not $B$ ', equivalently $A \cap B^{'}$

Two events $A, B$ are mutually exclusive if $A \cap B = \emptyset$ — they can't happen together. Events $E_{1}, \dots, E_{n}$ are exhaustive if $E_{1} \cup \dots \cup E_{n} = S$ — at least one of them must happen. A set of events that is both pairwise mutually exclusive and exhaustive is called mutually exclusive and exhaustive.

Worked, mirroring the textbook's own Example 3. A coin is tossed three times. $A$ : 'no head appears' $= {T T T}$ , $B$ : 'exactly one head' $= {H T T, T H T, T T H}$ , $C$ : 'at least two heads' $= {H H T, H T H, T H H, H H H}$ . Every outcome in $S$ falls into exactly one of $A, B, C$ , so $A \cup B \cup C = S$ (exhaustive) and each pair has empty intersection (mutually exclusive) — together, mutually exclusive and exhaustive.

4. The axiomatic approach, and probability of 'A or B' / 'not A'

Axioms. For a sample space $S = {ω_{1}, \dots, ω_{n}}$ , a probability assignment is any function $P$ with $P (ω_{i}) \geq 0$ for every outcome and $\sum P (ω_{i}) = 1$ over all of $S$ ; for any event $A$ , $P (A) = \sum P (ω_{i})$ over $ω_{i} \in A$ . Any assignment satisfying these two conditions is valid — there is no single "correct" way to assign probabilities beyond the axioms themselves.

When all outcomes are equally likely (each $ω_{i}$ gets probability $1/ n$ ), this reduces to the familiar formula connecting back to earlier classes:

$P (A) = \frac{n ( A )}{n ( S )} = \frac{number of outcomes favourable to A}{total number of outcomes}$

Addition rule, derived by noting that $A \cup B$ splits into the mutually exclusive pieces $A - B$ , $A \cap B$ , and $B - A$ :

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

which reduces to $P (A \cup B) = P (A) + P (B)$ exactly when $A, B$ are mutually exclusive (since then $P (A \cap B) = 0$ ).

Complement rule. Since $A$ and $A^{'}$ are mutually exclusive and exhaustive, $P (A) + P (A^{'}) = P (S) = 1$ , so:

$P (A^{'}) = 1 - P (A)$

Worked, mirroring the textbook's own Example 8. A committee of two is chosen from two men and two women. $P (no man)$ : choosing $2$ women from $2$ out of $4$ people total, $\frac{^{2} C _{2}}{^{4} C _{2}} = \frac{1}{6}$ . $P (one man) = \frac{^{2} C _{1} \times ^{2} C _{1}}{^{4} C _{2}} = \frac{2}{3}$ . $P (two men) = \frac{^{2} C _{2}}{^{4} C _{2}} = \frac{1}{6}$ — and these three probabilities sum to $1$ , since 'no man', 'one man', 'two men' are mutually exclusive and exhaustive.

Summary

An event is any subset of the sample space; occurrence means the actual outcome lies in that subset.
Impossible event $= \emptyset$ ; sure event $= S$ ; simple event has one sample point; compound event has more than one.
Set operations translate directly: $A^{'} =$ 'not $A$ ', $A \cup B =$ ' $A$ or $B$ ', $A \cap B =$ ' $A$ and $B$ ', $A - B =$ ' $A$ but not $B$ '.
Mutually exclusive: $A \cap B = \emptyset$ . Exhaustive: the events' union is all of $S$ .
Axiomatic probability: $P (ω_{i}) \geq 0$ for every outcome, $\sum P (ω_{i}) = 1$ over $S$ ; for equally likely outcomes this reduces to $P (A) = n (A) / n (S)$ .
Addition rule: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , becoming $P (A) + P (B)$ when mutually exclusive.
Complement rule: $P (A^{'}) = 1 - P (A)$ .
"Random experiment" and "sample space" are formative-only this year — recalled in one line, not re-derived, since the book treats them as already known from earlier classes.

Key formulas & results

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

Event algebra

A' = 'not A' (i.e. S-A); A union B = 'A or B'; A intersection B = 'A and B'; A-B = 'A but not B' = A intersection B'

Every event operation is exactly the matching set operation applied to subsets of the sample space S

Mutually exclusive events

A and B are mutually exclusive if A intersection B = empty set

They cannot occur together; simple events of a sample space are always mutually exclusive with each other

Exhaustive events

E1, E2, ..., En are exhaustive if E1 union E2 union ... union En = S

At least one of them must occur whenever the experiment is performed; combined with pairwise mutual exclusivity, they partition S

Axioms of probability

P(omega_i) >= 0 for each outcome; sum of P(omega_i) over all of S equals 1; P(A) = sum of P(omega_i) for omega_i in A

Any assignment satisfying these two conditions is a valid probability function — there is no single 'correct' assignment beyond satisfying the axioms

Probability of equally likely outcomes

P(A) = n(A)/n(S)

Valid only when every outcome in S has the same probability; this is the classical definition carried forward from earlier classes

Addition rule

P(A union B) = P(A) + P(B) - P(A intersection B)

Subtracting P(A intersection B) corrects for double-counting outcomes that are in both A and B

Addition rule, mutually exclusive case

If A intersection B = empty set, then P(A union B) = P(A) + P(B)

The special case of the addition rule when P(A intersection B) = 0

Complement rule

P(A') = 1 - P(A)

Follows since A and A' are mutually exclusive and exhaustive, so P(A)+P(A')=P(S)=1; the standard shortcut for 'at least one' problems

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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 union B) = P(A) + P(B) without subtracting P(A intersection B)

✓ This is only valid when A and B are mutually exclusive. In general, P(A union B) = P(A)+P(B)-P(A intersection B) — check for overlap before dropping the intersection term.

WATCH OUT

✗ Treating mutually exclusive and independent as the same concept

✓ Mutually exclusive means A intersection B = empty set (they cannot both happen). Independent means one event's occurrence doesn't affect the other's probability — a different, Class 12 concept. For events with positive probability, mutually exclusive events are actually never independent.

WATCH OUT

✗ Undercounting the sample space by ignoring order, e.g. treating HT and TH as the same outcome for two coin tosses

✓ Order matters in a sample space unless the experiment genuinely doesn't distinguish it. For two coin tosses, S={HH,HT,TH,TT} has 4 distinct outcomes, not 3.

WATCH OUT

✗ Applying P(A)=n(A)/n(S) when the outcomes are not equally likely

✓ The n(A)/n(S) formula is only valid under the equally-likely-outcomes assumption. For a weighted die or any experiment with unequal probabilities, use the axiomatic definition P(A)=sum of P(omega_i) for omega_i in A instead.

WATCH OUT

✗ Computing a probability greater than 1 or less than 0 without noticing

✓ Since 0 <= P(E) <= 1 always, an out-of-range result means a counting or arithmetic error upstream — recheck n(A) against n(S), or recheck an addition/subtraction of probabilities.

WATCH OUT

✗ Trying to re-derive 'random experiment' and 'sample space' as if they were new content in this chapter

✓ These are formative-only this year — the book itself only recalls them in one sentence before moving to events. Treat them as prior-class background, not something this chapter teaches from scratch.

WATCH OUT

✗ For 'at least one' problems, trying to enumerate every favourable case directly instead of using the complement

✓ P(at least one) = 1 - P(none) is almost always faster and less error-prone than listing every way to get one or more of something, especially in multi-draw or multi-toss problems.

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

Algebra of events, classifying and identifying mutually exclusive, exhaustive, simple, and compound events

Questions

Ex 14.2

Exercise 14.2

Axiomatic and classical probability, checking valid probability assignments, the addition rule, and the complement rule

Questions

Miscellaneous Ex

Miscellaneous Exercise

Applied problems on marbles, cards, lotteries, misplaced letters, and combined addition/complement rule questions

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Probability?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~12 marks in CBSE exams

5-minute revision

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

•An event is any subset of the sample space; it occurs when the actual outcome lies in that subset
•Impossible event = empty set; sure event = S; simple event has one sample point; compound event has more than one
•Set operations translate directly: A' = 'not A', A union B = 'A or B', A intersection B = 'A and B', A-B = 'A but not B'
•Mutually exclusive: A intersection B = empty set. Exhaustive: the events' union is all of S
•Axioms: P(omega_i) >= 0 for every outcome, and probabilities over all of S sum to 1
•Equally likely outcomes: P(A) = n(A)/n(S)
•Addition rule: P(A union B) = P(A)+P(B)-P(A intersection B), reducing to P(A)+P(B) when mutually exclusive
•Complement rule: P(A') = 1-P(A), the standard shortcut for 'at least one' problems
•'Random experiment' and 'sample space' are formative-only this year — recalled in one line, not re-taught from scratch

CBSE marks blueprint

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

Typical chapter weightage: Part of Unit V's 12-mark Statistics and Probability block (shared with Statistics; no chapter-wise split, per CBSE)

Question type	Marks each	Typical count	What it tests
Algebra of Events and Mutually Exclusive/Exhaustive Classification	2-3	1	Translating event descriptions into set operations, checking mutual exclusivity and exhaustiveness
Axiomatic and Classical Probability	2-4	1	Checking valid probability assignments, applying P(A)=n(A)/n(S) for dice, coins, and cards
Addition Rule and Complement Rule	4-6	1	P(A or B) computations, 'at least one' problems via the complement, applied word 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' questions, default to the complement rule first — it is almost always faster than direct enumeration
Before applying the addition rule, check whether the events are mutually exclusive — if so, the intersection term drops out entirely

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 pricing

Insurance premiums are set using the probability of a claim occurring; the addition rule helps compute the probability of at least one of several possible claim events happening within a policy period.

Weather forecasting

A '40% chance of rain' is a direct probability statement, computed from historical frequency data and atmospheric models — the same equally-likely-outcomes idea generalised to weighted, non-equally-likely events.

Quality control sampling

Manufacturers use probability to decide how many items to sample from a batch; the complement rule computes the probability that a random sample catches at least one defective item, guiding how large a sample needs to be.

Exam strategy

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

Always define the sample space (or at least its size) explicitly before computing any probability — this anchors n(S) in the denominator

For addition rule questions, write out P(A), P(B), and P(A and B) as three separate labelled values before combining them

For 'at least one' questions, try the complement first: compute P(none), then subtract from 1

In card and dice problems, double-check whether order matters in your sample space — this is the single most common source of a wrong n(S)

Going beyond the textbook

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

STRETCH

Geometric probability, where favourable and total outcomes are measured by length, area, or angle rather than counted discretely (P = favourable measure / total measure), is not part of the current book but extends the same equally-likely-outcomes idea to continuous sample spaces

STRETCH

Derangements (permutations with no element in its original position) generalise the book's own misplaced-letters Miscellaneous Exercise question beyond 3 objects — see the Advanced JEE problem above for the general formula

STRETCH

Bayes' Theorem and conditional probability (Class 12 topics) build directly on this chapter's addition and complement rules to handle probabilities that update based on new information

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainChecking a probability assignment against the axiomsDirect axiom verification

A sample space is $S = {ω_{1}, ω_{2}, ω_{3}, ω_{4}, ω_{5}}$ with $P (ω_{1}) = P (ω_{2}) = 0.15$ , $P (ω_{3}) = P (ω_{4}) = k$ , $P (ω_{5}) = 0.3$ . Find $k$ , and hence find $P (A)$ where $A = {ω_{1}, ω_{3}, ω_{5}}$ .

Stuck? Show the approach

Use the axiom that all probabilities in a sample space must sum to 1 to solve for $k$ , then sum the probabilities of the outcomes in $A$ .

Show the full solution

Sum of all probabilities: $0.15 + 0.15 + k + k + 0.3 = 1$ , so $0.6 + 2 k = 1$ , giving $k = 0.2$ . Then $P (A) = P (ω_{1}) + P (ω_{3}) + P (ω_{5}) = 0.15 + 0.2 + 0.3 = 0.65$ .

Answer: k = 0.2, P(A) = 0.65

The trap

Forgetting that $ω_{3}$ and $ω_{4}$ both equal $k$ (not just one of them) when setting up the sum-to-1 equation.

JEE MainAddition rule with three eventsExtending the two-event addition formula

If $A$ , $B$ , $C$ are events with $P (A) = 0.3$ , $P (B) = 0.4$ , $P (C) = 0.5$ , $P (A \cap B) = 0.15$ , $P (B \cap C) = 0.2$ , $P (A \cap C) = 0.1$ , and $P (A \cap B \cap C) = 0.05$ , find $P (A \cup B \cup C)$ .

Stuck? Show the approach

Use the three-event extension of the addition rule (derivable the same way as the two-event version, by splitting the union into mutually exclusive pieces).

Show the full solution

$P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (B \cap C) - P (A \cap C) + P (A \cap B \cap C) = 0.3 + 0.4 + 0.5 - 0.15 - 0.2 - 0.1 + 0.05 = 0.8$ .

Answer: 0.8

The trap

Forgetting to add back $P (A \cap B \cap C)$ at the end — subtracting all three pairwise intersections removes the triple-overlap region three times instead of the two times needed, so it must be added back once.

JEE MainComplement rule for 'at least one' across multiple independent-style drawsComplement applied to a repeated trial

A fair coin is tossed 5 times. Find the probability of getting at least one head.

Stuck? Show the approach

It is far easier to find the probability of the complementary event (no heads at all) than to enumerate every way of getting at least one head directly.

Show the full solution

P(no heads in 5 tosses) = P(all 5 are tails) = $(1/2)^{5} = 1/32$ .

Answer: P(at least one head) = 1 - 1/32 = 31/32

The trap

Attempting to add up P(exactly 1 head) + P(exactly 2 heads) + ... + P(exactly 5 heads) term by term is far more error-prone than the single complement calculation.

JEE MainClassical probability with combinationsCounting favourable and total outcomes with nCr

A bag contains 5 red and 7 black balls. Two balls are drawn at random. Find the probability that both are of the same colour.

Stuck? Show the approach

Count the total ways to draw 2 balls from 12, and separately the favourable ways (both red, or both black), using combinations.

Show the full solution

Total ways: $(2 12) = 66$ . Both red: $(2 5) = 10$ . Both black: $(2 7) = 21$ . Favourable total: $10 + 21 = 31$ .

Answer: 31/66

The trap

Forgetting that 'both red' and 'both black' are themselves mutually exclusive events, so their counts simply add — no intersection term is needed here, unlike the general addition rule.

JEE AdvancedDerangement-style probability, beyond a single applied formulaCombining counting principles with the complement rule

Four letters are written to four different people, and four envelopes are correctly addressed. If the letters are inserted into the envelopes at random (one per envelope), find the probability that at least one letter goes into its correct envelope.

Stuck? Show the approach

Use the complement: find the probability that no letter is in its correct envelope (a derangement of 4 objects), then subtract from 1. The number of derangements of n objects is $D_{n} = n! \sum_{k = 0}^{n} \frac{( - 1 ) ^{k}}{k !}$ .

Show the full solution

Total arrangements: $4! = 24$ . Derangements of 4 objects: $D_{4} = 24 (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 24 - 24 + 12 - 4 + 1 = 9$ . So $P (no letter correct) = 9/24 = 3/8$ .

Answer: P(at least one correct) = 1 - 3/8 = 5/8

The trap

This generalises the book's own Miscellaneous Exercise Q6 (the 3-letter version); assuming the same 2/3 answer applies here would be wrong, since the derangement count doesn't scale linearly with n — each n needs its own D_n calculation.

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.

Only in passing — the book's very first line recalls them as already known from earlier classes, matching their formative-only status in the CBSE syllabus. This chapter's own new content starts with events and their classification.

Mutually exclusive means the events cannot occur together (their intersection is empty). Exhaustive means their union covers the entire sample space (at least one must occur). Events can be one without the other, or both together.

Only if at least one of them has probability zero. For events with positive probability, mutually exclusive events are never independent, since P(A intersection B)=0 but P(A).P(B) would be positive if both events had positive probability.

Whenever the question involves 'at least one' of something, or when the complementary event is much simpler to count than the event itself — computing P(none) and subtracting from 1 is usually far less error-prone than listing every favourable case.

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