Probability – Class 11 Maths Chapter 14 Notes & NCERT PDF

Class 11MathematicsChapter 14NCERT book: Mathematics - Class 11

Notes and a simple summary of Chapter 14 of the NCERT Class 11 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 14: Probability

Introduction

Chapter 14, Probability, extends the idea of probability met in earlier classes. Probability measures the uncertainty of events. Earlier, it was defined in the classical way (favourable outcomes ÷ total outcomes, for equally likely outcomes) or by the statistical (experimental) approach. This chapter introduces the precise language of events and the axiomatic approach to probability developed by the Russian mathematician A.N. Kolmogorov in 1933.

Key Concepts

14.1 Event

A random experiment has more than one possible outcome, and the outcome cannot be predicted in advance. The set of all outcomes is the sample space S.

An event is any subset E of S. For a die, E = "a number less than 4" = {1, 2, 3}.

14.1.1 Occurrence of an event: event E occurs if the outcome ω belongs to E.

14.1.2 Types of events

Impossible event: the empty set φ (e.g. getting 7 on a die).

Sure event: the whole sample space S.

Simple event: exactly one sample point; compound event: more than one sample point.

14.1.3 Algebra of events

Complementary event A′ ('not A') = S − A.

A or B = A ∪ B; A and B = A ∩ B; A but not B = A − B = A ∩ B′.

14.1.4–14.1.5 Mutually exclusive and exhaustive events

Mutually exclusive: A ∩ B = φ — both cannot occur together (e.g. 'odd' and 'even' on a die).

Exhaustive: E₁ ∪ E₂ ∪ … ∪ Eₙ = S — at least one must occur.

Events that are both mutually exclusive and exhaustive partition the sample space.

Why a new approach?

The classical definition needs equally likely outcomes and a finite sample space, so it fails for many real situations.

The statistical approach needs the experiment to be repeated a very large number of times, which is not always possible.

The axiomatic approach avoids both limitations: any assignment of probabilities that obeys three simple rules (axioms) is valid, and the classical probabilities turn out to be a special case.

14.2 Axiomatic approach to probability

A probability P assigns to each event E a number P(E) such that:

P(E) ≥ 0 for every event E;

P(S) = 1;

If E and F are mutually exclusive, P(E ∪ F) = P(E) + P(F).

Consequences: P(φ) = 0, and for sample points ω₁, …, ωₙ, each 0 ≤ P(ωᵢ) ≤ 1 and ΣP(ωᵢ) = 1.

14.2.1 Probability of an event: P(A) = sum of the probabilities of the sample points in A.

14.2.2 Equally likely outcomes: if S has n equally likely outcomes and E has m of them, P(E) = m/n.

14.2.3 'A or B': P(A ∪ B) = P(A) + P(B) − P(A ∩ B); for mutually exclusive events this reduces to P(A) + P(B).

14.2.4 'not A': P(A′) = 1 − P(A).

Worked Examples

Mutually exclusive pairs (Example 2): two dice (36 outcomes); A: sum even (18 outcomes), B: sum a multiple of 3 (12), C: sum less than 4 (3), D: sum greater than 11 (1, namely (6, 6)). A ∩ B, A ∩ C, A ∩ D, B ∩ C and B ∩ D are all non-empty (for example (6, 6) is in A, B and D), so the only mutually exclusive pair is C and D.

Mutually exclusive and exhaustive (Example 3): a coin tossed three times gives 8 outcomes. A: no head = {TTT}; B: exactly one head = {HTT, THT, TTH}; C: at least two heads = {HHT, HTH, THH, HHH}. A ∪ B ∪ C = S and the events are pairwise disjoint, so they are mutually exclusive and exhaustive.

Cards (Example 5): one card from a deck of 52. P(diamond) = 13/52 = 1/4; P(not an ace) = 1 − 4/52 = 12/13; P(black card) = 26/52 = 1/2; P(not a diamond) = 3/4.

Discs (Example 6): 9 discs — 4 red, 3 blue, 2 yellow. P(red) = 4/9, P(yellow) = 2/9, P(blue) = 1/3, P(not blue) = 2/3, P(red or blue) = 7/9 (mutually exclusive, so add).

Examination (Example 7): P(Anil qualifies) = 0.05, P(Ashima) = 0.10, P(both) = 0.02.

- P(neither qualifies) = 1 − P(E ∪ F) = 1 − (0.05 + 0.10 − 0.02) = 0.87.

- P(at least one does not qualify) = 1 − P(both) = 0.98.

- P(only one qualifies) = P(E) + P(F) − 2P(E ∩ F) = 0.11.

Committee (Example 8): 2 persons from 2 men and 2 women (⁴C₂ = 6 ways). P(no man) = 1/6, P(one man) = (²C₁ × ²C₁)/6 = 2/3, P(two men) = 1/6.

Kings (Example 10): a hand of 7 cards. P(all 4 Kings) = ⁴C₄·⁴⁸C₃/⁵²C₇ = 1/7735; P(exactly 3 Kings) = ⁴C₃·⁴⁸C₄/⁵²C₇ = 9/1547; P(at least 3 Kings) = 46/7735.

Relay race (Example 12): five teams, first three places: ⁵P₃ = 60 equally likely orders. P(A, B, C finish 1st, 2nd, 3rd in that order) = 1/60; P(A, B, C are the first three in any order) = 3!/60 = 1/10.

Formula Sheet

P(E) = n(E)/n(S) for equally likely outcomes

0 ≤ P(E) ≤ 1; P(S) = 1; P(φ) = 0

P(A′) = 1 − P(A)

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Mutually exclusive: P(A ∪ B) = P(A) + P(B)

P(only one of E, F) = P(E) + P(F) − 2P(E ∩ F)

Counting: ⁿCᵣ = n!/[r!(n − r)!], ⁿPᵣ = n!/(n − r)!

Exam Focus

Write the sample space first (Exercise 14.1): for two dice there are 6² = 36 outcomes; for three coins, 2³ = 8.

- Outline: P(sum = 7 with two dice) = 6/36 = 1/6.

Use A ∪ B with the correction term (Exercise 14.2): given P(A) = 1/4, P(B) = 1/2, P(A ∩ B) = 1/8, P(A ∪ B) = 1/4 + 1/2 − 1/8 = 5/8.

Combination-based probability (Miscellaneous): count favourable selections with ⁿCᵣ and divide by total selections, exactly as in the committee and Kings examples.

Key Terms

Random experiment (यादृच्छिक परीक्षण)

Sample space (प्रतिदर्श समष्टि)

Event (घटना)

Impossible / Sure event (असंभव / निश्चित घटना)

Simple / Compound event (सरल / मिश्र घटना)

Complementary event (पूरक घटना)

Mutually exclusive events (परस्पर अपवर्जी घटनाएँ)

Exhaustive events (निःशेष घटनाएँ)

Equally likely outcomes (सम संभाव्य परिणाम)

Axiomatic approach (अभिगृहीतीय दृष्टिकोण)

Probability (प्रायिकता)

Combination / Permutation (संचय / क्रमचय)

Common Mistakes

Adding P(A) + P(B) for events that are not mutually exclusive — you must subtract P(A ∩ B).

Confusing mutually exclusive (cannot happen together) with exhaustive (together cover S).

Writing a probability greater than 1 or negative — always check 0 ≤ P ≤ 1.

Using permutations when order does not matter (committees, card hands need combinations).

Forgetting that 'at least one' is often easiest as 1 − P(none).

Treating non-equally-likely outcomes as equally likely.

💡 Key Learning Points

✓Events are subsets of the sample space; they may be impossible, sure, simple or compound.
✓Algebra of events: A′, A ∪ B, A ∩ B and A − B, plus mutually exclusive and exhaustive events.
✓The axiomatic approach (Kolmogorov, 1933): P(E) ≥ 0, P(S) = 1, and P(E ∪ F) = P(E) + P(F) for mutually exclusive events.
✓For equally likely outcomes P(E) = n(E)/n(S); P(A′) = 1 − P(A) and P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
✓Combinations and permutations are used to count favourable outcomes in card, committee and ordering problems.

👨‍🏫 Teaching Tips

→Begin by listing sample spaces for a die, two coins and two dice so students see that events are simply subsets.
→Use Venn diagrams to show A ∪ B, A ∩ B, A′ and A − B, and to explain why P(A ∩ B) is subtracted in the addition rule.
→Contrast 'odd and even' (mutually exclusive and exhaustive) with 'less than 4 and greater than 2' (neither) on a die.
→Work the Anil–Ashima example as a Venn diagram with regions 0.03, 0.02, 0.08 and 0.87 to make 'only one' and 'neither' visible.
→Have groups draw cards from a real deck to check P(diamond) and P(not ace) experimentally, then compare with the classical values.
→Teach a checklist for counting problems: Does order matter? → permutation; order irrelevant → combination.

📋 Assessment Questions

Can the student write the sample space and identify an event as a subset?
Can the student classify events as mutually exclusive and/or exhaustive?
Can the student state the three axioms of probability and derive P(A′) = 1 − P(A)?
Can the student apply P(A ∪ B) = P(A) + P(B) − P(A ∩ B) correctly?
Can the student solve probability problems that require combinations (cards, committees)?

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