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Class 11 Maths NCERT Solutions

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Probability Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 14, Probability — 28 questions from Ex 14.1 to Ex 14.2, each worked through step by step in the CBSE marking pattern. The addition theorem, conditional probability, multiplication, independence and total probability.

Class:11Subject:MathsChapter:14
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 14?

Chapter 14 carries 2 exercise questions, numbered Ex 14.1 to Ex 14.2. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

Probability assigns numbers between 0 and 1 to events built from a sample space. This chapter covers the language of events — mutually exclusive, exhaustive, simple and compound — the addition rule P(A∪B) = P(A) + P(B) − P(A∩B), and its complement form P(not A) = 1 − P(A). Counting (combinations) powers the probability questions on cards, coins, dice and lots.

Board pattern

Always identify the sample space first and count outcomes using ⁿCᵣ when order does not matter. A valid probability assignment needs every P(ωᵢ) ≥ 0 and their sum exactly 1. For 'at least one' always use 1 − P(none), and for any two events remember P(A∪B) = P(A) + P(B) − P(A∩B) — this one formula covers most of the exercise.
02

Exercise 14.1 — Events Built From a Sample Space

7Exercise questions

Step-by-step solution

  1. 1E = {4} and F = {2, 4, 6}.
  2. 2E ∩ F = {4} ≠ ∅, so they have a common outcome.

Final answer

No — not mutually exclusive, since 4 belongs to both.

Step-by-step solution

  1. 1A = {1,2,3,4,5,6} = S, B = ∅, C = {3,6}, D = {1,2,3}, E = {6}, F = {3,4,5,6}.
  2. 2A ∪ B = {1,2,3,4,5,6}; A ∩ B = ∅; B ∪ C = {3,6}.
  3. 3E ∩ F = {6}; D ∩ E = ∅.
  4. 4A − C = {1,2,4,5}; D − E = {1,2,3}.
  5. 5F′ = {1,2,5,6} and E ∩ F′ = {6} = E.

Final answer

A = S, B = ∅, C = {3,6}, D = {1,2,3}, E = {6}, F = {3,4,5,6}; A∪B = S, A∩B = ∅, B∪C = {3,6}, E∩F = {6}, D∩E = ∅, A−C = {1,2,4,5}, D−E = {1,2,3}, E∩F′ = {6}, F′ = {1,2,5,6}.

Step-by-step solution

  1. 1A = {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}.
  2. 2B = outcomes with a 2, e.g. (2,1)…(2,6),(1,2),(3,2),(4,2),(5,2),(6,2).
  3. 3C = {(3,6),(4,5),(5,4),(6,3)} (sum 9).
  4. 4A ∩ C = C ≠ ∅ and A ∩ B = {(2,6),(6,2)} ≠ ∅, but B ∩ C = ∅.

Final answer

Only the pair (B, C) is mutually exclusive.

Step-by-step solution

  1. 1A = {HHH}, B = {HHT, HTH, THH}, C = {TTT}, D = {HHH, HHT, HTH, HTT}.
  2. 2(i) Pairs with empty intersection: (A,B), (A,C), (B,C) and (C,D).
  3. 3(ii) Simple (single outcome): A = {HHH} and C = {TTT}.
  4. 4(iii) Compound (multiple outcomes): B and D.

Final answer

(i) (A,B), (A,C), (B,C), (C,D); (ii) A and C; (iii) B and D.

Step-by-step solution

  1. 1Take S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
  2. 2(i) A = {HHH} and B = {TTT} are mutually exclusive.
  3. 3(ii) 3 heads: {HHH}; exactly 2 heads: {HHT, HTH, THH}; at most 1 head: {HTT, THT, TTH, TTT} — exclusive and cover S.
  4. 4(iii) A = {HHH, HHT, HTH, HTT} (head first) and B = {HHH, HHT, THH, THT} (head second) share {HHH, HHT}.
  5. 5(iv) A = {HHH} and B = {TTT}: exclusive, union ≠ S.
  6. 6(v) {HHH}, {TTT}, {HHT, HTH, THH}: pairwise disjoint, union ≠ S.

Final answer

Any valid example is acceptable; the ones above use heads/tails patterns as listed.

Step-by-step solution

  1. 1A = {x even on first die}; B = {x odd}; C = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}.
  2. 2(i) A′ = odd on first die = B; (ii) not B = A.
  3. 3(iii) A or B = S; (iv) A and B = ∅.
  4. 4(v) A but not C = A ∩ C′ = 14 outcomes with first die even and sum ≥ 6: (2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1)…(6,6).
  5. 5(vi) B or C: first die odd or sum ≤ 5; (vii) B and C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}.
  6. 6(viii) B′ = A, so A ∩ B′ ∩ C′ = A ∩ C′ — the same 14 outcomes as (v).

Final answer

A′ = B; not B = A; A or B = S; A and B = ∅; A but not C = 14 listed outcomes; B and C = 6 listed outcomes; A∩B′∩C′ equals (v).

Step-by-step solution

  1. 1(i) True — even and odd on the first die cannot both happen.
  2. 2(ii) True — every outcome has first die either even or odd, and the two never overlap.
  3. 3(iii) True — B′ means first die is even, which is exactly A.
  4. 4(iv) False — A ∩ C = {(2,1),(2,2),(2,3),(4,1)} ≠ ∅.
  5. 5(v) False — B′ = A, so A and B′ coincide; their intersection is A, not empty.
  6. 6(vi) False — A′ = B, B′ = A and A ∪ B = S, but A ∩ C ≠ ∅, so they are not mutually exclusive.

Final answer

(i) T, (ii) T, (iii) T, (iv) F, (v) F, (vi) F — reasons above.

03

Exercise 14.2 — Axiomatic Probability and the Addition Rule

21Exercise questions

Step-by-step solution

  1. 1(a) Sum = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1, all non-negative → valid.
  2. 2(b) 7 × (1/7) = 1 → valid.
  3. 3(c) Sum = 2.8 ≠ 1 → invalid.
  4. 4(d) Contains negative values (−0.1, −0.2) → invalid.
  5. 5(e) Sum = 36/14 > 1 → invalid.

Final answer

Invalid: (c), (d), (e). Valid: (a), (b).

Step-by-step solution

  1. 1P(at least one tail) = 1 − P(no tail) = 1 − P(HH).
  2. 2P(HH) = 1/4 → answer = 3/4.

Final answer

3/4.

Step-by-step solution

  1. 1(i) Prime faces: 2, 3, 5 → 3/6 = 1/2.
  2. 2(ii) {3,4,5,6} → 4/6 = 2/3.
  3. 3(iii) {1} → 1/6.
  4. 4(iv) No face exceeds 6 → 0.
  5. 5(v) {1,2,3,4,5} → 5/6.

Final answer

(i) 1/2, (ii) 2/3, (iii) 1/6, (iv) 0, (v) 5/6.

Step-by-step solution

  1. 1(a) 52 cards → 52 points.
  2. 2(b) Exactly one ace of spades → 1/52.
  3. 3(c)(i) 4 aces → 4/52 = 1/13.
  4. 4(c)(ii) 26 black cards → 26/52 = 1/2.

Final answer

(a) 52, (b) 1/52, (c)(i) 1/13, (c)(ii) 1/2.

Step-by-step solution

  1. 1Sample space: (1,1)…(1,6),(6,1)…(6,6): 12 equally likely outcomes.
  2. 2(i) Sum 3 only from (1,2) → 1/12.
  3. 3(ii) Sum 12 only from (6,6) → 1/12.

Final answer

(i) 1/12, (ii) 1/12.

Step-by-step solution

  1. 1Total members = 10; women = 6.
  2. 2P(woman) = 6/10 = 3/5.

Final answer

3/5.

Step-by-step solution

  1. 1If H heads occur, T = 4 − H tails, so amount = H − 1.5(4 − H) = 2.5H − 6.
  2. 2H = 0, 1, 2, 3, 4 give 5 different amounts: −6, −3.5, −1, 1.5, 4.
  3. 3P(H = 0) = 1/16; P(H = 1) = 4/16 = 1/4; P(H = 2) = 6/16 = 3/8; P(H = 3) = 4/16 = 1/4; P(H = 4) = 1/16.

Final answer

Five amounts: Rs −6 (P 1/16), −3.5 (1/4), −1 (3/8), 1.5 (1/4), 4 (1/16).

Step-by-step solution

  1. 18 equally likely outcomes.
  2. 2(i) 1/8; (ii) 3/8 (HHT, HTH, THH); (iii) 4/8 = 1/2; (iv) 7/8; (v) 1/8.
  3. 3(vi) 1/8; (vii) 3/8; (viii) 1/8; (ix) 7/8.

Final answer

(i) 1/8, (ii) 3/8, (iii) 1/2, (iv) 7/8, (v) 1/8, (vi) 1/8, (vii) 3/8, (viii) 1/8, (ix) 7/8.

Step-by-step solution

  1. 1P(not A) = 1 − 2/11 = 9/11.

Final answer

9/11.

Step-by-step solution

  1. 1Word has 13 letters: A×3, S×4, I×2, N×2, T×1, O×1.
  2. 2(i) Vowels A, I, O → 6/13.
  3. 3(ii) Consonants S, N, T → 7/13.

Final answer

(i) 6/13, (ii) 7/13.

Step-by-step solution

  1. 1Total ways to choose 6 numbers from 20 = C(20, 6).
  2. 2C(20,6) = 38760; only one selection wins.
  3. 3P = 1/38760.

Final answer

1/38760.

Step-by-step solution

  1. 1(i) A ∩ B ⊆ A, so P(A ∩ B) ≤ P(A) is required; here 0.6 > 0.5 → inconsistent.
  2. 2(ii) P(A ∩ B) = 0.5 + 0.4 − 0.8 = 0.1 ≥ 0 → consistent.

Final answer

(i) Not consistent; (ii) consistent.

Step-by-step solution

  1. 1(i) P(A∪B) = 1/3 + 1/5 − 1/15 = 5/15 + 3/15 − 1/15 = 7/15.
  2. 2(ii) P(B) = 0.6 − 0.35 + 0.25 = 0.5.
  3. 3(iii) P(A∩B) = 0.5 + 0.35 − 0.7 = 0.15.

Final answer

(i) 7/15, (ii) 0.5, (iii) 0.15.

Step-by-step solution

  1. 1For mutually exclusive events, P(A ∩ B) = 0.
  2. 2P(A or B) = 3/5 + 1/5 = 4/5.

Final answer

4/5.

Step-by-step solution

  1. 1(i) P(E or F) = 1/4 + 1/2 − 1/8 = 2/8 + 4/8 − 1/8 = 5/8.
  2. 2(ii) P(not E and not F) = 1 − P(E or F) = 3/8.

Final answer

(i) 5/8, (ii) 3/8.

Step-by-step solution

  1. 1By De Morgan, E′ ∪ F′ = (E ∩ F)′, so P(E ∩ F) = 1 − 0.25 = 0.75.
  2. 2P(E ∩ F) ≠ 0, so E and F are not mutually exclusive.

Final answer

Not mutually exclusive (P(E ∩ F) = 0.75).

Step-by-step solution

  1. 1(i) P(not A) = 1 − 0.42 = 0.58.
  2. 2(ii) P(not B) = 1 − 0.48 = 0.52.
  3. 3(iii) P(A or B) = 0.42 + 0.48 − 0.16 = 0.74.

Final answer

(i) 0.58, (ii) 0.52, (iii) 0.74.

Step-by-step solution

  1. 1P(M or B) = P(M) + P(B) − P(M and B).
  2. 2= 0.4 + 0.3 − 0.1 = 0.6.

Final answer

0.6 (i.e. 60%).

Step-by-step solution

  1. 1P(both) = P(first) + P(second) − P(at least one).
  2. 2= 0.8 + 0.7 − 0.95 = 0.55.

Final answer

0.55.

Step-by-step solution

  1. 1P(E or H) = 1 − P(neither) = 1 − 0.1 = 0.9.
  2. 2P(H) = P(E or H) − P(E) + P(E and H) = 0.9 − 0.75 + 0.5 = 0.65.

Final answer

0.65.

Step-by-step solution

  1. 1(i) P(NCC or NSS) = (30 + 32 − 24)/60 = 38/60 = 19/30.
  2. 2(ii) Neither = 1 − 19/30 = 11/30.
  3. 3(iii) NSS only = (32 − 24)/60 = 8/60 = 2/15.

Final answer

(i) 19/30, (ii) 11/30, (iii) 2/15.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Addition theorem

Conditional probability

Multiplication rule

Mutually exclusive

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Add only when the events are mutually exclusive; if they overlap you must subtract the intersection or the count is doubled.
  • Independence means P(A ∩ B) = P(A)P(B), which is a statement about the probabilities, not a guess based on how the events look.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 14 (Probability)?

There are 2 exercise questions in this chapter, numbered Ex 14.1 to Ex 14.2. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Probability Class 11 Maths?

The formulas this chapter's questions actually turn on are: Addition theorem, Conditional probability, Multiplication rule, Mutually exclusive. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Probability important for JEE Main?

Very important — conditional probability and the addition theorem are used in Class 12 probability and in JEE Main every year, so this chapter pays off twice.

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