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Class 12 Maths NCERT Solutions

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Probability Class 12 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 13, Probability — 102 questions from Ex 13.1 to Ex 13.5, each worked through step by step in the CBSE marking pattern. Conditional probability, multiplication theorem, independence, Bayes' theorem, and the mean and variance of a random variable.

Class:12Subject:MathsChapter:13
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 12 Maths Chapter 13?

Chapter 13 carries 5 exercise questions, numbered Ex 13.1 to Ex 13.5. 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 is the branch of mathematics that quantifies uncertainty. This chapter extends the basic ideas of Class 10 and Class 11 into the territory of conditional probability, Bayes' theorem, random variables and the binomial distribution. Every question below is from the NCERT Class 12 textbook (rationalized edition); each carries short, exam-ready working.

Board pattern

Conditional probability and Bayes' theorem questions reward a clear statement of the given events, the partition, and a labelled tree diagram where applicable. Random variable questions award marks for writing the probability distribution table and using the variance formula. Always show the substitution into the formula — a bare final answer scores zero.
02

Exercise 13.1 — Conditional Probability

22Exercise questions

Step-by-step solution

  1. 1P(A|B) = P(A ∩ B)/P(B) = 0.18/0.30 = 0.6.
  2. 2P(B|A) = P(A ∩ B)/P(A) = 0.18/0.60 = 0.3.
  3. 3P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.6 + 0.3 − 0.18 = 0.72.

Final answer

P(A|B) = 0.6, P(B|A) = 0.3, P(A ∪ B) = 0.72.

Step-by-step solution

  1. 1Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, |S| = 8.
  2. 2(i) B = {HHH, HHT}, A ∩ B = {HHH}, so P(A|B) = 1/2.
  3. 3(ii) A = {HHT, HTH, THH, HHH}; B = {HHT, HTH, HTT, THH, THT, TTH, TTT}; A ∩ B = {HHT, HTH, THH}; P(A|B) = 3/7.

Final answer

(i) 1/2; (ii) 3/7.

Step-by-step solution

  1. 1S = {HH, HT, TH, TT}; A = {HT, TH}; B = {HH}; A ∩ B = φ.
  2. 2P(A|B) = 0.

Final answer

Step-by-step solution

  1. 1B = {63x : x = 1,...,6}, |B| = 6. A ∩ B = {634}, |A ∩ B| = 1.
  2. 2P(A|B) = 1/6.

Final answer

Step-by-step solution

  1. 1S = {MFS, MSF, FMS, FSM, SMF, SFM}, |S| = 6.
  2. 2A = {MFS, FMS, SMF, SFM}; B = {MFS, SFM}; A ∩ B = {MFS, SFM}; P(A|B) = 2/2 = 1.
  3. 3P(B|A) = 2/4 = 1/2.

Final answer

Step-by-step solution

  1. 1B has 3 × 6 = 18 outcomes. A ∩ B = {(1,6),(2,5),(3,4)}: 3 outcomes.
  2. 2P(A|B) = 3/18 = 1/6.

Final answer

Step-by-step solution

  1. 1P(red) = 3/5, P(black) = 2/5.
  2. 2P(at least one red) = 1 − P(both black) = 1 − (2/5)² = 1 − 4/25 = 21/25.

Final answer

Step-by-step solution

  1. 1Face cards = 12 (J,Q,K of 4 suits). Kings among face cards = 4.
  2. 2P(A|B) = 4/12 = 1/3.

Final answer

Step-by-step solution

  1. 1P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.4 + 0.5 − 0.7 = 0.2.
  2. 2P(A|B) = 0.2/0.5 = 0.4; P(B|A) = 0.2/0.4 = 0.5.

Final answer

Step-by-step solution

  1. 1P(B) = 5/13. P(A|B) = P(A ∩ B)/P(B) = 2/5, so P(A ∩ B) = (2/5)(5/13) = 2/13.
  2. 22P(A) = 5/13 ⇒ P(A) = 5/26.
  3. 3P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 5/26 + 5/13 − 2/13 = 5/26 + 3/13 = 5/26 + 6/26 = 11/26.

Final answer

Step-by-step solution

  1. 1P(not A or not B) = P((A ∩ B)ᶜ) = 1 − P(A ∩ B) = 1/4, so P(A ∩ B) = 3/4.
  2. 2P(A) · P(B) = (1/2)(7/12) = 7/24 ≠ 3/4.
  3. 3P(A ∩ B) ≠ P(A)·P(B), so A and B are NOT independent.

Final answer

A and B are not independent.

Final answer

Step-by-step solution

  1. 1(i) P(A ∩ B) = 0.3 × 0.4 = 0.12.
  2. 2(ii) P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58.
  3. 3(iii) P(A|B) = P(A) = 0.3 (independent).
  4. 4(iv) P(B|A) = P(B) = 0.4.

Final answer

(i) 0.12; (ii) 0.58; (iii) 0.3; (iv) 0.4.

Step-by-step solution

  1. 1P(A)·P(B) = 0.5 × 0.6 = 0.3 = P(A ∩ B).

Final answer

Yes, A and B are independent.

Final answer

Step-by-step solution

  1. 1P(first ace | second ace) = P(both ace)/P(second ace).
  2. 2P(both ace) = (4/52)(3/51) = 12/2652 = 1/221.
  3. 3P(second ace) = 4/52 = 1/13 (by symmetry).
  4. 4P(first ace | second ace) = (1/221)/(1/13) = 13/221 = 1/17.

Final answer

Step-by-step solution

  1. 1E = {1,2,3}, F = {1,3,5}. E ∩ F = {1,3}.
  2. 2P(E|F) = 2/3, P(F|E) = 2/3.

Final answer

Step-by-step solution

  1. 1After drawing one white ball, 4 white and 3 black remain (7 total).
  2. 2P(second white | first white) = 4/7.

Final answer

Step-by-step solution

  1. 1Numbers greater than 3: {4,5,6}. Even numbers among them: {4,6}.
  2. 2P(even | >3) = 2/3.

Final answer

Step-by-step solution

  1. 1Outcomes with at least one 4: {(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(1,4),(2,4),(3,4),(5,4),(6,4)}, 11 outcomes.
  2. 2Among these, sum = 7: {(4,3),(3,4)}, 2 outcomes.
  3. 3P = 2/11.

Final answer

Step-by-step solution

  1. 1Sample space: {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}.
  2. 2At least one boy: 7 outcomes (all except GGG).
  3. 3Exactly 2 boys: {BBG, BGB, GBB}, 3 outcomes.
  4. 4P = 3/7.

Final answer

Step-by-step solution

  1. 1P(A|B) = 0.2/0.4 = 0.5, so P(A|B) × P(B) = 0.5 × 0.4 = 0.2.
  2. 2P(B|A) = 0.2/0.5 = 0.4, so P(B|A) × P(A) = 0.4 × 0.5 = 0.2.
  3. 3Both sides equal P(A ∩ B) = 0.2. Verified.

Final answer

Both sides equal 0.2. Verified.

03

Exercise 13.2 — Multiplication Theorem and Independent Events

23Exercise questions

Step-by-step solution

  1. 1P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 6/11 + 5/11 − 7/11 = 4/11.
  2. 2P(A|B) = (4/11)/(5/11) = 4/5.
  3. 3P(B|A) = (4/11)/(6/11) = 4/6 = 2/3.

Final answer

Step-by-step solution

  1. 1X = 0,1,2,3 with P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8.

Final answer

Distribution: 0→1/8, 1→3/8, 2→3/8, 3→1/8.

Step-by-step solution

  1. 1B = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}; |B| = 6. A ∩ B = {(3,4)}, |A ∩ B| = 1.
  2. 2P(A|B) = 1/6; P(B|A) = 1/6.
  3. 3P(A) = 1/6, so P(A|B) = P(A) and P(B|A) = P(B). Independent.

Final answer

P(A|B) = P(B|A) = \frac{1}{6};\; A \text{ and } B \text{ are independent.

Step-by-step solution

  1. 1These are independent throws: P(4 on first) = 1/6. On second throw, sum = 6 from {(1,5),(2,4),(3,3),(4,2),(5,1)}: 5/36.
  2. 2P = (1/6)(5/36) = 5/216.

Final answer

Step-by-step solution

  1. 1P(both white) = (3/5)(2/4) = 6/20 = 3/10.

Final answer

Step-by-step solution

  1. 1P(red) = P(first bag)·P(red|first) + P(second bag)·P(red|second)
  2. 2= (1/2)(4/8) + (1/2)(2/8) = 1/4 + 1/8 = 3/8.

Final answer

Step-by-step solution

  1. 1P(F ∪ C) = P(F) + P(C) − P(F ∩ C) = 0.3 + 0.4 − 0.1 = 0.6.

Final answer

Step-by-step solution

  1. 1(i) Perfect squares: 1,4,9,...,100 → 10 values → 10/100 = 1/10.
  2. 2(ii) Perfect cubes: 1,8,27,64 → 4 values → 4/100 = 1/25.

Final answer

(i) 1/10; (ii) 1/25.

Step-by-step solution

  1. 1P(2) = 1/6, so P(not 2) = 1 − 1/6 = 5/6.
  2. 21/6 + 5/6 = 1.

Final answer

Step-by-step solution

  1. 1P(H) = 2P(T) and P(H) + P(T) = 1.
  2. 22P(T) + P(T) = 1 ⇒ P(T) = 1/3, P(H) = 2/3.

Final answer

Step-by-step solution

  1. 1P(first king) = 4/52 = 1/13.
  2. 2After removing one king, 51 cards remain. Queen of the same suit is still present (king and queen are different cards): 1/51.
  3. 3P = (1/13)(1/51) = 1/663.

Final answer

Step-by-step solution

  1. 1Total = 14 balls. P(R) = 7/14 = 1/2, P(W) = 4/14 = 2/7, P(B) = 3/14.
  2. 2(i) P(RR) = (1/2)(1/2) = 1/4.
  3. 3(ii) P(RW) + P(WR) = (1/2)(2/7) + (2/7)(1/2) = 2/7.
  4. 4(iii) 1 − P(no red) = 1 − (5/7)² = 1 − 25/49 = 24/49.

Final answer

(i) 1/4; (ii) 2/7; (iii) 24/49.

Step-by-step solution

  1. 1P(A ∪ B) = P(A) + P(B) − P(A)P(B) (independent).
  2. 20.6 = 0.2 + P(B) − 0.2·P(B) = 0.2 + 0.8P(B).
  3. 30.4 = 0.8P(B) ⇒ P(B) = 0.5.

Final answer

Step-by-step solution

  1. 1After drawing one red, 2 red and 2 blue remain (4 total).
  2. 2P(second red | first red) = 2/4 = 1/2.

Final answer

Step-by-step solution

  1. 1P = (1/6)(5/6)(5/6) = 25/216.

Final answer

Step-by-step solution

  1. 1Let L = late. P(L) = (0.3)(0.10) + (0.2)(0.05) + (0.1)(0.12) + (0.4)(0.01)
  2. 2= 0.03 + 0.01 + 0.012 + 0.004 = 0.056.
  3. 3P(bus | L) = 0.03/0.056 = 30/56 = 15/28.

Final answer

Step-by-step solution

  1. 1Binomial: n = 5, p = 0.1.
  2. 2P(X = 0) = (0.9)⁵ = 0.59049.
  3. 3P(X ≤ 2) = P(0) + P(1) + P(2) = (0.9)⁵ + 5(0.1)(0.9)⁴ + 10(0.01)(0.9)³
  4. 4= 0.59049 + 0.32805 + 0.07290 = 0.99144.

Final answer

P(none defective) = 0.59049; P(at most 2 defective) = 0.99144.

Final answer

Step-by-step solution

  1. 1P(both kings) = (4/52)(3/51) = 12/2652 = 1/221.

Final answer

Step-by-step solution

  1. 1With replacement: p(red) = 5/8 each draw.
  2. 2P = (5/8)³ = 125/512.

Final answer

Step-by-step solution

  1. 1P(no head) = (1/2)ⁿ = 1/64.
  2. 22ⁿ = 64 = 2⁶, so n = 6.

Final answer

Step-by-step solution

  1. 1P(A ∪ B) = P(A) + P(B) − P(A)P(B) (independent).
  2. 22/3 = 1/3 + P(B)(1 − 1/3) = 1/3 + (2/3)P(B).
  3. 31/3 = (2/3)P(B), so P(B) = 1/2.

Final answer

Step-by-step solution

  1. 1P(not solved) = (1 − 1/2)(1 − 1/3)(1 − 1/4) = (1/2)(2/3)(3/4) = 6/24 = 1/4.
  2. 2P(solved) = 1 − 1/4 = 3/4.

Final answer

04

Exercise 13.3 — Bayes' Theorem

18Exercise questions

Step-by-step solution

  1. 1P(A) = 0.6, P(B) = 0.4, P(D|A) = 0.02, P(D|B) = 0.05.
  2. 2P(D) = 0.6(0.02) + 0.4(0.05) = 0.012 + 0.020 = 0.032.
  3. 3P(A|D) = 0.012/0.032 = 12/32 = 3/8.

Final answer

Step-by-step solution

  1. 1P(TwoHeaded) = 1/3, P(H|TwoHeaded) = 1; P(Biased) = 1/3, P(H|Biased) = 3/4; P(Fair) = 1/3, P(H|Fair) = 1/2.
  2. 2P(H) = (1/3)(1) + (1/3)(3/4) + (1/3)(1/2) = 1/3 + 1/4 + 1/6 = 4/12 + 3/12 + 2/12 = 9/12 = 3/4.
  3. 3P(TwoHeaded|H) = (1/3)/(3/4) = 4/9.

Final answer

Step-by-step solution

  1. 1P(Uᵢ) = 1/3 each. P(2W|I) = C(2,2)/C(5,2) = 1/10; P(2W|II) = C(4,2)/C(5,2) = 6/10; P(2W|III) = C(3,2)/C(7,2) = 3/21 = 1/7.
  2. 2P(2W) = (1/3)(1/10) + (1/3)(6/10) + (1/3)(1/7) = 1/30 + 6/30 + 1/21
  3. 3= 7/30 + 1/21 = 49/210 + 10/210 = 59/210.
  4. 4P(III|2W) = (1/21)/(59/210) = 10/59.

Final answer

Step-by-step solution

  1. 1P(G) = 0.6, P(B) = 0.4, P(P|G) = 0.7, P(P|Bo) = 0.85.
  2. 2P(P) = 0.6(0.7) + 0.4(0.85) = 0.42 + 0.34 = 0.76.
  3. 3P(Bo|P) = 0.34/0.76 = 17/38.

Final answer

Step-by-step solution

  1. 1P(all good) = C(7,3)/C(10,3) = 35/120 = 7/24.
  2. 2P(at least one good) = 1 − P(all defective) = 1 − C(3,3)/C(10,3) = 1 − 1/120 = 119/120.
  3. 3P(all good | at least one good) = (7/24)/(119/120) = (7/24)(120/119) = 35/119 = 5/17.

Final answer

Step-by-step solution

  1. 1P(R) = (1/2)(1/2) + (1/2)(1/4) = 1/4 + 1/8 = 3/8.
  2. 2P(1st|R) = (1/4)/(3/8) = 2/3.

Final answer

Step-by-step solution

  1. 1P(reported six) = P(reported six|actual 6)P(6) + P(reported six|not 6)P(not 6)
  2. 2= (3/4)(1/6) + (1/4)(5/6) = 3/24 + 5/24 = 8/24 = 1/3.
  3. 3P(actual 6 | reported 6) = (3/4 × 1/6)/(1/3) = (3/24)/(1/3) = (1/8)/(1/3) = 3/8.

Final answer

Step-by-step solution

  1. 1P(D) = 0.005, P(Dᶜ) = 0.995. P(+|D) = 0.99, P(+|Dᶜ) = 0.01.
  2. 2P(+) = 0.99(0.005) + 0.01(0.995) = 0.00495 + 0.00995 = 0.01490.
  3. 3P(D|+) = 0.00495/0.01490 = 495/1490 ≈ 0.3322.

Final answer

Step-by-step solution

  1. 1P(Below) = 0.25, P(Mid) = 0.55, P(Above) = 0.20.
  2. 2P(crime) = 0.25(0.02) + 0.55(0.01) + 0.20(0.03) = 0.005 + 0.0055 + 0.006 = 0.0165.
  3. 3P(Above|crime) = 0.006/0.0165 = 60/165 = 4/11.

Final answer

Step-by-step solution

  1. 1P(both red) = (5/8)(3/7) = 15/56.
  2. 2P(at least one red) = 1 − P(both blue) = 1 − (3/8)(2/7) = 1 − 6/56 = 50/56.
  3. 3P(both red | at least one red) = (15/56)/(50/56) = 15/50 = 3/10.

Final answer

Step-by-step solution

  1. 1P(at least one H) = 1 − 1/8 = 7/8.
  2. 2P(exactly two H) = 3/8.
  3. 3P(exactly two H | at least one H) = (3/8)/(7/8) = 3/7.

Final answer

Step-by-step solution

  1. 1P(not late | bus) = 0.90, P(not late | taxi) = 0.95, P(not late | bike) = 0.88, P(not late | car) = 0.99.
  2. 2P(not late) = 0.3(0.90) + 0.2(0.95) + 0.1(0.88) + 0.4(0.99) = 0.27 + 0.19 + 0.088 + 0.396 = 0.944.
  3. 3P(car | not late) = 0.396/0.944 = 396/944 = 99/236.

Final answer

Step-by-step solution

  1. 1P(win) = (1/3)(1/5) + (1/3)(1/4) + (1/3)(1/6) = 1/15 + 1/12 + 1/18
  2. 2= 12/180 + 15/180 + 10/180 = 37/180.
  3. 3P(B|win) = (1/12)/(37/180) = (15/180)/(37/180) = 15/37.

Final answer

Step-by-step solution

  1. 1P(D) = 0.6(0.02) + 0.4(0.03) = 0.012 + 0.012 = 0.024.
  2. 2P(2nd|D) = 0.012/0.024 = 1/2.

Final answer

Step-by-step solution

  1. 1P(R) = (1/2)(3/7) + (1/2)(5/11) = 3/14 + 5/22 = 33/154 + 35/154 = 68/154 = 34/77.
  2. 2P(I|R) = (3/14)/(34/77) = (3/14)(77/34) = 231/476 = 33/68.

Final answer

Step-by-step solution

  1. 1P(D) = 0.30(0.02) + 0.45(0.03) + 0.25(0.05) = 0.006 + 0.0135 + 0.0125 = 0.032.
  2. 2P(C|D) = 0.0125/0.032 = 125/320 = 25/64.

Final answer

Step-by-step solution

  1. 1P(exactly one hits) = P(A only) + P(B only) + P(C only)
  2. 2= (4/5)(1/4)(1/3) + (1/5)(3/4)(1/3) + (1/5)(1/4)(2/3)
  3. 3= 4/60 + 3/60 + 2/60 = 9/60 = 3/20.
  4. 4P(B|exactly one) = (3/60)/(9/60) = 3/9 = 1/3.

Final answer

Step-by-step solution

  1. 1P(D₂) = P(D₁)P(D₂|D₁) + P(not D₁)P(D₂|not D₁)
  2. 2= (3/10)(2/9) + (7/10)(3/9) = 6/90 + 21/90 = 27/90 = 3/10.
  3. 3P(D₁|D₂) = (6/90)/(27/90) = 6/27 = 2/9.

Final answer

05

Exercise 13.4 — Random Variables and Probability Distributions

20Exercise questions

Step-by-step solution

  1. 1X = 0, 1, 2. P(X = 0) = 1/4, P(X = 1) = 2/4 = 1/2, P(X = 2) = 1/4.

Final answer

Distribution: 0 → 1/4, 1 → 1/2, 2 → 1/4.

Step-by-step solution

  1. 1X = 0,1,2,3. P(0) = 1/8, P(1) = 3/8, P(2) = 3/8, P(3) = 1/8.

Final answer

Distribution: 0→1/8, 1→3/8, 2→3/8, 3→1/8.

Step-by-step solution

  1. 1Sum of probabilities = 1 to find k. For NCERT standard: ΣP(X = xᵢ) = 1.
  2. 2Mean E(X) = Σ xᵢ P(X = xᵢ).

Final answer

Step-by-step solution

  1. 1P(success) = 2/6 = 1/3, P(failure) = 2/3.
  2. 2X ~ Binomial(2, 1/3). P(X = 0) = 4/9, P(X = 1) = 4/9, P(X = 2) = 1/9.

Final answer

Distribution: 0→4/9, 1→4/9, 2→1/9.

Step-by-step solution

  1. 1E(X) = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 0 + 0.3 + 0.4 + 0.3 = 1.0.
  2. 2E(X²) = 0(0.4) + 1(0.3) + 4(0.2) + 9(0.1) = 0 + 0.3 + 0.8 + 0.9 = 2.0.

Final answer

(i) E(X) = 1.0; (ii) E(X²) = 2.0.

Step-by-step solution

  1. 1(i) k(1+2+3+4+5+6) = 21k = 1 ⇒ k = 1/21.
  2. 2(ii) P(X > 4) = P(5) + P(6) = 5/21 + 6/21 = 11/21.
  3. 3(iii) P(X ≤ 4) = 1 − 11/21 = 10/21.

Final answer

(i) 1/21; (ii) 11/21; (iii) 10/21.

Step-by-step solution

  1. 1X ranges from 2 to 12. P(X = k) = (number of ways to sum to k)/36.
  2. 2P(2)=1/36, P(3)=2/36, P(4)=3/36, P(5)=4/36, P(6)=5/36, P(7)=6/36, P(8)=5/36, P(9)=4/36, P(10)=3/36, P(11)=2/36, P(12)=1/36.

Final answer

Distribution as above.

Step-by-step solution

  1. 1E(X) = Σ xᵢP(xᵢ).
  2. 2Var(X) = E(X²) − [E(X)]².

Final answer

Step-by-step solution

  1. 1(i) P(X < 2) = 0.09 + 0.15 = 0.24.
  2. 2(ii) P(X ≥ 3) = 0.25 + 0.20 + 0.11 = 0.56.
  3. 3(iii) E(X) = 0(0.09) + 1(0.15) + 2(0.20) + 3(0.25) + 4(0.20) + 5(0.11) = 0 + 0.15 + 0.40 + 0.75 + 0.80 + 0.55 = 2.65.

Final answer

(i) 0.24; (ii) 0.56; (iii) 2.65.

Step-by-step solution

  1. 1X ~ Binomial(4, 1/2).
  2. 2P(X = 0) = 1/16, P(1) = 4/16, P(2) = 6/16, P(3) = 4/16, P(4) = 1/16.
  3. 3E(X) = 4(1/2) = 2. Var(X) = 4(1/2)(1/2) = 1.

Final answer

E(X) = 2, Var(X) = 1.

Step-by-step solution

  1. 1P(X = 1) = 1/36, P(X = 2) = 3/36, P(X = 3) = 5/36, P(X = 4) = 7/36, P(X = 5) = 9/36, P(X = 6) = 11/36.
  2. 2E(X) = (1·1 + 2·3 + 3·5 + 4·7 + 5·9 + 6·11)/36 = (1+6+15+28+45+66)/36 = 161/36.

Final answer

Step-by-step solution

  1. 1(i) k + 2k + 3k + 4k = 10k = 1 ⇒ k = 1/10.
  2. 2(ii) E(X) = 0(1/10) + 1(2/10) + 2(3/10) + 3(4/10) = 0 + 0.2 + 0.6 + 1.2 = 2.0.

Final answer

(i) 1/10; (ii) 2.

Step-by-step solution

  1. 1P(X = 0) = C(2,2)/C(5,2) = 1/10; P(X = 1) = C(3,1)C(2,1)/C(5,2) = 6/10; P(X = 2) = C(3,2)/C(5,2) = 3/10.

Final answer

Distribution: 0→1/10, 1→6/10, 2→3/10.

Step-by-step solution

  1. 1(i) c[1/(1·2) + 1/(2·3) + 1/(3·4) + 1/(4·5)] = c[1/2 + 1/6 + 1/12 + 1/20]
  2. 2= c(30/60 + 10/60 + 5/60 + 3/60) = c(48/60) = 4c/5 = 1 ⇒ c = 5/4.
  3. 3(ii) P(X > 2) = P(3) + P(4) = (5/4)(1/12) + (5/4)(1/20) = 5/48 + 5/80 = 25/240 + 15/240 = 40/240 = 1/6.

Final answer

(i) c = 5/4; (ii) 1/6.

Step-by-step solution

  1. 1E(X) = −0.25 + 0 + 0.25 = 0.
  2. 2E(X²) = 1(0.25) + 0(0.50) + 1(0.25) = 0.50.
  3. 3Var(X) = 0.50 − 0² = 0.50.

Final answer

E(X) = 0, Var(X) = 0.50.

Step-by-step solution

  1. 1(i) a + 3a + 5a + 7a = 16a = 1 ⇒ a = 1/16.
  2. 2(ii) P(X < 2) = P(0) + P(1) = 1/16 + 3/16 = 4/16 = 1/4.
  3. 3(iii) P(X ≤ 2) = 1/16 + 3/16 + 5/16 = 9/16.
  4. 4(iv) P(X ≥ 2) = 5/16 + 7/16 = 12/16 = 3/4.

Final answer

(i) 1/16; (ii) 1/4; (iii) 9/16; (iv) 3/4.

Step-by-step solution

  1. 1(i) P(X ≥ 3) = 0.3 + 0.25 + 0.15 = 0.70.
  2. 2(ii) P(X ≤ 2) = 0.1 + 0.2 = 0.3.
  3. 3(iii) E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.25) + 5(0.15) = 0.1 + 0.4 + 0.9 + 1.0 + 0.75 = 3.15.

Final answer

(i) 0.70; (ii) 0.30; (iii) 3.15.

Step-by-step solution

  1. 1Each P(X = k) = 1/n.
  2. 2E(X) = (1/n)(1+2+...+n) = (1/n)·n(n+1)/2 = (n+1)/2.

Final answer

Step-by-step solution

  1. 1E(X) = (−2)(0.2) + (−1)(0.3) + 0(0.1) + 1(0.25) + 2(0.15) = −0.4 − 0.3 + 0 + 0.25 + 0.3 = −0.15.
  2. 2E(X²) = 4(0.2) + 1(0.3) + 0 + 1(0.25) + 4(0.15) = 0.8 + 0.3 + 0.25 + 0.6 = 1.95.

Final answer

E(X) = −0.15, E(X²) = 1.95.

Step-by-step solution

  1. 1(i) c(1+2+3+4) = 10c = 1, so c = 1/10.
  2. 2(ii) E(X) = (1/10)(1+4+9+16) = 30/10 = 3.
  3. 3(iii) E(X²) = (1/10)(1+16+81+256) = 354/10 = 35.4. Var(X) = 35.4 − 9 = 26.4.

Final answer

(i) 1/10; (ii) 3; (iii) 26.4.

06

Exercise 13.5 — Bernoulli Trials and Binomial Distribution

19Exercise questions

Final answer

No — trials are not independent (without replacement), so the conditions of a binomial distribution are not met.

Step-by-step solution

  1. 1X ~ Binomial(10, 1/2). P(X = 6) = C(10,6)(1/2)⁶(1/2)⁴ = 210/1024 = 105/512.

Final answer

Step-by-step solution

  1. 1(i) P(X = 3) = C(5,3)(3/4)³(1/4)² = 10 · 27/64 · 1/16 = 270/1024 = 135/512.
  2. 2(ii) P(X ≥ 3) = P(3) + P(4) + P(5)
  3. 3= 135/512 + 5·(81/256)(1/4) + 243/1024
  4. 4= 135/512 + 405/1024 + 243/1024 = 270/1024 + 405/1024 + 243/1024 = 918/1024 = 459/512.

Final answer

(i) 135/512; (ii) 459/512.

Step-by-step solution

  1. 1This is without replacement, so strictly hypergeometric, not binomial. Using hypergeometric:
  2. 2(i) P(2 defective) = C(4,2)C(16,3)/C(20,5) = 6·560/15504 = 3360/15504 = 70/323.
  3. 3(ii) P(at most 2) = [C(4,0)C(16,5) + C(4,1)C(16,4) + C(4,2)C(16,3)]/C(20,5)
  4. 4= [4368 + 4·1820 + 6·560]/15504 = [4368 + 7280 + 3360]/15504 = 15008/15504 = 938/969.

Final answer

(i) 70/323; (ii) 938/969.

Step-by-step solution

  1. 1X ~ Binomial(6, 0.05). p = 1/20, q = 19/20.
  2. 2(i) P(0) = (19/20)⁶ = 19⁶/20⁶ ≈ 0.7351.
  3. 3(ii) P(≤1) = P(0) + 6(1/20)(19/20)⁵ = 19⁶/20⁶ + 6·19⁵/20⁶ ≈ 0.7351 + 0.2321 = 0.9672.
  4. 4(iii) P(>1) = 1 − P(≤1) ≈ 0.0328.
  5. 5(iv) P(≥1) = 1 − P(0) ≈ 0.2649.

Final answer

(i) ≈0.7351; (ii) ≈0.9672; (iii) ≈0.0328; (iv) ≈0.2649.

Step-by-step solution

  1. 1X ~ Binomial(5, 1/4). P(X ≥ 2) = 1 − P(0) − P(1) = 1 − (3/4)⁵ − 5(1/4)(3/4)⁴
  2. 2= 1 − 243/1024 − 5·81/1024 = 1 − 243/1024 − 405/1024 = 1 − 648/1024 = 376/1024 = 47/128.

Final answer

Step-by-step solution

  1. 1P(doublet) = 6/36 = 1/6. X ~ Binomial(5, 1/6).
  2. 2(i) P(X = 2) = C(5,2)(1/6)²(5/6)³ = 10·(1/36)(125/216) = 1250/7776 = 625/3888.
  3. 3(ii) P(X ≥ 2) = 1 − P(0) − P(1) = 1 − (5/6)⁵ − 5(1/6)(5/6)⁴
  4. 4= 1 − 3125/7776 − 5(625/7776) = 1 − 3125/7776 − 3125/7776 = 1 − 6250/7776 = 1526/7776 = 763/3888.

Final answer

(i) 625/3888; (ii) 763/3888.

Step-by-step solution

  1. 1p = 2/8 = 1/4, q = 3/4. X ~ Binomial(3, 1/4).
  2. 2(i) P(3) = (1/4)³ = 1/64.
  3. 3(ii) P(0) = (3/4)³ = 27/64.
  4. 4(iii) P(≥1) = 1 − 27/64 = 37/64.

Final answer

(i) 1/64; (ii) 27/64; (iii) 37/64.

Step-by-step solution

  1. 1(i) C(3,4)/C(10,4) = 0 (only 3 defective available).
  2. 2(ii) C(7,4)/C(10,4) = 35/210 = 1/6.

Final answer

(i) 0; (ii) 1/6.

Step-by-step solution

  1. 1P(at least one hit) = 1 − (0.7)ⁿ > 0.95.
  2. 2(0.7)ⁿ < 0.05. Try n = 10: 0.7¹⁰ ≈ 0.02825 < 0.05. n = 9: 0.7⁹ ≈ 0.04036 < 0.05 too? 0.04036 < 0.05 ✓. n = 8: 0.7⁸ ≈ 0.05765 > 0.05 ✗.
  3. 3Minimum n = 9.

Final answer

Step-by-step solution

  1. 1p(safe) = 9/10. X ~ Binomial(5, 9/10).
  2. 2P(X ≥ 4) = P(4) + P(5) = 5(9/10)⁴(1/10) + (9/10)⁵
  3. 3= 5·6561/10000·(1/10) + 59049/100000 = 32805/100000 + 59049/100000 = 91854/100000 = 45927/50000.

Final answer

Step-by-step solution

  1. 1P(sum 7) = 6/36 = 1/6. X ~ Binomial(6, 1/6).
  2. 2P(X = 1) = 6(1/6)(5/6)⁵ = 6·5⁵/6⁶ = 6·3125/46656 = 18750/46656 = 3125/7776.

Final answer

Step-by-step solution

  1. 1X ~ Binomial(50, 0.01). P(X ≥ 1) = 1 − (0.99)⁵⁰ ≈ 1 − 0.6050 = 0.3950.

Final answer

Step-by-step solution

  1. 1p(red) = 5/8 per draw (with replacement). X ~ Binomial(4, 5/8).
  2. 2P(X = 2) = C(4,2)(5/8)²(3/8)² = 6·25/64·9/64 = 6·225/4096 = 1350/4096 = 675/2048.

Final answer

Step-by-step solution

  1. 1p = 1/2, X ~ Binomial(6, 1/2).
  2. 2(i) P(5) = 6(1/2)⁶ = 6/64 = 3/32.
  3. 3(ii) P(≥5) = P(5) + P(6) = 6/64 + 1/64 = 7/64.
  4. 4(iii) P(≤5) = 1 − P(6) = 1 − 1/64 = 63/64.

Final answer

(i) 3/32; (ii) 7/64; (iii) 63/64.

Step-by-step solution

  1. 1X ~ Binomial(5, 1/6).
  2. 2(i) P(X = 3) = C(5,3)(1/6)³(5/6)² = 10·(1/216)(25/36) = 250/7776 = 125/3888.
  3. 3(ii) P(X ≥ 3) = P(3) + P(4) + P(5)
  4. 4= 250/7776 + 5(1/1296)(5/6) + 1/7776
  5. 5= 250/7776 + 25/7776 + 1/7776 = 276/7776 = 23/648.

Final answer

(i) 125/3888; (ii) 23/648.

Step-by-step solution

  1. 1X ~ Binomial(12, 0.1).
  2. 2(i) P(2) = C(12,2)(0.1)²(0.9)¹⁰ = 66·0.01·0.3487 ≈ 0.2301.
  3. 3(ii) P(≤2) = P(0)+P(1)+P(2) ≈ 0.2824 + 0.3765 + 0.2301 ≈ 0.8891.
  4. 4(iii) P(>2) = 1 − 0.8891 ≈ 0.1109.

Final answer

(i) ≈ 0.2301; (ii) ≈ 0.8891; (iii) ≈ 0.1109.

Step-by-step solution

  1. 1P(doublet) = 6/36 = 1/6. X ~ Binomial(4, 1/6).
  2. 2P(X = 2) = C(4,2)(1/6)²(5/6)² = 6·(1/36)(25/36) = 150/1296 = 25/216.

Final answer

Step-by-step solution

  1. 1X ~ Binomial(15, 0.4).
  2. 2(i) P(X ≥ 10) = Σ from k=10 to 15 of C(15,k)(0.4)ᵏ(0.6)^{15−k} ≈ 0.0338.
  3. 3(ii) P(X ≤ 5) ≈ 0.4032.
  4. 4(iii) P(X = 6) = C(15,6)(0.4)⁶(0.6)⁹ ≈ 0.2066.

Final answer

(i) ≈ 0.0338; (ii) ≈ 0.4032; (iii) ≈ 0.2066.

Quick Revision

Key formulas at a glance

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

Conditional probability

Independence

Bayes' theorem

Total probability

Exam Strategy

How this chapter is asked

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

  • The conditioning event must have non-zero probability, and the whole probability space must be partitioned exactly once in the total probability formula or terms will be double-counted.
  • Bayes' theorem is just conditional probability applied the other way round, so set P(A|B) versus P(B|A) explicitly before substituting.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Maths Chapter 13 (Probability)?

There are 5 exercise questions in this chapter, numbered Ex 13.1 to Ex 13.5. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Probability Class 12 Maths?

The formulas this chapter's questions actually turn on are: Conditional probability, Independence, Bayes' theorem, Total probability. 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 — Bayes' theorem and the mean and variance of a random variable are major board and JEE Main topics and a stated NEET topic.

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