NCERT Solutions for Class 11 Mathematics Chapter Probability

This worksheet provides complete and accurate NCERT Solutions for Class 11 Mathematics Chapter Probability. The chapter introduces students to the concept of probability and explains how to calculate the likelihood of events in different real-life situations such as coin tosses, dice rolls, and card selections. It helps students understand how mathematical reasoning can be used to predict outcomes and make logical decisions. This chapter is important for building strong analytical skills and preparing students for examinations and higher-level mathematics topics. The solutions in this worksheet follow the NCERT pattern and structure, helping students learn the correct method of solving probability questions step by step. Parents and teachers can also use this worksheet as a reliable learning support tool for revision and practice.

Chapter summary: stories, poems & themes

This chapter is concept-based and focuses on mathematical problem solving rather than stories or poems. It includes numerical examples and probability experiments involving coins, dice, cards, and selections. Students learn how events occur in different situations and how probability values are calculated using mathematical formulas. The chapter emphasizes logical thinking, observation, and accurate calculation of outcomes. It is not picture-based but includes activity-style mathematical problems that require careful reasoning and computation.

What this NCERT chapter covers?

• Understanding the meaning of probability and how it is calculated 
• Identifying sample space and possible outcomes in experiments 
• Learning about mutually exclusive and exhaustive events 
• Applying probability formulas to coin tosses and dice problems 
• Solving problems involving cards, selections, and combinations 
• Using union, intersection, and complement of events 
• Interpreting probability values in real-life situations 
• Developing logical reasoning and mathematical accuracy 
• Practicing step-by-step problem solving methods 
• Building confidence for examinations and assessments 

How to use these NCERT solutions?

• Students should first read each question carefully and attempt to solve it independently 
• After solving, they can compare their answers with the solutions provided in this worksheet 
• Parents and teachers can guide students by explaining steps and checking calculations 
• The answers in this worksheet follow the same order as NCERT exercises 
• Using these solutions regularly helps in revision and improves understanding 
• Practicing these questions strengthens problem-solving skills and accuracy

Student tips & learning tricks

• Always identify the total number of outcomes before calculating probability 
• Carefully read the question to understand the event being asked 
• Write steps clearly and avoid skipping calculations 
• Check fractions and simplifications to prevent mistakes 
• Practice different types of probability problems regularly 
• Revise formulas and definitions before exams 
• Maintain neat and organized work while solving questions

Why NCERT solutions are important?

NCERT-aligned solutions help students understand mathematical concepts clearly and prepare effectively for school examinations. They support strong foundational learning and ensure that students follow the correct answering methods expected in assessments. Regular practice using these solutions improves confidence, accuracy, and readiness for tests.

Complete answer key – NCERT solutions

Exercise 14.1

1.  No, E and F are not mutually exclusive. 
E = {4}, F = {2, 4, 6} 
E ∩ F = {4} ≠ φ

2.  A = {1, 2, 3, 4, 5, 6} 
B = φ 
C = {3, 6} 
D = {1, 2, 3} 
E = {6} 
F = {3, 4, 5, 6} 
A ∪ B = {1, 2, 3, 4, 5, 6} 
A ∩ B = φ 
B ∪ C = {3, 6} 
E ∩ F = {6} 
D ∩ E = φ 
A – C = {1, 2, 4, 5} 
D – E = {1, 2, 3} 
E ∩ F′ = φ 
F′ = {1, 2} 

3.  A = {(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)} 
B = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (1,2), (3,2), (4,2), (5,2), (6,2)} 
C = {(3,6), (4,5), (5,4), (6,3), (6,6)} 
Mutually exclusive pairs: 
A and B 
B and C 

4.  A = {HHH} 
B = {HHT, HTH, THH} 
C = {TTT} 
D = {HHH, HHT, HTH, HTT} 
(i) Mutually exclusive: 
A and B, A and C, B and C, C and D 
(ii) Simple: 
A, C 
(iii) Compound: 
B, D 

5.   (i) Two mutually exclusive events: 
A = {HHH} 
B = {TTT} 
(ii) Three mutually exclusive and exhaustive events: 
A = {TTT} 
B = {HTT, THT, TTH} 
C = {HHT, HTH, THH, HHH} 
(iii) Two events which are not mutually exclusive: 
A = {HHT, HTH, THH, HHH} 
B = {HTT, THT, TTH, HHT, HTH, THH, HHH} 
(iv) Two events which are mutually exclusive but not exhaustive: 
A = {HHH} 
B = {TTT} 
(v) Three events which are mutually exclusive but not exhaustive: 
A = {HHH} 
B = {HHT, HTH, THH} 
C = {TTT} 

6.  A = {(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), 
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), 
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
B = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), 
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), 
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)} 
C = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)} 

7.  (i) True 
(ii) True 
(iii) True 
(iv) False 
(v) False 
(vi) False 

Exercise 14.2
1. Invalid assignments: 
(c), (d), (e) 

2.  P(at least one tail) = 3/4 

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

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

5.  (i) 1/12 
(ii) 1/12 

6.  P(woman) = 6/10 = 3/5 

7. Possible amounts of money after 4 tosses: 
Rs –6, Rs –3.50, Rs –1, Rs 1.50, Rs 4 
Probabilities: 
P(Rs –6) = 1/16 
P(Rs –3.50) = 4/16 = 1/4 
P(Rs –1) = 6/16 = 3/8 
P(Rs 1.50) = 4/16 = 1/4 
P(Rs 4) = 1/16 

8.  (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 

9.  P(not A) = 1 – 2/11 = 9/11 

10. Total letters = 13 
Vowels = 6 
Consonants = 7 
(i) 6/13 
(ii) 7/13

11. P(winning) = 1 / 20C6 = 1/38760

 12. (i) Not consistently defined 
(ii) Consistently defined 

13. (i) P(A ∪ B) = 7/15 
(ii) P(B) = 0.50 
(iii) P(A ∩ B) = 0.15

14. P(A or B) = 3/5 + 1/5 = 4/5 

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

16. No, E and F are not mutually exclusive. 

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

18. P(Mathematics or Biology) = 0.40 + 0.30 – 0.10 = 0.60 

19. P(both) = 0.80 + 0.70 – 0.95 = 0.55 

20. Let H be the event of passing Hindi. 
0.90 = 0.75 + P(H) – 0.50 
P(H) = 0.65 

21. (i) P(NCC or NSS) = (30 + 32 – 24)/60 = 38/60 = 19/30 
(ii) P(neither NCC nor NSS) = 1 – 19/30 = 11/30 
(iii) P(NSS but not NCC) = (32 – 24)/60 = 8/60 = 2/15 

Miscellaneous exercise

1. (i) P(all blue) = 20C5 / 60C5 
(ii) P(at least one green) = 1 – 30C5 / 60C5

2.  P(3 diamonds and 1 spade) = (13C3 × 13C1) / 52C4 

3. (i) P(2) = 3/6 = 1/2 
(ii) P(1 or 3) = (2 + 1)/6 = 1/2 
(iii) P(not 3) = 5/6 

4. (a) P(not getting a prize with 1 ticket) = 9990/10000 
(b) P(not getting a prize with 2 tickets) = 9998C10 / 10000C10

(c) P(not getting a prize with 10 tickets) = 9990C10 / 10000C10                                  

5.  (a) P(same section) = (40/100 × 39/99) + (60/100 × 59/99) = 17/33 
(b) P(different sections) = 1 – 17/33 = 16/33 

6.  P(at least one letter in proper envelope) = 1 – 2/6 = 2/3 

7. (i) P(A ∪ B) = 0.54 + 0.69 – 0.35 = 0.88 
(ii) P(A′ ∩ B′) = 1 – 0.88 = 0.12 
(iii) P(A ∩ B′) = 0.54 – 0.35 = 0.19 
(iv) P(B ∩ A′) = 0.69 – 0.35 = 0.34 

8. Required persons = male or over 35 years 
= {Harish, Rohan, Sheetal, Salim} 
P(required event) = 4/5 

9. (i) With repetition allowed: 
P(divisible by 5) = 100/250 = 2/5 
(ii) Without repetition: 
P(divisible by 5) = 18/48 = 3/8 

10.  Total possible sequences = 10P4 = 5040 
P(right sequence) = 1/5040 

Build strong mathematical problem-solving skills with structured learning support designed to help students succeed confidently in exams. 
Book a free trial!