NCERT Solutions for Class 11 Maths Chapter 14

NCERT Solutions for Class 11 Maths Chapter 14
Last Updated At: 7 Apr 2026
8 min read

NCERT solutions for Class 11 Mathematics Chapter Probability – complete answers & explanations

Class 11 Mathematics Chapter Probability is an important chapter that introduces students to the concept of chance and uncertainty in everyday situations. In this chapter, students learn how to calculate the likelihood of events using mathematical methods and logical reasoning. Understanding probability helps students build strong analytical skills and prepares them for higher-level mathematics and real-life decision-making. This chapter is widely used in exams and competitive tests, making it essential for scoring well and developing confidence in problem solving. These clear and reliable NCERT solutions support students in understanding the correct approach to answering questions step by step. Parents and teachers can also use these solutions to guide students during revision and practice. Download the worksheet and practice alongside solutions for better clarity. Book a free trial now to get expert guidance and strengthen mathematical understanding.

NCERT Solutions for Class 11 Maths Chapter 14.png

What this NCERT chapter covers?

1. This chapter introduces the basic idea of probability and explains how to measure the likelihood of events in simple experiments. 
2. Students learn about events such as mutually exclusive events, simple events, compound events, and exhaustive events. 
3. The chapter explains how to represent outcomes using sets and sample spaces in probability problems. 
4. It helps students understand the relationship between events using operations like union, intersection, and complement. 
5. Students learn how to calculate probabilities for coin tosses, dice throws, and card selection problems. 
6. The chapter also covers real-life applications of probability in games, selections, and predictions. 
7. Learners understand the use of formulas and logical reasoning to find probabilities accurately. 
8. It develops problem-solving skills and improves mathematical thinking through step-by-step calculations. 
9. The concepts in this chapter form the foundation for advanced topics in statistics and probability. 
10. Practicing these questions helps students perform confidently in school exams and competitive assessments.

How to use these NCERT solutions?

1. First, read the question carefully and attempt to solve it on your own before checking the answer. 
2. Use these solutions to verify your calculations and understand the correct method of solving probability problems. 
3. Compare your steps with the given answers to identify mistakes and improve accuracy. 
4. Practice similar questions regularly to build confidence and speed in solving probability questions. 
5. Parents and teachers can use these answers to guide students and explain difficult concepts clearly. 
6. Follow the sequence of questions exactly as presented to maintain proper learning flow. 
7. Use these solutions during revision to quickly recall formulas and important methods. 
8. Refer to the answers while preparing for exams to ensure complete understanding of probability concepts.

Important tips & tricks for students

1. Always identify the total number of possible outcomes before calculating probability. 
2. Carefully check whether events are mutually exclusive or not before applying formulas. 
3. Use proper mathematical notation such as union, intersection, and complement symbols. 
4. Avoid skipping steps in calculations, especially in complex probability problems. 
5. Double-check fractions and simplifications to prevent calculation errors. 
6. Practice different types of probability questions regularly to strengthen problem-solving skills. 
7. Read each question carefully to understand what event is being asked. 
8. Keep formulas and definitions clear in your mind for quick recall during exams. 
9. Maintain neat and organized work to avoid confusion while solving problems.

NCERT solutions – complete answer key

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

Why NCERT solutions help students?

NCERT solutions help students prepare effectively for exams by providing clear and accurate answers that match expected formats. They improve understanding of concepts, build confidence in solving problems, and support consistent revision. Regular practice with correct answers helps students strengthen their mathematical skills and perform better in assessments.

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