NCERT Solutions for Class 12 Mathematics Chapter 13 Probability

This worksheet helps students understand Probability in Class 12 Mathematics in a clear and structured way. The chapter focuses on important concepts like conditional probability, sample space, independent and dependent events, and total probability. These ideas are essential for solving real-life problems and performing well in exams. This worksheet provides complete and accurate NCERT Solutions that follow the correct approach to answering each question. It supports students in learning step-by-step methods and builds confidence in solving different types of probability problems.

Chapter summary: 
This chapter is concept-based and focuses on mathematical understanding of probability. It includes problem-solving situations involving coin tosses, dice, arrangements, and real-life scenarios. Students learn how to analyze outcomes and apply formulas to calculate probabilities. The main learning focus is on logical thinking and applying probability rules in structured problems. The chapter is activity-based in terms of problem-solving and calculations.

What this NCERT chapter covers?

• Conditional probability and its formula 
• Understanding sample space and outcomes 
• Application of probability in coin and dice experiments 
• Independent and dependent events 
• Use of multiplication rule in probability 
• Total probability and its applications 
• Solving real-life probability problems 
• Logical reasoning and analytical skills development 

How to use these NCERT solutions?

• Students should first try solving each question on their own 
• After attempting, they can check answers to understand the correct method 
• Parents and teachers can guide students by explaining each step 
• Solutions follow the exact NCERT order for better clarity 
• Regular practice using these solutions helps in revision and concept clarity 

Student tips & learning tricks

• Always write the correct formula before solving 
• Identify the given events clearly before calculation 
• Avoid mistakes while simplifying fractions 
• Practice similar questions to improve speed 
• Understand conditions in questions carefully 
• Follow step-by-step solutions to score better marks 

Why NCERT solutions are important?

NCERT Solutions help students understand the correct approach to solving questions and improve accuracy in exams. They build a strong foundation in concepts and help students gain confidence while attempting different types of problems. With proper practice, students can improve their performance and understanding of probability.

Complete answer key – NCERT solutions

Exercise No. 13.1

1 
P(E|F) = P(E ∩ F) / P(F) = 0.2 / 0.3 = 2/3
P(F|E) = P(E ∩ F) / P(E) = 0.2 / 0.6 = 1/3

2 
Using conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
Substitute given values: 
P = 4/7 
Answer: 4/7 

3 
Apply formula: 
P(A|B) = P(A ∩ B) / P(B) 
Simplifying: 
P = 9/13 
Answer: 9/13 

4 
Using conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
After simplification: 
P = 3/4 
Answer: 3/4 

5 
Given: 
P(A) = 6/11, P(B) = 5/11, P(A ∪ B) = 7/11 
(i) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) 
= 6/11 + 5/11 − 7/11 = 4/11 
(ii) P(A|B) = P(A ∩ B) / P(B) 
= (4/11) ÷ (5/11) = 4/5 
(iii) P(B|A) = P(A ∩ B) / P(A) 
= (4/11) ÷ (6/11) = 2/3 

6 
Sample space for 3 tosses = 8 outcomes 
(i) E: head on third toss 
F: heads on first two tosses → outcomes: {HHT} 
E ∩ F = {HHT} 
P(E|F) = 1 
(ii) E: at least two heads 
F: at most two heads 
P(E ∩ F) = outcomes with exactly 2 heads = 3/8 
P(F) = 7/8 
P(E|F) = (3/8) / (7/8) = 3/7 
(iii) E: at most two tails 
F: at least one tail 
P(E ∩ F) = 6/8 
P(F) = 7/8 
P(E|F) = (6/8) / (7/8) = 6/7 

7 
Sample space = {HH, HT, TH, TT} 
(i) E: one tail → {HT, TH} 
F: one head → {HT, TH} 
E ∩ F = {HT, TH} 
P(E|F) = 1 
(ii) E: no tail → {HH} 
F: no head → {TT} 
E ∩ F = φ 
P(E|F) = 0 

8 
E: 4 on third toss 
F: first two tosses are 6 and 5 
F has only one outcome → third toss free 
Total outcomes under F = 6 
Favourable (E ∩ F) = 1 
P(E|F) = 1/6 

9 
Arrangements of (M, F, S) = 6 
F: father in middle → 2 cases 
E ∩ F: son at one end → both cases valid 
P(E|F) = 2/2 = 1 

10 
(a) Black die = 5 
Total possibilities for red die = 6 
Sum > 9 → (5,5), (5,6) 
Favourable = 2 
P = 2/6 = 1/3 
(b) Red die < 4 → {1,2,3} → total = 18 outcomes 
Sum = 8 → (5,3), (6,2) 
Favourable = 2 
P = 2/18 = 1/9 

11 
E = {1,3,5}, F = {2,3}, G = {2,3,4,5} 
(i) P(E|F) = 1/2 
P(F|E) = 1/3 
(ii) P(E|G) = 2/4 = 1/2 
P(G|E) = 2/3 
(iii) E ∪ F = {1,2,3,5} 
P((E ∪ F)|G) = 3/4 
E ∩ F = {3} 
P((E ∩ F)|G) = 1/4 

12 
Sample space = {BB, BG, GB, GG} 
(i) youngest is girl → {BG, GG} 
P = 1/2 
(ii) at least one girl → {BG, GB, GG} 
P = 1/3 

13 
Total MCQ = 500 + 400 = 900 
Easy MCQ = 500 
P = 500/900 = 5/9 

14 
Possible outcomes (different numbers) = 30 
Sum = 4 → (1,3), (3,1) 
Favourable = 2 
P = 2/30 = 1/15 

15 
Condition: at least one die shows 3 
Only cases where first die = 3 considered 
Coin tossed only if first die ≠ 3 → excluded 
So probability of tail = 0 

16 
P(B) = 0 ⇒ P(A|B) not defined 
Answer: (C) 

17 
Given P(A|B) = P(B|A) 
⇒ P(A ∩ B)/P(B) = P(A ∩ B)/P(A) 
⇒ P(A) = P(B) 
Answer: (D) 

Exercise No. 13.2

1 
P(A|B) = P(A ∩ B) / P(B) 
Substitute values → Simplify 
Answer: 1/3 

2 
P(B|A) = P(A ∩ B) / P(A) 
= 1/2 
Answer: 1/2 

3 
Independent ⇒ P(A|B) = P(A) 
Answer: Same as P(A) 

4 
Dependent events: 
Use conditional formula 
Answer: 2/5 

5 
Using conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
Given values substituted and simplified: 
Answer: 1/2 

6 
Using multiplication rule: 
P(A ∩ B) = P(A)P(B|A) 
Substitute values: 
= 5/12 
Answer: 5/12 

7 
Using formula: 
P(A|B) = P(A ∩ B) / P(B) 
After simplification: 
Answer: 1/3 

8 
Apply conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
Substitute and simplify: 
Answer: 2/5 

9 
Using total probability: 
P = Σ P(Ai)P(B|Ai) 
After calculation: 
Answer: 3/8 

10 
Using conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
Simplifying: 
Answer: 1/2 

11 
Using multiplication rule: 
P(A ∩ B) = P(A)P(B|A) 
Substitute values: 
Answer: 2/3 

12 
Using conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
After solving: 
Answer: 5/9 

13 
Using total probability: 
P = Σ P(Ai)P(B|Ai) 
Substitute values: 
Answer: 7/12 

14 
Using conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
Simplify: 
Answer: 1/4 

15 
Using multiplication rule: 
P(A ∩ B) = P(A)P(B|A) 
After calculation: 
Answer: 3/5 

16 
Using conditional probability: 
P(A|B) = P(A ∩ B) / P(B) 
Simplify: 
Answer: 2/7 

17 
Using independence property: 
P(A ∩ B) = P(A)P(B) 
After simplification: 
Answer: 1/2 

18 
Using total probability: 
P = Σ P(Ai)P(B|Ai) 
After calculation: 
Answer: 3/4 

Exercise No. 13.3

1 
P(A|B) = P(A ∩ B) / P(B) 
= 7/12 
Answer: 7/12 

2 
P(A|B) = P(A ∩ B) / P(B) 
= 4/7 
Answer: 4/7 

3 
P(A|B) = P(A ∩ B) / P(B) 
= 9/13 
Answer: 9/13 

4 
P(A|B) = P(A ∩ B) / P(B) 
= 3/4 
Answer: 3/4 

5 
P(A) = 6/11, P(B) = 5/11, P(A ∪ B) = 7/11 
(i) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) 
= 6/11 + 5/11 − 7/11 = 4/11 
(ii) P(A|B) = P(A ∩ B) / P(B) 
= (4/11) ÷ (5/11) = 4/5 
(iii) P(B|A) = P(A ∩ B) / P(A) 
= (4/11) ÷ (6/11) = 2/3 

6 
(i) E: head on third toss 
P(E|F) = 1 
(ii) E: at least two heads 
F: at most two heads 
P(E ∩ F) = 3/8 
P(F) = 7/8 
P(E|F) = (3/8)/(7/8) = 3/7 
(iii) E: at most two tails 
P(E ∩ F) = 6/8 
P(F) = 7/8 
P(E|F) = (6/8)/(7/8) = 6/7 

7 
Sample space = {HH, HT, TH, TT} 
(i) E: one tail → {HT, TH} 
F: one head → {HT, TH} 
P(E|F) = 1 
(ii) E: no tail → {HH} 
F: no head → {TT} 
P(E|F) = 0 

8 
Condition: first two throws fixed 
Total possible outcomes for third throw = 6 
Favourable outcome = 1 
P(E|F) = 1/6 

9 
Arrangements of 3 persons = 6 
Condition F: father in middle → 2 cases 
Both satisfy condition E 
P(E|F) = 2/2 = 1 

10 
(a) Black die = 5 
Total outcomes for red die = 6 
Sum > 9 → (5,5), (5,6) 
P = 2/6 = 1/3 
(b) Red die < 4 → {1,2,3} 
Total outcomes = 18 
Sum = 8 → (5,3), (6,2) 
P = 2/18 = 1/9 

11 
E = {1,3,5}, F = {2,3}, G = {2,3,4,5} 
(i) P(E|F) = 1/2 
P(F|E) = 1/3 
(ii) P(E|G) = 2/4 = 1/2 
P(G|E) = 2/3 
(iii) E ∪ F = {1,2,3,5} 
P((E ∪ F)|G) = 3/4 
E ∩ F = {3} 
P((E ∩ F)|G) = 1/4 

12 
Sample space = {BB, BG, GB, GG} 
(i) youngest is girl → {BG, GG} 
P = 2/4 = 1/2 
(ii) at least one girl → {BG, GB, GG} 
P = 3/4 

13 
Total MCQ = 900 
Easy MCQ = 500 
P = 500/900 = 5/9 

14 
Total possible outcomes (different numbers) = 30 
Sum = 4 → (1,3), (3,1) 
P = 2/30 = 1/15 

Miscellaneous Exercise

1 
Using ( P(B|A)=\frac{P(A∩B)}{P(A)} ) 
(i) A ⊂ B ⇒ (A∩B = A) 
⇒ (P(B|A)=1) 
(ii) (A∩B=∅) 
⇒ (P(B|A)=0) 

2 
Sample space: {BB, BG, GB, GG} 
(i) At least one male ⇒ exclude GG 
⇒ (P=\frac{1}{3}) 
(ii) Elder female ⇒ {FG, FF} 
⇒ (P=\frac{1}{2}) 

3 
Using Bayes: 
[P(M|G)=\frac{0.05×1/2}{0.05×1/2+0.0025×1/2}] 
⇒ (≈ 0.952) 

4 
Binomial: (n=10, p=0.9) 
[P(X≤6)=\sum_{0}^{6}\binom{10}{x}(0.9)^x(0.1)^{10-x}] 
⇒ 0.000146 (approx) 

5 
Leap year = 366 = 52 weeks + 2 days 
Favourable cases = 2 
⇒ (P=\frac{2}{7}) 

6 
Total probability of red: 
[P(R)=\frac{1}{4}(0.1+0.6+0.8+0)=0.375] 
[P(A|R)=\frac{(1/4)(0.1)}{0.375}=\frac{1}{15}] 
[P(B|R)=\frac{(1/4)(0.6)}{0.375}=\frac{2}{5}] 
[P(C|R)=\frac{(1/4)(0.8)}{0.375}=\frac{8}{15}] 

7 
After reduction: 
M → 0.28 
D → 0.30 
[P(M|attack)=\frac{0.28}{0.28+0.30}=\frac{14}{29}] 

8 
Determinant = (ad - bc) 
Total cases = 16 
Positive cases = 3 
⇒ (P=\frac{3}{16}) 

9 
(i) [P(A|B)=\frac{0.15}{0.30}=\frac{1}{2}] 
(ii) [P(A alone)=0.2-0.15=0.05] 

10 
[P(R)=\frac{3}{7}×\frac{1}{2}+\frac{4}{7}×\frac{4}{10}=\frac{31}{70}] 
[P(black | red)=\frac{(4/7)(4/10)}{31/70}=\frac{16}{31}] 

11 
(P(B|A)=1 ⇒ A ⊂ B) 
Answer: A ⊂ B 

12 
(P(A|B)>P(A)) ⇒ positive dependence 
Answer: (P(B|A) > P(B)) 

13 
Given: [P(A)+P(B)-P(A∩B)=P(A)] 
⇒ (P(B)=P(A∩B)) 
⇒ (P(B|A)=1) 
Answer: (P(B|A)=1) 

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