NCERT Solutions for class 12 Mathematics Chapter 13
NCERT solutions for Class 12 Mathematics Chapter 13 Probability – complete answers & explanations
This chapter in Class 12 Mathematics focuses on Probability, an important topic that helps students understand how to calculate the likelihood of events. It introduces concepts like conditional probability, independent events, and total probability, which are essential for solving real-life and exam-based problems. Mastering this chapter builds strong analytical and logical thinking skills that are useful not only in mathematics but also in everyday decision-making. This blog provides clear and reliable NCERT solutions that help students follow the correct approach to answering each question. Download the worksheet and practice alongside solutions for better clarity. With step-by-step answers and structured solutions, students can gain confidence in solving even complex probability problems. Parents can also use this as a helpful guide to support their child’s learning at home. Book a free trial now to get expert guidance and improve performance in exams.
What this NCERT chapter covers?
1. The chapter explains the concept of conditional probability and how to calculate it using formulas.
2. It covers the relationship between events, including intersection and union of events.
3. Students learn how to apply probability rules in different real-life situations.
4. The chapter introduces independent and dependent events and their properties.
5. It explains the multiplication rule and total probability theorem in detail.
6. Learners understand how to analyze sample spaces and outcomes in probability experiments.
7. The chapter includes problems based on coin tosses, dice rolls, and real-world scenarios.
8. It strengthens logical reasoning and analytical thinking skills.
9. Students learn how to interpret probability values and apply them correctly.
10. The chapter prepares students for board exams with structured and application-based questions.
How to use these NCERT solutions?
1. Start by attempting each question from the worksheet on your own before checking the answers.
2. Carefully compare your solution with the given answer to understand any mistakes.
3. Focus on the steps used in solving probability problems to improve your approach.
4. Use these solutions to revise important formulas and concepts regularly.
5. Parents and teachers can guide students by discussing the logic behind each answer.
6. Practice similar questions after reviewing the solutions to strengthen understanding.
7. Follow the sequence of questions as given to maintain clarity and structure.
8. Revisit difficult questions multiple times to build confidence.
9. Use these solutions as a reference during exam preparation.
10. Ensure you understand the method used rather than just memorizing answers.
Important tips & tricks for students
1. Always write the correct formula before solving probability questions.
2. Clearly identify events and sample spaces before calculating probabilities.
3. Avoid calculation errors by simplifying fractions carefully.
4. Practice different types of problems like dice, coins, and real-life cases.
5. Understand the difference between independent and dependent events.
6. Read questions carefully to avoid misunderstanding conditions.
7. Show proper steps in exams to score full marks.
8. Revise important formulas regularly to improve speed and accuracy.
9. Pay attention to conditional probability concepts as they are frequently asked.
10. Stay consistent with practice to master the chapter.
NCERT solutions – complete answer key
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)
Why NCERT solutions help students?
NCERT solutions help students build strong conceptual clarity and prepare effectively for exams. They provide the correct approach to solving questions, improve accuracy, and boost confidence. With structured answers aligned to exam expectations, students can easily understand how to present their solutions and score better marks.
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