Probability Class 12 Exemplar Solutions - Class 12 Maths Ch 13

The Probability Class 12 Exemplar Solutions page compiles NCERT Class 12 Mathematics Chapter 13 into a single download-ready resource, aligned to the 2026-27 NCERT syllabus. The page covers definitions, solved examples, exam-weightage data and common mistakes, with every formula matched to the CBSE marking scheme used in recent board papers.

Compiled by Collegedunia subject specialists against the 2026-27 NCERT Exemplar print, the latest CBSE 2025 board paper and the JEE Main 2025 January plus April sessions.

Also Check:

• Probability Class 12 Maths NCERT Exemplar Book PDF
• these notes Maths NCERT Solutions
• the PDF Maths Formula Sheet

Problem-Type Split in the Exemplar Class 12 Maths Solutions Chapter 13 Probability

The Exemplar bundles its roughly 40 problems across six formats. Tracking the split helps you allocate practice time: MCQ-II and LA are the slow burners, while VSA and Fill in the Blanks are the quick warm-ups before a CBSE pre-board.

In CBSE 2025, two of the five 4-mark questions from the this chapter Exemplar Solutions were near-verbatim Exemplar LA problems.

Probability NCERT Exemplar Video Solutions

Source: Magnet Brains on YouTube

Concept Coverage of the Exemplar Probability Solutions PDF

Chapter 13 is part of Unit VI: Probability. The Exemplar drills four concept blocks more aggressively than the main textbook does, and our solutions tag every problem with the block it tests so revision is targeted.

• Conditional probability and the multiplication rule: P(A ∩ B) = P(A) · P(B|A). Used in MCQ Q1 to Q9 and most VSA questions.
• Independence of events: pairwise vs mutual, distinguishing the two through the MCQ-II problems.
• Bayes theorem: P(E_i|A) = P(E_i) P(A|E_i)∑ P(E_j) P(A|E_j) . The single most JEE-cited formula from these notes Exemplar Solutions.
• Random variables and binomial distribution: mean np , variance npq , and the binomial probability mass function P(X = r) = nr p^r q^n-r .

How Collegedunia Builds Each Exemplar Solution

Every problem in our PDF follows a fixed five-line structure so a student can scan the page during revision and find the load-bearing step in under ten seconds.

1. Concept Used: the formula or theorem stated in full before any substitution.
2. Given / To Find: rewritten in symbolic form so the question's English does not confuse the algebra.
3. Solution: formula, substitution and arithmetic on three separate lines. No step is compressed.
4. Boxed Answer: the final value highlighted so it is unambiguous.
5. Expert Solution: an alternate angle, often a tree diagram or a complementary-event shortcut that saves about 90 seconds in JEE Main.

Tip callouts sit either inside the teal solution container or directly below it, flagging the classic traps. Common ones include reversing the conditional (writing P(A|B) when the question asks for P(B|A)), forgetting that binomial trials must be independent, and skipping the prior probabilities in a Bayes setup.

JEE Main Question Patterns Drawn From the Probability Exemplar

Of the four to six probability questions JEE Main asked across 2024 and 2025, almost every Bayes and binomial question traces back to an Exemplar archetype. The table below maps the dominant patterns latest first.

Around 70% of Probability questions in JEE Main 2024 and 2025 mapped to an Exemplar Bayes or binomial archetype.

Common Mistakes the Exemplar Class 12 Maths Solutions Chapter 13 Probability Flags

Tip callouts in the this Class 12 page Exemplar Solutions call out five recurring traps. These are the same five that account for most lost marks in CBSE and JEE Main probability sections.

• Reversing the conditional: solving for P(A|B) when the question wants P(B|A). Always rewrite the conditional in symbols before computing.
• Treating pairwise independence as mutual: three events can be pairwise independent yet not mutually independent. The Exemplar MCQ-II problems test exactly this.
• Forgetting binomial trial conditions: trials must be independent and have a fixed success probability. A question about drawing without replacement is NOT binomial.
• Dropping priors in Bayes: P(E_i) must appear in both the numerator and the denominator.
• Mean and variance mix-up: variance is npq, NOT np.

Top Formulas From the Probability Exemplar (Quick Recall)

Five formulas account for almost every Exemplar problem. The full list with derivations lives on the the resource Exemplar Solutions formula sheet, but these five are the ones to lock in before solving the Exemplar.

Full PYQ map: the chapter notes Maths Formula Sheet

How Collegedunia's Exemplar Solutions Will Help You

The Probability Exemplar is dense, and a flat answer key does not help in a timed setting. Collegedunia's PDF tags every problem by type, lists the Concept Used up front, and adds an Expert Solution next to each main solution.

That structure mirrors the way a JEE Main aspirant scans during revision, and the way a CBSE student double-checks a 4-mark working two days before the board.

Related Resources

• the PDF Maths NCERT Solutions
• this chapter Maths Notes
• these notes Maths Formula Sheet
• this Class 12 page Maths NCERT Exemplar Book PDF
• the resource Maths Handwritten Notes

NCERT Exemplar Solutions for Class 12 Maths: All Chapters

The table below summarises the recent CBSE Class 12 pattern for this chapter and is a quick pre-exam reference.

the chapter notes Exemplar Solutions: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.

Exercise-wise Breakdown of the Probability Chapter

The Probability chapter splits into 3 numbered exercises plus a Miscellaneous Exercise. The table below maps every exercise to the specific concept it tests, so students can plan revision per exercise and click straight into the worked solutions.

PDF Download Formats and Languages for the Probability Chapter

The Probability Class 12 PDF on this page is available in three formats - each suited to a different revision style. The table below summarises what each format is best for:

The probability class 12 ncert pdf and the parallel Hindi-medium edition both follow the same notation and equation numbering as the printed NCERT 2026-27 release. Key points students should know:

• NCERT-faithful: Every definition, theorem and exercise on the probability class 12 ncert pdf matches the printed textbook line for line.
• Hindi-medium edition: The probability class 12 pdf is also available in Hindi - same page numbering, same equation labels.
• Formula PDF separate: The probability class 12 formulas pdf is a one-page A4 reference sheet listing every identity used in the chapter.
• Solutions PDF separate: The probability class 12 solutions pdf gives every NCERT exercise worked out step by step.
• State-board alignment: Students on the Maharashtra board, HSC, or any state-board syllabus will find the same definitions in this this chapter - only the exercise numbers differ.

Tip: Many toppers keep two parallel copies - a printed formula sheet on A4 for desk revision (the probability class 12 formulas pdf), and the full these notes on a phone for commute revision. Both files are free and linked above.

Important Questions and Previous Year Trends for the Probability Chapter

The most repeated question patterns in CBSE Class 12 Maths for the Probability chapter have settled into a stable cluster across 2019 to 2024 boards. Three question templates account for over 80% of the marks this chapter contributes:

Walking through one example of each template before the exam covers most of the predictable probability class 12 important questions you will see on board day.

• this Class 12 page previous year questions for 2019-2024 are linked from the PYQ block at the bottom of this page - the exact CBSE phrasings.
• The probability class 12 important questions with solutions set is reused by toppers in the last fortnight of revision.
• For NCERT Exemplar practice, the matching the resource extra questions set adds advanced problems suitable for JEE Main and JEE Advanced.
• The MCQ pattern in CBSE has stabilised around 1-2 questions per shift from this chapter - mostly short calculations or assertion-reason items.

Year-wise PYQ Distribution

The table below maps the dominant question type asked from the Probability chapter across recent CBSE Class 12 Maths boards:

The full probability class 12 important questions with solutions set (every year, every paper, every question type) is linked from the PYQ page at the bottom of this article.

How the Probability Notes Pair with NCERT Solutions and the Formula Sheet

The Probability Class 12 notes work best when paired with two sister resources from the Class 12 Maths hub. The table below shows how each resource fits into a typical revision week:

Around 60 percent of the chapter's scoring vocabulary appears on all three pages, so cross-resource use reinforces recall without adding study time.

• The probability class 12 ncert solutions cover every back-of-chapter exercise plus the miscellaneous exercise.
• The probability class 12 solutions for each individual exercise are indexed by exercise number on the sister NCERT Solutions page (see the Exercise-wise Breakdown table above for direct links).
• The probability class 12 formulas reference sheet is the same A4 file students sometimes refer to as probability class 12 all formulas - it lists every identity used in the chapter.
• State-board references: RD Sharma, ML Aggarwal, Teachoo and the Maharashtra board the chapter notes textbook PDF all share the same core definitions.
• For class-first search phrasings - class 12 probability solutions, class 12 probability ncert solutions, ncert class 12 probability solutions - the same files cover the request.

Reference Books and State-Board Mapping

Students using reference books beyond NCERT, or studying under a state board, can map this chapter cleanly:

How to Use the Probability Notes Page Most Effectively

The recommended study plan for these notes chapter splits across three sittings. The table below outlines what to do in each:

For students preparing for both CBSE board and JEE Main:

• 60 percent of revision time on NCERT - irreplaceable for board marking-scheme phrasings.
• 40 percent of revision time on JEE-style problem sets - sharpens speed and conceptual depth.
• The probability class 12 important questions set on the previous-year page is the closest free analogue to a JEE-style problem set for this chapter.
• For CUET (UG) Mathematics, focus on definitions and one-step applications - CUET's MCQ pattern rewards reflexive recall.

All NCERT Exemplar Questions for Probability with Step-by-Step Solutions

Every question of the NCERT Exemplar set for Class 12 Mathematics Chapter 13 Probability is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.

I. Multiple Choice Questions (MCQ)

Q 13.1

If P(A)=45 and P(A∩ B)=710, then P(BA) is equal to
(A) 110      (B) 18      (C) 78      (D) 1720

Check SolutionExpert Solution

Correct option: (C) 78.

Concept used. The conditional probability of event B given that event A has occurred is defined as P(BA) = P(A∩ B)P(A), P(A) > 0. This is the basic formula introduced in 13.2 of the NCERT chapter.

1. Identify the given values: P(A) = 45 and P(A∩ B) = 710.
2. Substitute into the conditional formula: P(BA) = P(A∩ B)P(A) = 7/104/5.
3. Simplify the complex fraction: 7/104/5 = 710×54 = 7× 510× 4 = 3540 = 78.
4. Sanity check: 0≤ P(BA) = 7/8 ≤ 1 .

P(BA) = 78; option (C).

AS

Aarav Sharma

M.Sc Mathematics, IIT Kanpur

Verified Expert

Direct substitution.

Concept used. P(BA) = P(A∩ B)P(A).

1. Plug in: 7/104/5.
2. Re-write as multiplication by reciprocal: 710·54=3540.
3. Reduce: gcd(35,40)=5, so =78.

78; option (C).

Q 13.2

If P(A∩ B) = 710 and P(B)=1720, then P(AB) equals
(A) 1417      (B) 1720      (C) 78      (D) 18

Check SolutionExpert Solution

Correct option: (A) 1417.

Concept used. Conditional probability formula: P(AB) = P(A∩ B)/P(B) provided P(B)>0.

1. Substitute: P(AB) = 7/1017/20.
2. Re-write as multiplication: 710·2017 = 7× 2010× 17 = 140170.
3. Reduce: gcd(140,170)=10: 140170 = 1417.
4. Sanity: 14/17 ≈ 0.824 ∈ [0,1] .

P(AB) = 1417; option (A).

PI

Priya Iyer

Ph.D Mathematics, IISc Bangalore

Verified Expert

Cross-multiplication angle. P(AB) = P(A∩ B)/P(B): 7/1017/20 = 7· 2010· 17 = 140170 = 1417.

1417; option (A).

Q 13.3

If P(A) = 310, P(B)=25 and P(A∪ B)=35, then P(BA) + P(AB) equals
(A) 14      (B) 13      (C) 512      (D) 712

Check SolutionExpert Solution

Correct option: (D) 712.

Concept used. We need P(A∩ B) first (from the addition theorem), then use conditional probability twice.

1. Find P(A∩ B) using the addition theorem P(A∪ B) = P(A)+P(B)-P(A∩ B): 35 = 310+25-P(A∩ B).
2. Combine right-hand side: 310+25= 310+410=710.
3. So 35 = 710 - P(A∩ B), giving P(A∩ B) = 710 - 35 = 710 -610 = 110.
4. Now compute the two conditional probabilities: P(BA) = P(A∩ B)P(A) = 1/103/10 = 13, P(AB) = P(A∩ B)P(B) = 1/102/5 = 110·52 = 14.
5. Add them: P(BA) + P(AB) = 13 + 14 = 4+312 = 712.

P(BA) + P(AB) = 712; option (D).

VR

Vikram Rao

M.Sc Mathematics, Delhi University

Verified Expert

Symmetry angle. Note that P(BA) + P(AB) = P(A∩ B)(1P(A) + 1P(B)) = P(A∩ B) ·P(A)+P(B)P(A)P(B).

Concept used. Pulling out the common factor P(A∩ B).

1. Compute P(A∩ B) = 1/10 as before.
2. P(A)+P(B)P(A)P(B) = 3/10+2/5(3/10)(2/5) = 7/106/50 = 710·506 = 356.
3. Product: 110·356 = 3560 = 712.

712; option (D).

Q 13.4

If P(A) = 25, P(B)=310 and P(A∩ B)=15, then P(A'B')· P(B'A') is equal to
(A) 56      (B) 57      (C) 2542      (D) 1

Check SolutionExpert Solution

Correct option: (C) 2542.

Concept used. For complements: P(A')=1-P(A), and P(A'∩ B') = P((A∪ B)') = 1 - P(A∪ B).

1. Compute P(A∪ B) = P(A)+P(B)-P(A∩ B) = 25+310-15 = 410+310-210 = 510 = 12.
2. P(A') = 1 - 25 = 35. P(B') = 1 - 310 = 710. P(A'∩ B') = 1 - P(A∪ B) = 1 - 12 = 12.
3. Compute the two conditional probabilities: P(A'B') = P(A'∩ B')P(B') = 1/27/10 = 12·107 = 1014=57. P(B'A') = P(A'∩ B')P(A') = 1/23/5 = 12·53 = 56.
4. Product: 57·56 = 2542.

P(A'B')· P(B'A') = 2542; option (C).

AS

Aarav Sharma

M.Sc Mathematics, IIT Kanpur

Verified Expert

Complement-trick angle. The denominator of both conditionals is a complement probability; the numerator P(A'∩ B') = 1-P(A∪ B) saves three steps.

Concept used. P(A'∩ B') = P((A∪ B)') –- De Morgan.

1. P(A∪ B) = 0.4+0.3-0.2 = 0.5, so P(A'∩ B') = 0.5.
2. Conditionals: P(A'B')=0.5/0.7=5/7; P(B'A')=0.5/0.6=5/6.
3. Product = (5/7)(5/6) = 25/42.

2542; option (C).

Q 13.5

Three persons A, B and C, fire at a target in turn, starting with A. Their probabilities of hitting the target are 0.4, 0.3 and 0.2 respectively. The probability of two hits is
(A) 0.024      (B) 0.188      (C) 0.336      (D) 0.452

Check SolutionExpert Solution

Correct option: (B) 0.188.

Concept used. The shots are independent, so we use P(specific outcome) = ∏ P_i^hit· ∏ (1-P_j)^miss. ``Exactly two hits'' means one of the three persons misses; we sum over all three ``one-misser'' cases.

1. Let a=P(A hits)=0.4, b=0.3, c=0.2 and a'=0.6, b'=0.7, c'=0.8.
2. Three cases of ``exactly two hits'':
   1. [label=()]
   2. A and B hit, C misses: a· b· c' = 0.4(0.3)(0.8) = 0.096.
   3. A and C hit, B misses: a· b'· c = 0.4(0.7)(0.2) = 0.056.
   4. B and C hit, A misses: a'· b· c = 0.6(0.3)(0.2) = 0.036.
3. Add all three (mutually exclusive cases): P(exactly two hits) = 0.096 + 0.056 + 0.036 = 0.188.

0.188; option (B).

PI

Priya Iyer

Ph.D Mathematics, IISc Bangalore

Verified Expert

Generating-function angle. The probability of exactly k hits is the coefficient of z^k in (0.6+0.4z)(0.7+0.3z)(0.8+0.2z).

Concept used. Multiply out and collect coefficients.

1. Coefficient of z²: pick z from two factors and the constant from one.
2. Three ways: (0.4)(0.3)(0.8)+(0.4)(0.7)(0.2)+(0.6)(0.3)(0.2) = 0.096 + 0.056 + 0.036 = 0.188.

0.188; option (B).

Q 13.6

In a college, 30% of the students fail in Physics, 25% fail in Mathematics and 10% fail in both. One student is selected at random. The probability that the student fails in Physics given that he has failed in Mathematics is
(A) 110      (B) 25      (C) 920      (D) 13

Check SolutionExpert Solution

Correct option: (B) 25.

Concept used. Translate percentages to probabilities, then apply conditional probability P(F_PF_M) = P(F_P∩ F_M)/P(F_M).

1. Let F_P = fails Physics, F_M = fails Mathematics. Given P(F_P) = 0.30, P(F_M)=0.25, P(F_P∩ F_M)=0.10.
2. Apply conditional formula: P(F_PF_M) = P(F_P∩ F_M)P(F_M) = 0.100.25.
3. Compute: 0.100.25 = 1025 = 25.

P(F_PF_M) = 25; option (B).

VR

Vikram Rao

M.Sc Mathematics, Delhi University

Verified Expert

Reduced-sample-space angle. ``Given failed in Math'' restricts the universe to the 25% who failed Math. Of these, 10% also failed Physics. Fraction =10/25 = 2/5.

25; option (B).

Q 13.7

If A and B are independent events with P(A)=13 and P(B)=14, then P(A'∩ B') equals
(A) 112      (B) 14      (C) 12      (D) 23

Check SolutionExpert Solution

Correct option: (C) 12.

Concept used. Two events A,B are independent iff P(A∩ B)=P(A)P(B). A standard Theorem (NCERT 13.4): if A,B are independent, then so are A',B'. Therefore P(A'∩ B') = P(A')· P(B').

1. Compute the complements: P(A') = 1-P(A) = 1-13 = 23. P(B') = 1-14 = 34.
2. Use independence of A',B': P(A'∩ B') = P(A')· P(B') = 23·34 = 612 = 12.

P(A'∩ B') = 12; option (C).

AS

Aarav Sharma

M.Sc Mathematics, IIT Kanpur

Verified Expert

Direct-multiplication angle. For independent events, complements are independent. So P(A'∩ B') = (1-P(A))(1-P(B)) = 23·34 = 12.

12; option (C).

Q 13.8

A die is thrown. Let A be the event ``odd number turns up'' and B the event ``a number ≤ 4 turns up''. Then P(AB) is
(A) 12      (B) 13      (C) 23      (D) 34

Check SolutionExpert Solution

Correct option: (A) 12.

Concept used. For equiprobable finite sample spaces, conditional probability can be computed by counting: P(AB) = |A∩ B|/|B|.

1. Sample space S=1,2,3,4,5,6.
2. Event A (odd): A=1,3,5, |A|=3.
3. Event B (≤ 4): B=1,2,3,4, |B|=4.
4. A∩ B = 1,3, |A∩ B|=2.
5. P(AB) = |A∩ B||B| = 24 = 12.

P(AB) = 12; option (A).

PI

Priya Iyer

Ph.D Mathematics, IISc Bangalore

Verified Expert

Counting angle.

Concept used. In an equiprobable sample space, P(AB) = (favorable outcomes in B) / (total outcomes in B).

1. Restrict universe to B=1,2,3,4: 4 outcomes.
2. Within B, the odd ones are 1,3: 2 outcomes.
3. Probability = 2/4 = 1/2.

12; option (A).

Q 13.9

If A and B are two events such that P(A)=12, P(B)=13, and P(AB)=14, then P(A'∩ B') equals
(A) 112      (B) 316      (C) 34      (D) 14

Check SolutionExpert Solution

Correct option: (D) 14.

Concept used. Use P(A∩ B) = P(B)P(AB), then P(A∪ B) = P(A)+P(B)-P(A∩ B), and finally P(A'∩ B') = 1 - P(A∪ B).

1. Compute P(A∩ B): P(A∩ B) = P(B)· P(AB) = 13·14 = 112.
2. Compute P(A∪ B): P(A∪ B) = P(A)+P(B)-P(A∩ B) = 12+13-112.
3. Common denominator 12: 612+412-112=912 =34.
4. Apply De Morgan: P(A'∩ B') = P((A∪ B)') = 1-P(A∪ B) = 1 - 34 = 14.

P(A'∩ B') = 14; option (D).

VR

Vikram Rao

M.Sc Mathematics, Delhi University

Verified Expert

Walk-the-chain.

1. P(A∩ B) = 1/12.
2. P(A∪ B) = 6/12+4/12-1/12 = 9/12 = 3/4.
3. P(A'∩ B') = 1-3/4 = 1/4.

14.

Q 13.10

If X is a random variable with the probability distribution

tabularc|cccc X & 0 & 1 & 2 & 3
P(X) & k & 2k & 3k & 4k tabular

then E(X) equals
(A) 1      (B) 32      (C) 2      (D) 52

Check SolutionExpert Solution

Correct option: (C) 2.

Concept used. A discrete probability distribution requires ∑ p_i = 1. Use this to find k. The mean (or expectation) is E(X) = ∑ x_i p_i.

1. Find k from the normalisation ∑ p_i = 1: k+2k+3k+4k = 10k = 1 ⇒ k = 110.
2. Verify each p_i∈[0,1]: p₀=0.1, p₁=0.2, p₂=0.3, p₃=0.4 .
3. Compute E(X) = ∑ x_i p_i: E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4) = 0 + 0.2 + 0.6 + 1.2 = 2.0.

E(X) = 2; option (C).

AS

Aarav Sharma

M.Sc Mathematics, IIT Kanpur

Verified Expert

Normalise + sum.

1. 10k=1 ⇒ k=0.1.
2. Weighted sum: 0.0+0.2+0.6+1.2 = 2.0.

E(X)=2; option (C).

II. Short Answer Questions (SA)

Q 13.11

An unbiased coin is tossed 5 times. Find the probability of getting exactly 3 heads.

Check SolutionExpert Solution

Concept used. Binomial distribution. The number of heads X in n independent tosses of a fair coin follows X∼ B(n,p=12) with PMF P(X=r) = nr p^r(1-p)^n-r.

1. Identify parameters: n=5, p=12, q=1-p=12, r=3.
2. Substitute: P(X=3) = 53(12)³ (12)².
3. Compute 53: 53=5!3! 2!=5· 42=10.
4. Combine: (12)³(12)² = (12)⁵ = 132.
5. Multiply: P(X=3) = 10·132 = 1032 = 516.

P(X=3) = 516 = 0.3125.

PI

Priya Iyer

Ph.D Mathematics, IISc Bangalore

Verified Expert

Sample-space counting angle. For a fair coin, all 2⁵=32 sequences are equally likely. Number of sequences with exactly 3 heads: 53=10. So P=10/32=5/16.

Concept used. Equiprobable sample-space approach (works because p=1/2).

516.

Q 13.12

Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. Find the probability that both cards are kings.

Check SolutionExpert Solution

Concept used. Multiplication theorem for dependent events: P(A∩ B) = P(A)· P(BA). The two draws are dependent because the first king is not replaced.

1. Let A = first card is a king, B = second card is a king. P(A) = 452 (four kings in 52).
2. Given A, the deck has 51 cards, 3 kings remaining. So P(BA) = 351.
3. Multiplication theorem: P(A∩ B) = P(A)· P(BA) = 452· 351 = 122652.
4. Simplify: 122652 = 1221 (divide num and denom by 12).
5. Verification by alternative count: 42/522 = 6/1326 = 1/221 .

P(both kings) = 1221.

VR

Vikram Rao

M.Sc Mathematics, Delhi University

Verified Expert

Combinations angle. P(both kings) = 42522 = 61326=1221.

Concept used. Symmetric counting –- ``ways to choose 2 kings out of 4'' over ``ways to choose any 2 cards out of 52''.

1221.

Q 13.13

The probability distribution of a random variable X is given by

tabularc|ccccc X & 0 & 1 & 2 & 3 & 4
P(X) & 0.1 & k & 0.3 & 2k & 0.1 tabular

Find k and the variance of X.

Check SolutionExpert Solution

Concept used. Use ∑ p_i = 1 to find k; then compute E(X) and E(X²); finally use Var(X) = E(X²) - [E(X)]².

1. Normalisation: 0.1 + k + 0.3 + 2k + 0.1 = 1 ⇒ 0.5 + 3k = 1 ⇒ k = 0.53 = 16≈ 0.1667. So k = 16 and 2k = 13.
2. Re-write the distribution with numerical probabilities:
   tabularc|ccccc X & 0 & 1 & 2 & 3 & 4
   P & 0.1 & 16 & 0.3 & 13 & 0.1 tabular
   Or with common denominator 30: 330, 530, 930, 1030, 330. Sum = 30/30 = 1 .
3. Compute E(X) = ∑ x_i p_i: E(X) = 0·330+1·530+2·930 +3·1030+4·330. Compute each: 0 + 530+1830+3030+1230 = 6530 = 136.
4. Compute E(X²) = ∑ x_i² p_i: E(X²) = 0+1·530+4·930+9·1030 +16·330. Compute each: 0+530+3630+9030+4830 = 17930.
5. Variance: Var(X) = E(X²) - [E(X)]² = 17930 - (136)².
6. Compute (13/6)² = 169/36. Common denominator 180: 17930 = 1074180, 16936= 845180. Subtract: 1074-845180 = 229180≈ 1.2722.

k=16; Var(X) = 229180 ≈ 1.27.

AS

Aarav Sharma

M.Sc Mathematics, IIT Kanpur

Verified Expert

Tabular angle. Build a single table with columns x, p, xp, x²p and total at the bottom.

Concept used. Direct tabular evaluation of E(X) and E(X²) side by side.

1. ∑ p = 1 ⇒ k = 1/6.
2. ∑ xp = 13/6 ≈ 2.1667.
3. ∑ x² p = 179/30 ≈ 5.9667.
4. Var = 5.9667 - (2.1667)² = 5.9667 - 4.6944 = 1.2722.

k=1/6; Var(X)≈ 1.272.

Q 13.14

In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper, and 20% read both. A student is selected at random. Find the probability that he reads neither Hindi nor English newspapers.

Check SolutionExpert Solution

Concept used. Use complement of the union: P(neither) = 1 - P(at least one) = 1 - P(H∪ E), with P(H∪ E) = P(H) + P(E) - P(H∩ E).

1. Let H = reads Hindi, E = reads English. Given P(H)=0.6, P(E)=0.4, P(H∩ E)=0.2.
2. Compute P(H∪ E) by the addition theorem: P(H∪ E) = 0.6 + 0.4 - 0.2 = 0.8.
3. ``Neither'' is the complement: P(H'∩ E') = 1 - P(H∪ E) = 1 - 0.8 = 0.2.

P(neither) = 0.2.

PI

Priya Iyer

Ph.D Mathematics, IISc Bangalore

Verified Expert

Inclusion–exclusion + De Morgan. One line: P(neither) = 1-(0.6+0.4-0.2) = 0.2.

0.2.

Q 13.15

Two dice are thrown together. What is the probability that the sum of the numbers on the two dice is 8, if it is known that the second die always exhibits an odd number?

Check SolutionExpert Solution

Concept used. Conditional probability on the reduced sample space defined by ``second die odd''.

1. Sample space restricted by ``second die odd'': second die can be 1,3,5, so the restricted sample space is (i,j) : i∈1,…,6, j∈1,3,5 with 6× 3 = 18 equally likely outcomes.
2. Sum equals 8 on this restricted space: need i+j=8 with j∈1,3,5.
   1. [label=()]
   2. j=1⇒ i=7: impossible (i≤ 6).
   3. j=3⇒ i=5: (5,3) .
   4. j=5⇒ i=3: (3,5) .
   Two outcomes.
3. Conditional probability: P(sum=8second die odd) = 218 = 19.

P = 19.

VR

Vikram Rao

M.Sc Mathematics, Delhi University

Verified Expert

Pair-spotter angle. Need ordered pair (i,j) with i+j=8 and j odd. Two ordered pairs: (5,3), (3,5). Restricted total =18.

Concept used. Count favourable, divide by restricted total.

19.

III. Long Answer Questions (LA)

Q 13.16

In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4, 2 per cent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by machine B?

Check SolutionExpert Solution

Concept used. Bayes' theorem. Let E₁= bolt is from A, E₂= from B, E₃= from C (a partition of the sample space). Let D= bolt is defective. We want P(E₂D): P(E₂D) = P(E₂) P(DE₂)P(E₁)P(DE₁) + P(E₂)P(DE₂) + P(E₃)P(DE₃).

1. Tabulate priors and likelihoods:
   tabularc|cc Machine & P(E_i) & P(DE_i)
   A & 0.25 & 0.05
   B & 0.35 & 0.04
   C & 0.40 & 0.02 tabular
2. Compute joint probabilities P(E_i)P(DE_i): aligned P(E₁)P(DE₁) &= 0.25× 0.05 = 0.0125,
   P(E₂)P(DE₂) &= 0.35× 0.04 = 0.0140,
   P(E₃)P(DE₃) &= 0.40× 0.02 = 0.0080. aligned
3. Total probability of defect: P(D) = 0.0125 + 0.0140 + 0.0080 = 0.0345.
4. Apply Bayes': P(E₂D) = 0.01400.0345.
5. Multiply numerator and denominator by 10 000: 140345. Reduce by gcd(140,345) = 5: 2869.
6. Numerical value: 28/69 ≈ 0.4058.
7. Sanity check. Verify P(E₁D)+P(E₂D)+P(E₃D) = 1: 125345+140345+80345 = 345345 = 1 .

P(E₂D) = 2869 ≈ 0.406.

AS

Aarav Sharma

M.Sc Mathematics, IIT Kanpur

Verified Expert

Probability-tree angle. Branch out: machine choice (A/B/C) at the first level, defect / no-defect at the second. The posterior P(Bdefect) is the conditional weight of the B branch among the defect leaves.

Concept used. Bayes' theorem as a normalised reweighting of priors by likelihoods.

1. Defect leaf weights: A→ 25· 5 = 125, B→ 35· 4 = 140, C→ 40· 2 = 80. Sum =345.
2. Posterior for B: 140/345 = 28/69.

2869.

Q 13.17

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the mean and standard deviation of the number of aces.

Check SolutionExpert Solution

Concept used. Hypergeometric-style discrete distribution. X = number of aces in two cards drawn from a 52-card deck, with 4 aces and 48 non-aces. So X∈0,1,2.

1. Total ways to draw 2 cards: 522 = 1326.
2. Compute P(X=0): 2 non-aces from 48. P(X=0) = 482522 = 11281326 = 188221.
3. Compute P(X=1): 1 ace from 4 and 1 non-ace from 48. P(X=1) = 41481522 = 4· 481326 = 1921326 = 32221.
4. Compute P(X=2): 2 aces from 4. P(X=2) = 42522 = 61326 = 1221.
5. Verification: 188+32+1221=221221 = 1 .
6. Distribution table:
   tabularc|ccc X & 0 & 1 & 2
   P(X) & 188221 & 32221 & 1221 tabular
7. Compute mean E(X): E(X) = 0·188221+1·32221 +2·1221 = 0+32+2221 = 34221 = 213. (Divide: 34=2· 17, 221=13· 17, cancel 17.)
8. Compute E(X²): E(X²) = 0+1·32221+4·1221 = 36221.
9. Variance: Var(X) = E(X²)-[E(X)]² = 36221 - (213)². Compute (2/13)² = 4/169. Common denominator: 221 = 13· 17, 169 = 13· 13. LCM =13· 13· 17 = 2873. So 36221 = 36· 132873 = 4682873. 4169 = 4· 172873 = 682873. Subtract: 468-682873 = 4002873. Reduce: gcd(400,2873)=1, so 4002873≈ 0.1393.
10. Standard deviation: σ = √Var(X) = √4002873 ≈ 0.373.

E(X) = 213; σ = √4002873 ≈ 0.373.

PI

Priya Iyer

Ph.D Mathematics, IISc Bangalore

Verified Expert

Trial-by-trial angle. Compute mean via linearity of expectation: X = I₁ + I₂ where I_k = 1 if the kth card is an ace.

Concept used. E(I_k) = P(kth card is an ace) = 4/52 = 1/13 by symmetry (every position is equally likely to be an ace).

1. E(X) = E(I₁) + E(I₂) = 2 · 113 = 213. Match!
2. For variance, use E(X²) from the distribution (the indicators are dependent, so Var is not just a sum): Var(X) = 36/221 - 4/169 = 400/2873.
3. Standard deviation σ = √Var(X) = √400/2873 ≈ 0.373.

E(X) = 213; σ = √4002873 ≈ 0.373.

Q 13.18

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.

Check SolutionExpert Solution

Concept used. Binomial distribution with n=4 independent trials, where ``success'' is a doublet (both dice show the same number) on a single throw of two dice.

1. Probability of a doublet on one throw. Sample space of throwing two dice has 36 outcomes. Doublets are (1,1),(2,2),(3,3),(4,4),(5,5),(6,6): 6 outcomes. So p = 636 = 16, and q = 1-p = 56.
2. Binomial setup. X = number of doublets in n=4 independent throws. X∼ B(n=4, p=1/6) with PMF P(X=r) = 4r(16)^r (56)^4-r.
3. Compute P(X=2). Plug r=2: P(X=2) = 42(16)² (56)². Compute 42=6. Compute (1/6)² = 1/36 and (5/6)² = 25/36. So P(X=2) = 6·136·2536 = 6· 2536· 36 = 1501296.
4. Reduce: gcd(150,1296) = 6: 1501296 = 25216.
5. Decimal check: 25/216 ≈ 0.1157.

P(X=2) = 25216≈ 0.116.

VR

Vikram Rao

M.Sc Mathematics, Delhi University

Verified Expert

Step-by-step substitution.

1. Doublet probability p = 6/36 = 1/6.
2. Apply Binomial: P(X=2) = 42(1/6)²(5/6)².
3. = 6· (1/36)· (25/36) = 150/1296 = 25/216.

25216.

Class 12 Mathematics Revision Strategy and Exam Practice Routines

Most CBSE Class 12 students benefit from a three-pass revision rhythm: the first pass is slow and definition-by-definition, the second works through every back-of-chapter problem, and the third uses past board papers at exam pace. JEE and CUET aspirants should add a fourth pass focused on the JEE-specific question bank, because the same chapter content gets tested under different time pressure. Within these passes, a few habits separate students who hit the 85+ band from the rest:

• Read two previous-year marking schemes before the exam — marking-scheme phrasings reward exact wording, which pays off more than another mock paper.
• Write a one-page formula recall sheet per chapter that fits on one side of A4; the night before the exam should be spent only on this sheet and a single full-length mock.
• Solve the CBSE 2026-27 sample paper twice — it is the highest-fidelity guide to question difficulty and lifts mock-paper accuracy by 8 to 12 percent.
• Self-evaluate every two hours by writing the chapter's key results from memory, rather than reading passively.
• Finish back-of-chapter exercises once and revisit the miscellaneous exercise twice — past-board data shows this is worth roughly 2 extra marks.

Common arithmetic slips cost most students at least one mark per paper, and most marks lost in long-answer questions go to incomplete working, not wrong answers. Write every intermediate step in full, even on questions that feel straightforward — method marks are claimed step by step even when the final number is off. The case-study format introduced in recent CBSE boards now appears regularly, framing a real-world scenario that tests definitions plus one-step applications, so practising case studies from the CBSE sample paper translates directly into marks.

Time allocation in the last fortnight matters most. Two thirds of revision time should go to weak chapters, the remaining third to maintaining strong ones; students who revise this chapter twice in the last 10 days score 1.5 to 2 marks higher on past boards. The night before the exam is best spent on:

• The one-page formula recall sheet built earlier in revision.
• A single full-length mock paper at exam timing.
• Avoid learning any new material the night before — sleep matters more.

Mock papers serve two distinct purposes — subject mocks build chapter-level recall while full-paper mocks build time-management discipline. Tracking your own mock-paper scores week by week is the single best predictor of board outcome; a simple spreadsheet with date, paper, score, and one note on a recurring mistake is enough. For students using only one reference, the printed NCERT remains the highest-yield resource — books beyond NCERT add depth but rarely change board outcomes, since the marking scheme rewards NCERT phrasing first. Hindi-medium students can keep the bilingual NCERT edition handy because it follows the same notation, and group study works best when each student picks one sub-topic to explain.

Past CBSE marking schemes from 2020 to 2024 show that average board marks for Class 12 Maths have settled around the 75 to 82 percent band. Students who hit the upper end usually share the same revision rhythm: NCERT first, mock papers second, and previous-year papers third.