NCERT Solutions for Class 12 Math Chapter 13 Probability

Chapter 13, Probability, of NCERT Class 12 Mathematics develops probability beyond the basic concepts studied in earlier classes. The chapter covers conditional probability, the multiplication rule of probability, independent events, Bayes' theorem, random variables and probability distributions, mean and variance, and the binomial distribution.

These topics are important for Class 12 board preparation as well as competitive-exam preparation. The chapter requires students to apply probability formulas systematically, particularly in conditional-probability, independence, Bayes' theorem and distribution-based problems.

Infinity Learn's NCERT Solutions for Class 12 Maths Chapter 13 provide step-by-step solutions to the NCERT exercise questions, with the calculations arranged clearly so that students can follow the required method and present their answers effectively in examinations.

Download NCERT Solutions for Class 12 Maths Chapter 13 Probability

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Key Concepts in Class 12 Maths Chapter 13 Probability

The chapter covers:

• Conditional probability and its properties
• Multiplication theorem on probability
• Independent events
• Partition of a sample space
• Theorem of total probability
• Bayes' theorem
• Random variables
• Probability distributions
• Mean and variance of a probability distribution
• Binomial distribution
• Bernoulli trials

NCERT Solutions for Class 12 Maths Chapter 13: Probability

The main topics covered in Chapter 13 are:

Important Formulas

Conditional Probability

P(E | F) = P(E ∩ F) / P(F), P(F) ≠ 0

Multiplication Theorem

P(E ∩ F) = P(E)·P(F | E) or P(E ∩ F) = P(F)·P(E | F)

Independent Events

P(E ∩ F) = P(E)·P(F)

Theorem of Total Probability

If E1, E2, …, En form a partition of the sample space:

P(A) = Σi=1n P(Ei)·P(A | Ei)

Bayes' Theorem

P(Ei | A) = [P(Ei)·P(A | Ei)] / Σj=1n [P(Ej)·P(A | Ej)]

Mean of a Random Variable

E(X) = Σi xi·P(X = xi)

Variance

Var(X) = E(X2) − [E(X)]2

Binomial Distribution

If X follows a binomial distribution with n trials, probability of success p, and q = 1 − p:

P(X = r) = nCr pr qn−r

CBSE Weightage for Probability

In the CBSE Class 12 Mathematics 2026–27 course structure, Unit VI: Probability carries 8 marks out of the 80-mark theory paper.

Students should therefore prepare the complete prescribed Probability syllabus rather than relying on a particular question type or assumed chapter-specific distribution of marks.

NCERT Class 12 Maths Chapter 13: All Exercises

The current NCERT 2026–27 reprint supplied for this page contains the following exercise structure:

The current reprint does not contain separately numbered Exercises 13.4 or 13.5. The chapter itself, however, continues to cover Bayes' theorem, random variables, probability distributions, mean and variance, and binomial distribution.

NCERT Class 12 Maths Chapter 13 Probability: Step-by-Step Solutions

Question 1: Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P(E|F) and P(F|E).

Step 1: Formula for conditional probability

P(E | F) = P(E ∩ F) / P(F)

Step 2: Substitute the given values

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

Similarly,

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

Answer

P(E | F) = 2/3, P(F | E) = 1/3

Question 2: A coin is tossed three times, where 

(i) E: head on third toss, F: heads on first two tosses 

(ii) E: at least two heads, F: at most two heads 

(iii) E: at most two tails, F: at least one tail. Find P(E|F) in each case.

The sample space for three tosses is:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

There are 8 equally likely outcomes.

(i) E = {HHH, THH, HTH, TTH}; F (first two heads) = {HHH, HHT}

Therefore E ∩ F = {HHH}, so:

P(E | F) = (1/8) / (2/8) = 1/2

(ii) E (at least two heads) = {HHH, HHT, HTH, THH}

F (at most two heads) = {HHT, HTH, THH, HTT, THT, TTH, TTT}

Therefore E ∩ F = {HHT, HTH, THH}, so P(E ∩ F) = 3/8 and P(F) = 7/8:

P(E | F) = (3/8) / (7/8) = 3/7

(iii) E (at most two tails) = all outcomes except TTT

F (at least one tail) = all outcomes except HHH

Therefore P(E ∩ F) = 6/8 and P(F) = 7/8:

P(E | F) = (6/8) / (7/8) = 6/7

Answer

(i) P(E | F) = 1/2

(ii) P(E | F) = 3/7

(iii) P(E | F) = 6/7

Question 3: A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all three are good, the box is approved for sale. Find the probability that a box of 15 oranges, 12 good and 3 bad, will be approved for sale.

For the box to be approved, all three selected oranges must be good.

Step 1–3: Probabilities at each draw

P(G1) = 12/15, P(G2 | G1) = 11/14, P(G3 | G1G2) = 10/13

Step 4: Apply the multiplication theorem

P(G1 ∩ G2 ∩ G3) = (12/15) × (11/14) × (10/13) = 44/91

Answer

44/91

Question 4: Probability of solving a specific problem independently by A and B are 1/2 and 1/3 respectively. If both try independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

Given

P(A) = 1/2, P(B) = 1/3

Therefore P(A′) = 1/2 and P(B′) = 2/3. Since A and B work independently, P(A ∩ B) = P(A)·P(B).

Fig. 1 — Venn diagram: the problem is solved if the outcome falls in A ∪ B.

(i) Probability that the problem is solved

The problem is solved if at least one of A or B solves it:

P(A ∪ B) = 1 − P(A′)·P(B′) = 1 − (1/2 × 2/3) = 1 − 1/3 = 2/3

(ii) Probability that exactly one solves the problem

Exactly one solves the problem when A solves and B does not, or B solves and A does not:

P(exactly one) = P(A ∩ B′) + P(A′ ∩ B) = (1/2 × 2/3) + (1/2 × 1/3) = 1/3 + 1/6 = 1/2

Answer

(i) P(problem solved) = 2/3

(ii) P(exactly one) = 1/2

Question 5: An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour noted and returned, with 2 additional balls of that colour added. A second ball is then drawn. Find the probability that the second ball is red.

Let R₁ = red drawn first, B₁ = black drawn first, R₂ = red drawn second. Initially P(R₁) = P(B₁) = 1/2.

Case 1: First ball is red (2 red balls added → 7 red, 5 black)

P(R2 | R1) = 7/12

Case 2: First ball is black (2 black balls added → 5 red, 7 black)

P(R2 | B1) = 5/12

Apply the theorem of total probability

P(R2) = P(R1)P(R2|R1) + P(B1)P(R2|B1)

= (1/2 × 7/12) + (1/2 × 5/12) = 7/24 + 5/24 = 1/2

Answer: 1/2

Question 6: In a college, 60% of students reside in the hostel and 40% are day scholars. 30% of hostellers and 20% of day scholars attain an A grade. A student chosen at random has an A grade — what is the probability that the student is a hosteller?

Given

P(H) = 0.6, P(D) = 0.4

P(A | H) = 0.3, P(A | D) = 0.2

We need P(H | A). Apply Bayes' theorem:

P(H | A) = [P(H)·P(A|H)] / [P(H)·P(A|H) + P(D)·P(A|D)]

= (0.6×0.3) / (0.6×0.3 + 0.4×0.2) = 0.18 / (0.18 + 0.08) = 0.18/0.26 = 9/13

Answer: 9/13

Question 7: A factory has two machines A and B. Machine A produces 60% of output, machine B produces 40%. 2% of A's items and 1% of B's items are defective. An item chosen at random is found defective — what is the probability that it was produced by machine B?

Given

P(A) = 0.6, P(B) = 0.4

P(D | A) = 0.02, P(D | B) = 0.01

Fig. 2 — Probability tree: machine choice, then defective / not-defective outcomes.

We need P(B | D). Apply Bayes' theorem:

P(B | D) = [P(B)·P(D|B)] / [P(A)·P(D|A) + P(B)·P(D|B)]

= (0.4×0.01) / (0.6×0.02 + 0.4×0.01) = 0.004 / (0.012 + 0.004) = 0.004/0.016 = 1/4

Answer: 1/4

Question 8: A couple has two children. 

(i) Find the probability that both children are males, given that at least one is male. 

(ii) Find the probability that both are females, given that the elder child is female.

Assume each child is equally likely to be male or female. The sample space is {MM, MF, FM, FF}, each with probability 1/4.

(i) Both males, given at least one male

Outcomes with at least one male: {MM, MF, FM}

P(both male | at least one male) = P(MM) / [P(MM)+P(MF)+P(FM)] = (1/4) / (3/4) = 1/3

(ii) Both females, given the elder child is female

If the elder child is female, possible outcomes are {FM, FF}; only FF has both female:

P(FF | elder is female) = 1/2

Answer

(i) 1/3

(ii) 1/2

Question 9: Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Given

n = 10, p = 0.9, q = 1 − p = 0.1

Using the binomial probability formula:

P(X = r) = 10Cr (0.9)r (0.1)10−r

We need P(X ≤ 6) = P(X=0) + P(X=1) + ⋯ + P(X=6):

P(X ≤ 6) = Σr=0610Cr(0.9)r(0.1)10−r

= 7996999/625000000 ≈ 0.0128

Answer: P(X ≤ 6) ≈ 0.0128

Question 10: If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

A leap year contains:

366 = 52 × 7 + 2

Every weekday occurs at least 52 times, and two consecutive weekdays occur 53 times. For 53 Tuesdays, Tuesday must be one of the two “extra” weekdays. There are 7 equally likely possibilities for the first extra weekday, and Tuesday occurs in 2 of them:

P(53 Tuesdays) = 2/7

Answer: 2/7

Question 11: Bag I contains 3 red and 4 black balls; Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II. It is found to be red. Find the probability that the transferred ball was black.

Let B = transferred ball is black, R = transferred ball is red, D = ball drawn from Bag II is red. Initially P(B) = 4/7 and P(R) = 3/7.

Case 1: A black ball is transferred (Bag II → 4 red, 6 black)

P(D | B) = 4/10 = 2/5

Case 2: A red ball is transferred (Bag II → 5 red, 5 black)

P(D | R) = 5/10 = 1/2

Apply Bayes' theorem

P(B | D) = [P(B)·P(D|B)] / [P(B)·P(D|B) + P(R)·P(D|R)]

= (4/7 × 2/5) / (4/7 × 2/5 + 3/7 × 1/2) = (8/35) / (8/35 + 3/14) = 16/(16+15) = 16/31

Answer: 16/31

Key Features of These NCERT Solutions

• Exact NCERT question statements for the selected questions
• Step-by-step calculations for numerical problems
• Formula-first approach for probability questions
• Clear working for conditional probability and Bayes' theorem
• Separate treatment of independent-event problems
• Structured probability calculations for repeated trials
• Solutions designed for clear examination presentation
• Coverage of questions from the current NCERT exercise structure
• Useful for CBSE board preparation and competitive-exam practice

Why Choose Infinity Learn NCERT Solutions for This Chapter?

Probability problems can involve several stages of calculation, particularly when conditional probability, independent events and Bayes' theorem are involved. Infinity Learn's solutions present the calculation in a structured sequence so students can identify the appropriate formula, substitute the given values and reach the final answer systematically.

The solutions also help students revise the chapter through NCERT exercise questions rather than relying only on formula memorisation.

Students can use Infinity Learn's Class 12 Maths resources for NCERT solutions, practice material, video lessons and additional exam preparation. Infinity Learn also provides Class 12 learning resources and competitive-exam preparation through its platform.

Download NCERT Solutions for Class 12 Maths Chapter 13 Probability

Download the free PDF of the NCERT Solutions for Class 12 Maths Chapter 13 Probability to revise the complete exercise set and study the step-by-step solutions offline.

For additional Class 12 Maths learning resources, practice material and exam preparation, explore Infinity Learn's Class 12 resources.