RD Sharma Solutions for Class 12 Maths Chapter 31 – Probability (Free PDF Updated for 2025–26)

RD Sharma Solutions for Class 12 Chapter 31 – Probability are provided here to help students effectively resolve doubts while working through textbook exercises. These step-by-step solutions are curated by expert subject faculty and follow the latest CBSE syllabus and exam guidelines.

In this chapter, students explore the axiomatic approach to probability, introduced by Russian mathematician A.N. Kolmogorov. This method is compared with the classical theory of probability, especially in cases involving equally likely outcomes. Using this foundation, students learn to calculate probabilities for events within discrete sample spaces.

As an extension of the addition theorem, the concept of conditional probability is introduced, leading to the multiplication rule of probability. The RD Sharma solutions for Class 12 also cover key theorems that help compute the probability of simultaneous events—an important application in complex problem-solving.

Towards the end of the chapter, students will gain a strong understanding of the total probability theorem and Bayes’ theorem through detailed solved examples and practice problems. These RD Sharma Solutions offer a structured and simplified learning path, enabling students to confidently master all core probability concepts for their Class 12 board exams.

RD Sharma Probability PDF Download – (2025–26) Edition

Click below to download the updated PDF of RD Sharma Class 12 Chapter 31 – Probability:

• Step-by-step CBSE-compliant answers
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Access answers to RD Sharma Solutions For Class 12 Chapter 31 Probability

Question: If P(A) = 0.6, P(B) = 0.3, and P(A ∩ B) = 0.2, find P(A|B) and P(B|A).

Solution: P(A|B) = P(A ∩ B) / P(B) 

= 0.2 / 0.3 

= 2/3

P(B|A) = P(A ∩ B) / P(A) 

= 0.2 / 0.6 

= 1/3

Question: Given P(A) = 7/13, P(B) = 9/13, and P(A ∩ B) = 4/13, find P(A|B).

Solution: P(A|B) = P(A ∩ B) / P(B)

= (4/13) / (9/13)

= 4/9

Question: If P(A ∩ B) = 0.32 and P(B) = 0.5, find P(A|B).

Solution: P(A|B) = P(A ∩ B) / P(B)

= 0.32 / 0.5 

= 0.64

Question: Given P(A) = 0.4, P(B) = 0.8, and P(B|A) = 0.6, find P(A ∩ B), P(A|B), and P(A ∪ B).

Solution: P(A ∩ B) = P(B|A) × P(A) = 0.6 × 0.4 = 0.24

P(A|B) = P(A ∩ B) / P(B) = 0.24 / 0.8 = 0.3

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.8 − 0.24 = 0.96

Question: If P(A) = 1/3, P(B) = 1/4, and P(A ∪ B) = 5/12, find P(A ∩ B).

Solution: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = (1/3) + (1/4) − (5/12)

= (4/12 + 3/12 − 5/12) = 2/12 = 1/6

Question: Given P(A) = 0.5, P(B) = 0.6, and P(A ∩ B) = 0.3, find P(A ∪ B).

Solution: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

= 0.5 + 0.6 − 0.3 = 0.8

Question: If P(A) = 0.7, P(B) = 0.5, and P(A ∩ B) = 0.35, find P(A ∪ B).

Solution: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) 

= 0.7 + 0.5 − 0.35 

= 0.85

Question: A bag contains 3 red and 2 black balls. Another bag contains 2 red and 3 black balls. One bag is chosen at random, and a ball is drawn. Find the probability that the ball is red.

Solution: Let E₁: Choosing Bag 1, E₂: Choosing Bag 2

P(E₁) = P(E₂) = 1/2

P(Red|E₁) = 3/5, P(Red|E₂) = 2/5

P(Red) = P(E₁) × P(Red|E₁) + P(E₂) × P(Red|E₂)

= (1/2 × 3/5) + (1/2 × 2/5) 

= (3/10) + (1/5) 

= 1/2

Question: A factory has two machines A and B. Machine A produces 60% of items, and B produces 40%. 5% of items from A and 10% from B are defective. If an item is selected at random and found defective, what is the probability it was produced by machine A?

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

P(D|A) = 0.05, P(D|B) = 0.10

P(D) = P(A)×P(D|A) + P(B)×P(D|B) = (0.6×0.05) + (0.4×0.10) = 0.03 + 0.04 = 0.07

P(A|D) = (P(A)×P(D|A)) / P(D) 

= (0.6×0.05)/0.07 

= 0.03/0.07 ≈ 0.4286

Question: A test for a disease gives a positive result with 99% probability if the person has the disease and with 5% probability if the person doesn't have the disease. If 0.1% of the population has the disease, what is the probability that a person has the disease given that the test result is positive?

Solution: Let D: has disease, T⁺: test positive

P(D) = 0.001, P(¬D) = 0.999

P(T⁺|D) = 0.99, P(T⁺|¬D) = 0.05

P(T⁺) = P(D)×P(T⁺|D) + P(¬D)×P(T⁺|¬D) = (0.001×0.99) + (0.999×0.05) = 0.00099 + 0.04995 = 0.05094

P(D|T⁺) = (P(D)×P(T⁺|D)) / P(T⁺) 

= 0.00099 / 0.05094 ≈ 0.0194

Question: A student has a 70% chance of passing Mathematics and an 80% chance of passing Physics. The probability of passing both is 60%. What is the probability that the student passes at least one subject?

Solution: P(M ∪ P) = P(M) + P(P) − P(M ∩ P) 

= 0.7 + 0.8 − 0.6 

= 0.9

Question: In a class, 60% of students are girls. 70% of girls and 50% of boys pass an exam. What is the probability that a student who passed is a girl?

Solution: P(G) = 0.6, P(B) = 0.4

P(Pass|G) = 0.7, P(Pass|B) = 0.5

P(Pass) = (0.6×0.7) + (0.4×0.5) = 0.42 + 0.2 = 0.62

P(G|Pass) = (0.6×0.7) / 0.62 

= 0.42 / 0.62 ≈ 0.677

Question: A box contains 5 red and 3 blue balls. Two balls are drawn without replacement. What is the probability that both are red?

Solution: P(First red) = 5/8

P(Second red | First red) = 4/7

P(Both red) = (5/8) × (4/7) 

= 20/56 

= 5/14

Question: A card is drawn from a well-shuffled deck. What is the probability that it is a king or a heart?

Solution: P(King) = 4/52, P(Heart) = 13/52

P(King ∩ Heart) = 1/52

P(King ∪ Heart) = P(King) + P(Heart) − P(King ∩ Heart) 

= (4 + 13 − 1)/52 

= 16/52 

= 4/13

Question: Two dice are thrown. What is the probability that the sum is 8 given that the first die shows a 3?

Solution: Given first die is 3, second die can be 1 to 6

Favorable outcome: second die is 5 (since 3+5=8)

P = 1/6

Question: A factory has three machines A, B, and C producing 30%, 45%, and 25% of items respectively. The defect rates are 2%, 3%, and 4%. If an item is found defective, what is the probability it was produced by machine B?

Solution: P(A) = 0.3, P(B) = 0.45, P(C) = 0.25

P(D|A) = 0.02, P(D|B) = 0.03, P(D|C) = 0.04

P(D) = (0.3×0.02) + (0.45×0.03) + (0.25×0.04) = 0

Why Use Infinity Learn RD Sharma Solutions for Probability?

• Concept Clarity: Perfect for clearing doubts while solving exercises
• Exam Readiness: Aligned with the latest CBSE exam pattern
• Time-saving Resource: Helps revise complex theories quickly
• Prepared by Experts: Solutions crafted by experienced Maths faculty.
• Boosts Confidence: Ideal for board exam practice and competitive prep

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