Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is 2/3. Reason (R) : For any two events A and B, P(A B) = (P(A B))/(P(B)) (A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true and Reason (R) is false. (D) Assertion (A) is false and Reason (R) is true.

MathsCBSE Class 12 · Board exam 2026, set 65/1/1 · 1 mark · Assertion-reason

CBSE Class 12 Mathematics 2026: question 19

Set in the CBSE Class 12 board exam of 2026, paper 65/1/1, section A, question 19, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q19 · 1 mark · Assertion-reasonCompetency-basedProbability

Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is $\frac{2}{3}$ .
Reason (R) : For any two events A and B, $P (A ∣ B) = \frac{P ( A \cup B )}{P ( B )}$

(A)Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true and Reason (R) is false.
(D)Assertion (A) is false and Reason (R) is true.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 65/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

List odd outcomes and which of them are prime.

Hint 2

Check whether conditional probability uses union or intersection.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Odd: ${1, 3, 5}$ ; prime and odd: ${3, 5}$ . $P = \frac{2}{3}$ , so A is true.
Correct formula is $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ , not with $\cup$ , so R is false. Option (C).
1 mark

Answer: Option (C)

Where the marks are

Computes $\frac{2}{3}$ and identifies wrong formula in R (1 mark)

Revise the chapter

Chapter: Probability (notes, formulas, where marks are lost and every past question)
Topic: Conditional Probability

Practise this chapterBoard-style questions from Probability, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 65/1/1: every question, by section.

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CBSE Class 12 Mathematics 2026: question 19

Hints

Step-by-step solution

Revise the chapter

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