Question 191 markSection A

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 23\frac{2}{3}. Reason (R): For any two events A and B, P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}.

Probability
Conditional Probability

Options

(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.

Official answer

The correct option is (A) — both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).


Verification of Assertion: For an unbiased die, Odd={1,3,5}\text{Odd} = \{1,3,5\} so P(Odd)=36=12P(\text{Odd}) = \frac{3}{6} = \frac{1}{2}, and Prime∩Odd={3,5}\text{Prime} \cap \text{Odd} = \{3,5\} so P(Prime∩Odd)=26=13P(\text{Prime} \cap \text{Odd}) = \frac{2}{6} = \frac{1}{3}. Then P(Prime∣Odd)=1/31/2=23P(\text{Prime}|\text{Odd}) = \frac{1/3}{1/2} = \frac{2}{3}, confirming the Assertion.


Verification of Reason: P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A\cap B)}{P(B)} is the standard definition of conditional probability, and it is precisely this formula that was used above to compute 2/3 — so R correctly explains A.

Answer keywords:conditional probabilityP(A|B)sample spaceprime numbersodd numbersunbiased dieassertion reason

Marking scheme

  1. 1.1 mark: correct option (A) — full marks require correctly identifying both statements as true AND that R explains A; no partial credit for assertion-reason MCQs.

Hint

List the sample space, identify the 'odd' and 'prime' events, then directly apply P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A\cap B)}{P(B)}.

Quick oral answer

Odd numbers on a die are {1,3,5}\{1,3,5\} and primes among them are {3,5}\{3,5\}, so P(Prime∣Odd)=P(Prime∩Odd)P(Odd)=1/31/2=23P(\text{Prime}|\text{Odd}) = \frac{P(\text{Prime}\cap\text{Odd})}{P(\text{Odd})} = \frac{1/3}{1/2} = \frac{2}{3}, which is exactly what the Reason's formula gives — so both statements are true and R explains A.

Common mistakes

  1. 1.Listing 2 as an odd prime number by mistake (2 is prime but even, so it is excluded from Prime∩Odd\text{Prime} \cap \text{Odd}).
  2. 2.Computing P(Odd∣Prime)P(\text{Odd}|\text{Prime}) instead of P(Prime∣Odd)P(\text{Prime}|\text{Odd}), which reverses the conditioning event.
  3. 3.Assuming the Reason is only a 'reference formula' rather than recognising it is the exact method used to derive the Assertion's value.

Analysis and explanation

This question tests the definition and direct application of conditional probability.


Why (A) is correct: The sample space for a fair die is {1,2,3,4,5,6}\{1,2,3,4,5,6\}. The event 'odd' = {1,3,5}\{1,3,5\} and the event 'prime' = {2,3,5}\{2,3,5\}. Their intersection is {3,5}\{3,5\}. Applying the Reason's formula: P(Prime∣Odd)=P(Prime∩Odd)P(Odd)=2/63/6=23P(\text{Prime}|\text{Odd}) = \frac{P(\text{Prime}\cap\text{Odd})}{P(\text{Odd})} = \frac{2/6}{3/6} = \frac{2}{3}, which matches the Assertion exactly. Since the Assertion's numeric claim is derived directly from the Reason's formula, R is the correct explanation of A.


Why the distractors are wrong:

  • (B) would apply if the Assertion were true but derived independently of the Reason's formula — not the case here, since the formula is exactly what produces the 2/3 result.
  • (C) would require the Reason to be false, but P(A|B) = P(A∩B)/P(B) is the universally accepted definition of conditional probability, so R is true.
  • (D) would require the Assertion to be false, but direct calculation confirms P(Prime∣Odd)=23P(\text{Prime}|\text{Odd}) = \frac{2}{3} is correct.

Interesting facts

Among the six faces of a standard die, exactly three numbers (2,3,5)(2, 3, 5) are prime — a fact often re-used across CBSE probability questions.

The formula P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A\cap B)}{P(B)}, formalized in the work of Thomas Bayes and later Pierre-Simon Laplace in the 18th century, underlies all of modern Bayesian statistics and machine learning classifiers.

Question details

Q19 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section A
Marks
1
Chapter
Probability
Topic
Conditional Probability
Bloom's level
analyze
Syllabus topic
Unit 5: Probability — Conditional Probability, P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A\cap B)}{P(B)}
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is 2 excluded from the intersection of 'prime' and 'odd'?

Because 2 is an even number, even though it is prime. The intersection of 'prime' {2,3,5}\{2,3,5\} and 'odd' {1,3,5}\{1,3,5\} is only {3,5}\{3,5\}, since 2 fails the 'odd' condition.

How do you quickly verify an assertion-reason question in probability?

Independently calculate the exact numeric value asked in the Assertion using the formula given in the Reason. If they match and the formula was the actual method used, the Reason correctly explains the Assertion.