Board 2026 Class 12 Maths Q19 Solution: Prime number given an odd number on a die

The question

Select the correct answer from the codes (A), (B), (C) and (D) as given below.

(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.

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, $P(A|B) = \frac{P(A \cup B)}{P(B)}$

Exam answer

Let E: prime number $= \{2, 3, 5\}$ and F: odd number $= \{1, 3, 5\}$ ⁠. Then $E \cap F = \{3, 5\}$ ⁠.

Why?

The sample space is $\{1, 2, 3, 4, 5, 6\}$ ⁠, with 6 equally likely outcomes.

The primes up to 6 are 2, 3 and 5 (1 is not prime). The odd numbers are 1, 3 and 5. The outcomes in both are 3 and 5.

\begin{aligned} P(E|F) &= \frac{P(E \cap F)}{P(F)} \\ &= \frac{2/6}{3/6} = \frac{2}{3} \end{aligned}

Why?

Conditional probability, for $P(F) \ne 0$ ⁠:

P(E|F) = \frac{P(E \cap F)}{P(F)}

$E \cap F$ has 2 of the 6 outcomes and $F$ has 3, so the probabilities are $\frac{2}{6}$ and $\frac{3}{6}$ ⁠.

To divide by a fraction, multiply by it upside down:

\begin{aligned} \frac{2/6}{3/6} &= \frac{2}{6} \times \frac{6}{3} \\ &= \frac{2 \times 6}{6 \times 3} \\ &= \frac{12}{18} = \frac{12 \div 6}{18 \div 6} = \frac{2}{3} \end{aligned}

Shortcut: once we know the number is odd, only 1, 3, 5 remain, and 2 of them are prime.

So Assertion (A) is true. Reason (R) is false, since $P(A|B) = \frac{P(A \cap B)}{P(B)}$ ⁠, not $\frac{P(A \cup B)}{P(B)}$ ⁠.

For example, $E \cup F = \{1, 2, 3, 5\}$ ⁠, and $\frac{4/6}{3/6} = \frac{4}{3} > 1$ ⁠, which no probability can be.

Why?

$E \cup F$ has 4 of the 6 outcomes, so $P(E \cup F) = \frac{4}{6}$ ⁠. Then

\begin{aligned} \frac{4/6}{3/6} &= \frac{4}{6} \times \frac{6}{3} = \frac{4}{3} \end{aligned}

Watch out: the formula uses $\cap$ (both happen), not $\cup$ (either happens). A probability is never more than 1, so the Reason's formula cannot be right.

A false Reason cannot explain the Assertion, so the code is (C).

Answer: (C) Assertion (A) is true and Reason (R) is false.

Where this comes from

Chapter 13, Probability: Conditional probability.

Question type: Dice, coins and cards

How to spot it: click to see

Type 2. Dice, coins and cards: shrink the sample space

Spot it: equally likely outcomes, and the words "given that" or "it is known that".

Here: "Given that the number is odd" makes $F = \{1, 3, 5\}$ the new sample space. Two of these, 3 and 5, are prime, so $P(E|F) = \frac23$ and the Assertion is true; the Reason has $\cup$ where $\cap$ belongs.

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Q19: Prime number given an odd number on a die