Question Analysis: Assertion (A) and Reason (R) based on Conditional Probability.

Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]

The sample space for throwing an unbiased die is $S = \{1, 2, 3, 4, 5, 6\}$.

Let $A$ be the event of getting a prime number: $A = \{2, 3, 5\}$.

Let $B$ be the event of getting an odd number: $B = \{1, 3, 5\}$.

The formula for conditional probability is:

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

• Intersection of events: $A \cap B = \{3, 5\}$, so $n(A \cap B) = 2$.
• Probability of intersection: $P(A \cap B) = \frac{2}{6}$.
• Probability of event $B$: $P(B) = \frac{3}{6}$.
• Calculating $P(A|B)$:

Thus, Assertion (A) is True.

Now, evaluating Reason (R): The formula provided is $P(A|B) = \frac{P(A \cup B)}{P(B)}$. This is mathematically incorrect as the standard definition is $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Thus, Reason (R) is False.

Step 3 (Final Concluded Answer): [0.25 Mark]

Conclusion: Assertion (A) is True, but Reason (R) is False.

💡 EL10 Concept Insight: Conditional probability is like shrinking your world; since we already know the number is odd, we ignore the even numbers and only look at the odd ones (1, 3, 5) to see how many of those are prime (3 and 5), giving us 2 out of 3.