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The sample space for an unbiased die is $S = \{1, 2, 3, 4, 5, 6\}$. Let event $A$ be getting a prime number, so $A = \{2, 3, 5\}$. Let event $B$ be getting an odd number, so $B = \{1, 3, 5\}$.
The probability of event $B$ is $P(B) = \frac{3}{6} = \frac{1}{2}$. The intersection $A \cap B$ (prime and odd) is $\{3, 5\}$, so $P(A \cap B) = \frac{2}{6} = \frac{1}{3}$.
The conditional probability is $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/3}{1/2} = \frac{2}{3}$. Thus, the Assertion is True.
The provided formula in the context is $P(A|B) = \frac{P(A \cup B)}{P(B)}$. This is mathematically incorrect. The correct definition of conditional probability is $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Thus, the Reason is False.
Final Answer: Assertion is True, but Reason is False.