GATE CS 1994 Q1.4 — Conditional Probability and Bayes Theorem

For any two arbitrary events $A$ and $B$, we can analyze the given options using the fundamental laws of probability:

• Option A: $P(A \cap B) = P(A)P(B)$ holds true only if the events $A$ and $B$ are independent. It is not true for arbitrary events.
• Option B: $P(A \cup B) = P(A) + P(B)$ holds true only if the events $A$ and $B$ are mutually exclusive (disjoint). It is not true for arbitrary events.
• Option C: By the definition of conditional probability, $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ (assuming $P(B) > 0$), which is not equal to $P(A \cap B)P(B)$ in general.
• Option D: By the addition rule of probability: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Since probability is non-negative, we have $P(A \cap B) \geq 0$. This directly implies: $$P(A \cup B) \leq P(A) + P(B)$$ This inequality (known as Boole's Inequality or the Union Bound) always holds true for any arbitrary events $A$ and $B$.