Q31: Let A and B are any two events and P(A)=... - Statistical Methods & Experimental Designs - ICAR AIEEA PG Statistical Sciences 09 July 2023 (Shift 1) Paper | Agrigyan

Derivation using probability axioms and De Morgan's laws:

• Incorrect: The denominator must be $P(\bar{B}) = 1 - p_2$, not $1 - p_3$.
• Incorrect: $p_1 / p_2$ applies only in trivial independence or ratio cases.
• Correct: By definition of conditional probability:
  $$P(\bar{A} / \bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}$$
  Applying De Morgan's law, $\bar{A} \cap \bar{B} = \overline{A \cup B}$, so:
  $$P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(A \cap B)] = 1 - p_1 - p_2 + p_3$$
  Since $P(\bar{B}) = 1 - P(B) = 1 - p_2$, we obtain:
  $$P(\bar{A} / \bar{B}) = \frac{1 - p_1 - p_2 + p_3}{1 - p_2}$$
• Incorrect: The denominator contains $1 - p_1$, which corresponds to conditioning on $\bar{A}$.