Question

For two events A and B
Match List I with List II

List I	List II
A. \(P (\bar{\bigwedge} \cap \bar{B}) = P(\bar{\bigwedge}) . P(\bar{B})\)	I. \(P(A/B) \geq P(A)\)
B. \(A \subset B \) and \(P(B) \neq0\)	II. \(P(A) = P(B)\)
C. \(P(A \bigcup B) = P(A) + P(B)\)	III. A and B are independent events
D. \(P(A|B) = P (B|A)\)	IV. A and B are mutually exclusive events

Choose the correct answer from the options given below:

• A. A-III, В-II, C-IV, D-I
• B. А-III, В-I, C-IV, D-II
• C. A-IV, B-II, C-III, D-I
• D. A-IV, B-I, C-III, D-II

Answer 1

The Correct Option is B

Let's match List I with List II using the properties of probability:

List I	List II
A. \(P (\bar{\bigwedge} \cap \bar{B}) = P(\bar{\bigwedge}) . P(\bar{B})\)	III. A and B are independent events
B. \(A \subset B\) and \(P(B) \neq 0\)	I. \(P(A/B) \geq P(A)\)
C. \(P(A \bigcup B) = P(A) + P(B)\)	IV. A and B are mutually exclusive events
D. \(P(A|B) = P(B|A)\)	II. \(P(A) = P(B)\)

Explanation:

• Option A: For two events A and B, if \(P (\bar{\bigwedge} \cap \bar{B}) = P(\bar{\bigwedge}) . P(\bar{B})\), this describes the condition for two independent events.
• Option B: If \(A \subset B\) and \(P(B) \neq 0\), then \(P(A/B) \geq P(A)\) because all parts of A are within B.
• Option C: If \(P(A \bigcup B) = P(A) + P(B)\), this indicates A and B are mutually exclusive events, as they do not overlap.
• Option D: \(P(A|B) = P(B|A)\) means both events have the same probability, indicating \(P(A) = P(B)\).

Therefore, the matching is А-III, В-I, C-IV, D-II.