Match List-I with List-II
| List-I | List-II | ||
| (A) | \( P(A\cap B)=P(A).P(B) \) | (I) | \( \displaystyle\frac{P(B\cap A)}{P(A)} \), \( P(A)\ne 0 \) |
| (B) | \( \displaystyle\frac{P(S\cap B)}{P(B)} \), S = Sample space | (II) | \( A \), \( B \) are independent events |
| (C) | P(B/A) = | (III) | \( P(\bar{A}\cap\bar{B}) \) |
| (D) | \( P(\overline{A\cup B}) \) | (IV) | 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
(A) \( P(A\cap B)=P(A).P(B) \)
This matches (II) A, B are independent events. This product rule is the definition of independent events.
(B) \( \frac{P(S\cap B)}{P(B)} \), S = Sample space
This matches (IV) 1. Every event is a subset of the sample space, so \( S\cap B=B \):
\( \displaystyle \frac{P(S\cap B)}{P(B)}=\frac{P(B)}{P(B)} \)
\( \displaystyle =1 \)
(C) P(B/A) =
This matches (I) \( \frac{P(B\cap A)}{P(A)} \), \( P(A)\ne 0 \). This is the definition of conditional probability.
(D) \( P(\overline{A\cup B}) \)
This matches (III) \( P(\bar{A}\cap\bar{B}) \). By De Morgan’s law, the complement of a union is the intersection of the complements:
\( \displaystyle \overline{A\cup B}=\bar{A}\cap\bar{B} \)
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).