Match List-I with List-II
Let A and B be any two events
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(P(A')\) | (I) | \(\displaystyle\frac{P(A \cap B)}{P(A)}; P(A) \ne 0\) |
| (B) | \(P(\Phi)\) | (II) | \(\displaystyle\frac{P(A \cap B)}{P(B)}; P(B) \ne 0\) |
| (C) | \(P(A|B)\) | (III) | \(1 - P(A)\) |
| (D) | \(P(B|A)\) | (IV) | 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(A) \(P(A')\)
This matches (III) \(1 - P(A)\). A and its complement A’ are mutually exclusive and together make up the sample space S:
\(P(A) + P(A') = P(S) = 1\)
\(P(A') = 1 - P(A)\)
(B) \(P(\Phi)\)
This matches (IV) 0. The empty set Φ is the impossible event, and its probability is 0.
(C) \(P(A|B)\)
This matches (II). By the definition of conditional probability of A given B:
\(\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)},\quad P(B) \ne 0\)
(D) \(P(B|A)\)
This matches (I). In the same way, for B given A:
\(\displaystyle P(B|A) = \frac{P(A \cap B)}{P(A)},\quad P(A) \ne 0\)
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).