Match List-I with List-II
Let A and B be two events such that \(P(A) = 0.2, P(B) = 0.4, P(B|A) = 0.5\)
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(P(A \cap B)\) | (I) | 0.5 |
| (B) | \(P(A|B)\) | (II) | 0.8 |
| (C) | \(P(A \cup B)\) | (III) | 0.25 |
| (D) | \(P(A')\) | (IV) | 0.1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \(P(A \cap B)\)
This matches (IV) 0.1. By the multiplication rule:
\(P(A \cap B) = P(A)\,P(B|A) = 0.2 \times 0.5 = 0.1\)
(B) \(P(A|B)\)
This matches (III) 0.25.
\( \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.1}{0.4} = 0.25 \)
(C) \(P(A \cup B)\)
This matches (I) 0.5. By the addition rule:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(= 0.2 + 0.4 - 0.1 = 0.5\)
(D) \(P(A')\)
This matches (II) 0.8.
\(P(A') = 1 - P(A) = 1 - 0.2 = 0.8\)
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).