Match List-I with List-II
Let A and B are two events such that \(P(A) = 0.8, P(B) = 0.5, P(B|A) = 0.4\)
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(P(A \cap B)\) | (I) | 0.2 |
| (B) | \(P(A | B)\) | (II) | 0.32 |
| (C) | \(P(A \cup B)\) | (III) | 0.64 |
| (D) | \(P(A')\) | (IV) | 0.98 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A) \(P(A \cap B)\)
This matches (II) 0.32, by the multiplication rule:
\(P(A \cap B) = P(A)\,P(B|A)\)
\(= 0.8 \times 0.4 = 0.32\)
(B) \(P(A | B)\)
This matches (III) 0.64, by the definition of conditional probability:
\(\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)}\)
\(\displaystyle = \frac{0.32}{0.5} = 0.64\)
(C) \(P(A \cup B)\)
This matches (IV) 0.98, by the addition rule:
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(= 0.8 + 0.5 - 0.32 = 0.98\)
(D) \(P(A')\)
This matches (I) 0.2:
\(P(A') = 1 - P(A) = 1 - 0.8 = 0.2\)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).