Let A and B be independent events such that P(A) = 0.3 and P(B) = 0.4, then
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(P(A \cap B)\) | (I) | 0.3 |
| (B) | \(P(A \cup B)\) | (II) | 0.4 |
| (C) | \(P(A|B)\) | (III) | 0.12 |
| (D) | \(P(B|A)\) | (IV) | 0.58 |
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)-(I), (D)-(II)
For independent events, \(P(A \cap B) = P(A)P(B)\), and conditioning on one event does not change the probability of the other.
(A) \(P(A \cap B)\)
This matches (III) 0.12.
\(P(A \cap B) = 0.3 \times 0.4 = 0.12\)
(B) \(P(A \cup B)\)
This matches (IV) 0.58.
\(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
\(= 0.3 + 0.4 - 0.12 = 0.58\)
(C) \(P(A|B)\)
This matches (I) 0.3.
\( \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = 0.3 \)
(D) \(P(B|A)\)
This matches (II) 0.4.
\( \displaystyle P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.3} = 0.4 \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
NCERT (2026-27 edition):
Independent Events
