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 ∩ B) |
(I) 0.3 |
|
(B) P(A ∪ 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)
Given:
$P(A) = 0.3$
$P(B) = 0.4$
$A$ and $B$ are independent events.
Calculations:
(A) $P(A \cap B) = P(A) \cdot P(B) = 0.3 \cdot 0.4 = 0.12$
(B) $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.3 + 0.4 - 0.12 = 0.58$
(C) $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.4} = 0.3$
(D) $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.3} = 0.4$
Matching:
| List-I | List-II |
|---|---|
| (A) $P(A \cap B)$ | (III) 0.12 |
| (B) $P(A \cup B)$ | (IV) 0.58 |
| (C) $P(A|B)$ | (I) 0.3 |
| (D) $P(B|A)$ | (II) 0.4 |