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) $P(A \mid B)$	(III) 0.12
(D) $P(B \mid A)$	(IV) 0.58

Choose the correct answer from the options given below:

Given: $P(A) = 0.3$, $P(B) = 0.4$, and $A$ and $B$ are independent events.

For independent events, $P(A \cap B) = P(A) \times P(B)$.

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For (A) $P(A \cap B)$:

$P(A \cap B) = P(A) \times P(B)$

$P(A \cap B) = 0.3 \times 0.4$

$P(A \cap B) = 0.12$

(A) matches with (III). We know enough—can mark the option right away and save time!

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For (B) $P(A \cup B)$:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$P(A \cup B) = 0.3 + 0.4 - 0.12$

$P(A \cup B) = 0.58$

(B) matches with (IV)

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For (C) $P(A|B)$:

Since $A$ and $B$ are independent, $P(A|B) = P(A)$.

$P(A|B) = 0.3$

(C) matches with (I)

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For (D) $P(B|A)$:

Since $A$ and $B$ are independent, $P(B|A) = P(B)$.

$P(B|A) = 0.4$

(D) matches with (II)

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Therefore, the correct answer is: (A) - (III), (B) - (IV), (C) - (I), (D) - (II)