If A and B are independent events and $P(A) = \frac{1}{2}$ $P(B) = \frac{1}{3}$, then

Match List-I with List-II

List-I	List-II
(A) $P(A \cap B)$	(I) $\frac{1}{2}$
(B) $P(\bar{A})P(B) + P(A)P(\bar{B})$	(II) $\frac{1}{3}$
(C) $P(A \mid B) + P(B \mid A)$	(III) $\frac{1}{6}$
(D) $P(A \cap \bar{B})$	(IV) $\frac{5}{6}$

Choose the correct answer from the options given below:

Given that A and B are independent events with $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$.

For independent events:

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

$P(A|B) = P(A)$

$P(B|A) = P(B)$

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Finding $P(A \cap B)$:

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

$= \frac{1}{2} \times \frac{1}{3}$

$= \frac{1}{6}$

(A) matches with (III)

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Finding $P(\bar{A})P(B) + P(A)P(\bar{B})$:

$P(\bar{A}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2}$

$P(\bar{B}) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$

$P(\bar{A})P(B) + P(A)P(\bar{B}) = \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) + \left(\frac{1}{2}\right)\left(\frac{2}{3}\right)$

$= \frac{1}{6} + \frac{2}{6}$

$= \frac{3}{6}$

$= \frac{1}{2}$

(B) matches with (I)

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Finding $P(A \mid B) + P(B \mid A)$:

Since A and B are independent:

$P(A|B) = P(A) = \frac{1}{2}$

$P(B|A) = P(B) = \frac{1}{3}$

$P(A|B) + P(B|A) = \frac{1}{2} + \frac{1}{3}$

$= \frac{3}{6} + \frac{2}{6}$

$= \frac{5}{6}$

(C) matches with (IV)

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Finding $P(A \cap \bar{B})$:

Since A and B are independent, A and $\bar{B}$ are also independent.

$P(A \cap \bar{B}) = P(A) \times P(\bar{B})$

$= \frac{1}{2} \times \frac{2}{3}$

$= \frac{2}{6}$

$= \frac{1}{3}$

(D) matches with (II)

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Therefore, the correct matching is:

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)