If A and B are two events such that $P(A) = \frac{1}{2}, P(B) =\frac{1}{3}$ and $P(A∩B) =\frac{1}{4}$ then which of the following statements are true?

(A) A and B are independent events
(B) $P(A | B) = \frac{3}{4}$
(C) $P(A' | B') = \frac{5}{8}$
(D) $P(A' | B) = \frac{1}{4}$

Choose the correct answer from the options given below:

(B), (C) and (D) only

The correct answer is Option (2) → (B), (C) and (D) only

(A) A and B are independent events  (False)
(B) $P(A | B) = \frac{3}{4}$ (True)
(C) $P(A' | B') = \frac{5}{8}$ (True)
(D) $P(A' | B) = \frac{1}{4}$ (True)

Given:

$P(A) = \frac{1}{2},\quad P(B) = \frac{1}{3},\quad P(A \cap B) = \frac{1}{4}$

(A): A and B are independent if:

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

$\frac{1}{4} \overset{?}{=} \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$

False. So, A and B are not independent.

(B): $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{4}}{\frac{1}{3}} = \frac{3}{4}$

True.

(C): $P(A'|B') = \frac{P(A' \cap B')}{P(B')}$

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

$P(A' \cap B') = 1 - P(A \cup B)$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{6 + 4 - 3}{12} = \frac{7}{12}$

$P(A' \cap B') = 1 - \frac{7}{12} = \frac{5}{12}$

$P(A'|B') = \frac{\frac{5}{12}}{\frac{2}{3}} = \frac{5}{12} \cdot \frac{3}{2} = \frac{15}{24} = \frac{5}{8}$

True.

(D): $P(A'|B) = \frac{P(A' \cap B)}{P(B)}$

$P(A' \cap B) = P(B) - P(A \cap B) = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}$

$P(A'|B) = \frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{12} \cdot \frac{3}{1} = \frac{3}{12} = \frac{1}{4}$

True.