It is given that $$P(A \cup B) < \frac{3}{4}, P(A) > \frac{1}{8}, P\left(\frac{A}{B}\right) < \frac{1}{2}$$. Which of the following is true?

It is given that, $$P(A \cup B) < \frac{3}{4}, P(A) > \frac{1}{8}, P\left(\frac{A}{B}\right) < \frac{1}{2}$$

$$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P\left(B\right)}< \frac{1}{2}$$

$$2\ P(A \cap B) < P(B)$$

 $$P(A \cap B) < P(B)-P(A \cap B)$$.........(1)

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) < \frac{3}{4}$$

$$ P(A) + P(B) - P(A \cap B) < \frac{3}{4}$$

$$ P(A) + (P(A \cap B) <) < \frac{3}{4}$$      From Eq (1),

$$ P(A) + P(A \cap B) < \frac{3}{4}$$

Hence Options (A), (B) & (D) are false.

$$ P(A) + P(A \cap B) < \frac{3}{4}$$

$$ (> \frac{1}{8}) + P(A \cap B) < \frac{3}{4}$$       From  $$(P(A) > \frac{1}{8})$$

$$P(A \cap B) < \frac{3}{4}-\frac{1}{8}$$

$$P(A \cap B) < \frac{5}{8}$$

Option (C) is correct.