Question

If A and B are two events such that \( P(A \cap B) = 0.1 \), and \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \), then the value of \( \frac{P(A \cup B)}{P(A \cap B)} \) is:

• A. \( \frac{4}{3} \)
• B. \( \frac{7}{4} \)
• C. \( \frac{5}{3} \)
• D. \( \frac{9}{4} \)

Hint:

For problems involving conditional probabilities: - Recall that \( P(A|B) = \frac{P(A \cap B)}{P(B)} \) and \( P(B|A) = \frac{P(A \cap B)}{P(A)} \). - Use these relationships to form equations based on the given conditions, and then solve for the required ratio of probabilities.

Answer 1

The Correct Option is A

We are given that \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \). By Vieta's formulas, we know the sum and product of the roots: \[ P(A|B) + P(B|A) = \frac{7}{12}, \quad P(A|B) \cdot P(B|A) = \frac{1}{12}. \] Using the relationships for conditional probabilities, we can compute \( \frac{P(A \cup B)}{P(A \cap B)} \). This yields the value \( \frac{4}{3} \).