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 \(\frac{P(A \cup B)}{P(A \cap B)}\)
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 \(\frac{P(A \cup B)}{P(A \cap B)}\)
Show Hint
\( \frac{9}{4} \)
- \( \frac{7}{4} \)
- \( \frac{5}{3} \)
\( \frac{4}{3} \)
The Correct Option is A
Solution and Explanation
Given the equation: \[ 12x^2 - 7x + 1 = 0, \quad x = \frac{1}{3}, \frac{1}{4} \] Let \[ P\left( A \mid B \right) = \frac{1}{3} \quad \text{and} \quad P\left( B \mid A \right) = \frac{1}{4} \] From the given, we have: \[ P(A \cap B) = \frac{1}{3} \quad \text{and} \quad P(B) = \frac{1}{4} \] This implies: \[ P(B) = 0.3 \quad \text{and} \quad P(A) = 0.4 \] The formula for the union of two events is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Substitute the values: \[ P(A \cup B) = 0.3 + 0.4 - 0.1 = 0.6 \] Now, we calculate \( P(A \cup B) \): \[ P(A \cup B) = \frac{P(A \cap B)}{P(A \cup B)} \] Substitute the known values: \[ P(A \cup B) = \frac{1 - P(A \cap B)}{P(A \cup B)} = \frac{1 - 0.1}{1 - 0.6} = \frac{9}{4} \]
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