Question

If \(E\) and \(F\) are two independent events such that \[ P(E)=\frac{3}{10}, \qquad P(E\cup F)=\frac{1}{2}, \] then \[ P(E\mid F)-P(F\mid E) \] is equal to \[ \_\_\_\_. \]

• A. \( \frac{2}{7} \)
• B. \( \frac{3}{35} \)
• C. \( \frac{1}{70} \)
• D. \( \frac{1}{7} \)

Hint:

For independent events, conditional probabilities collapse immediately: \( P(A|B) = P(A) \). This avoids wasting time setting up complex fractions for the final subtraction step.

Answer 1

The Correct Option is C

Concept: This problem involves independent events and conditional probability. Key properties used:
• For independent events, \( P(E \cap F) = P(E) \cdot P(F) \)
• Conditional probabilities simplify to: \( P(E|F) = P(E) \) and \( P(F|E) = P(F) \)
• Probability of union: \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \)

Step 1: Determine the value of \( P(F) \).
Using the formula for the union of two events: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \] Since \( E \) and \( F \) are independent, substitute \( P(E \cap F) = P(E) \cdot P(F) \): \[ P(E \cup F) = P(E) + P(F) - P(E) \cdot P(F) \] Substitute the given values \( P(E) = \frac{3}{10} \) and \( P(E \cup F) = \frac{1}{2} \): \[ \frac{1}{2} = \frac{3}{10} + P(F) - \frac{3}{10}P(F) \] \[ \frac{1}{2} - \frac{3}{10} = P(F) \left(1 - \frac{3}{10}\right) \] \[ \frac{5 - 3}{10} = P(F) \cdot \frac{7}{10} \quad \Rightarrow \quad \frac{2}{10} = \frac{7}{10} P(F) \] \[ P(F) = \frac{2}{7} \]

Step 2: Evaluate the conditional probability expression.
Because \( E \) and \( F \) are independent events, the occurrence of one does not affect the probability of the other: \[ P(E|F) = P(E) = \frac{3}{10} \] \[ P(F|E) = P(F) = \frac{2}{7} \] Now compute the required difference: \[ P(E|F) - P(F|E) = \frac{3}{10} - \frac{2}{7} = \frac{3 \times 7 - 2 \times 10}{70} = \frac{21 - 20}{70} = \frac{1}{70} \] This value directly matches option (C).