Question

Let \( E \) and \( F \) be two events such that \( P(E) = 0.1 \), \( P(F) = 0.3 \), \( P(E \cup F) = 0.4 \). Then \( P(F|E) \) is:

• A. \( 0.6 \)
• B. \( 0.4 \)
• C. \( 0.5 \)
• D. \( 0 \)

Hint:

When \( P(E \cap F) = 0 \), the events \( E \) and \( F \) are mutually exclusive, meaning they cannot occur simultaneously.

Answer 1

The Correct Option is D

Step 1: Recall the formula for conditional probability
The conditional probability \( P(F|E) \) is given by: \[ P(F|E) = \frac{P(E \cap F)}{P(E)}. \] Step 2: Determine \( P(E \cap F) \)
Utilize the formula for the probability of the union of two events: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F). \] Substitute the provided values: \( P(E \cup F) = 0.4 \), \( P(E) = 0.1 \), \( P(F) = 0.3 \): \[ 0.4 = 0.1 + 0.3 - P(E \cap F). \] Solve for \( P(E \cap F) \): \[ P(E \cap F) = 0.1 + 0.3 - 0.4 = 0. \] Step 3: Compute \( P(F|E) \)
Insert \( P(E \cap F) = 0 \) and \( P(E) = 0.1 \) into the \( P(F|E) \) formula: \[ P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{0}{0.1} = 0. \] Step 4: State the conclusion
The conditional probability \( P(F|E) \) is \( 0 \). This implies that events \( E \) and \( F \) are mutually exclusive.