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:

For mutually exclusive events \( E \) and \( F \), \( P(E \cap F) = 0 \), leading to a zero conditional probability.

Answer 1

The Correct Option is D

The probability of the union of events E and F is calculated using the formula: \[P(E \cup F) = P(E) + P(F) - P(E \cap F).\] Given the values, we substitute them into the equation: \[0.4 = 0.1 + 0.3 - P(E \cap F) \quad \Rightarrow \quad P(E \cap F) = 0.\] The conditional probability of F given E is then determined by: \[P(F \,|\, E) = \frac{P(E \cap F)}{P(E)} = \frac{0}{0.1} = 0.\]
Final Answer: \( \boxed{0} \)