Question

E and F are two independent events such that \( P(\overline{E}) = 0.6 \) and \( P(E \cup F) = 0.6 \). Find \( P(F) \) and \( P(\overline{E} \cup \overline{F}) \).

Hint:

For independent events, use \( P(E \cap F) = P(E) \cdot P(F) \).

Answer 1

Step 1: Find \( P(E) \) \[ P(E) = 1 - P(\overline{E}) = 1 - 0.6 = 0.4. \] Step 2: Use the formula for \( P(E \cup F) \) \[ P(E \cup F) = P(E) + P(F) - P(E \cap F). \] For independent events: \[ P(E \cap F) = P(E) \cdot P(F). \] Substituting the given values: \[ 0.6 = 0.4 + P(F) - (0.4 \cdot P(F)). \] \[ 0.6 - 0.4 = P(F)(1 - 0.4). \] \[ 0.2 = 0.6P(F). \] \[ P(F) = \frac{0.2}{0.6} = \frac{1}{3}. \] Step 3: Find \( P(\overline{E} \cup \overline{F}) \) \[ P(\overline{E} \cup \overline{F}) = P(\overline{E}) + P(\overline{F}) - P(\overline{E} \cap \overline{F}). \] For independent events: \[ P(\overline{E} \cap \overline{F}) = P(\overline{E}) \cdot P(\overline{F}). \] Substituting \( P(\overline{F}) = 1 - P(F) = 1 - \frac{1}{3} = \frac{2}{3} \): \[ P(\overline{E} \cup \overline{F}) = 0.6 + \frac{2}{3} - (0.6 \cdot \frac{2}{3}). \] \[ = 0.6 + 0.6667 - 0.4 = 0.8667. \] Final Answer: \[ \boxed{P(F) = \frac{1}{3}, \quad P(\overline{E} \cup \overline{F}) \approx 0.867.} \]