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 \(E\) and \(F\), always use the formula \(P(E \cap F) = P(E) \cdot P(F)\), and for unions and complements, apply the addition rule and complement rule carefully.

Answer 1

Step 1: Determine \(P(E)\) from \(P(\overline{E})\). The relationship is \( P(E) + P(\overline{E}) = 1 \). Given \(P(\overline{E}) = 0.6\), we calculate \( P(E) = 1 - 0.6 = 0.4 \).

Step 2: Calculate \(P(F)\) using the union formula for independent events. The formula for the union of two events is \( P(E \cup F) = P(E) + P(F) - P(E \cap F) \). For independent events, \( P(E \cap F) = P(E) \cdot P(F) \). Thus, the formula becomes \( P(E \cup F) = P(E) + P(F) - P(E) \cdot P(F) \). Substituting \(P(E) = 0.4\) and \(P(E \cup F) = 0.6\): \( 0.6 = 0.4 + P(F) - (0.4 \cdot P(F)) \). Simplifying the equation: \( 0.6 - 0.4 = P(F)(1 - 0.4) \), which leads to \( 0.2 = 0.6P(F) \). Solving for \(P(F)\) yields \( P(F) = \frac{0.2}{0.6} = \frac{1}{3} \). 

Step 3: Compute \(P(\overline{E} \cup \overline{F})\). Using the complement rule, \( P(\overline{E} \cup \overline{F}) = 1 - P(E \cap F) \). Since \(E\) and \(F\) are independent, \( P(E \cap F) = P(E) \cdot P(F) \). Substituting the values \(P(E) = 0.4\) and \(P(F) = \frac{1}{3}\): \( P(E \cap F) = 0.4 \cdot \frac{1}{3} = \frac{2}{15} \). Therefore, \( P(\overline{E} \cup \overline{F}) = 1 - \frac{2}{15} = \frac{13}{15} \). 

Final Answer: \( P(F) = \frac{1}{3}, \quad P(\overline{E} \cup \overline{F}) = \frac{13}{15} \).