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: Relationship between \(P(E)\) and \(P(\overline{E})\) We know that: \[ P(E) + P(\overline{E}) = 1. \] 

Substituting \(P(\overline{E}) = 0.6\): \[ P(E) = 1 - 0.6 = 0.4. \] 

Step 2: Use the formula for \(P(E \cup F)\) For any two events \(E\) and \(F\), we have: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F). \] Since \(E\) and \(F\) are independent, \(P(E \cap F) = P(E) \cdot P(F)\). 

Substituting this: \[ P(E \cup F) = P(E) + P(F) - P(E) \cdot P(F). \] Substitute \(P(E) = 0.4\) and \(P(E \cup F) = 0.6\): \[ 0.6 = 0.4 + P(F) - (0.4 \cdot P(F)). \] Simplify: \[ 0.6 = 0.4 + P(F) - 0.4P(F). \] \[ 0.6 - 0.4 = P(F)(1 - 0.4). \] \[ 0.2 = 0.6P(F). \] \[ P(F) = \frac{0.2}{0.6} = \frac{1}{3}. \] 

Step 3: Find \(P(\overline{E} \cup \overline{F})\) Using the complement rule: \[ P(\overline{E} \cup \overline{F}) = 1 - P(E \cap F). \] From the formula for complements: \[ P(E \cap F) = P(E) \cdot P(F). \] 

Substitute \(P(E) = 0.4\) and \(P(F) = \frac{1}{3}\): \[ P(E \cap F) = 0.4 \cdot \frac{1}{3} = \frac{2}{15}. \] 

Thus: \[ P(\overline{E} \cup \overline{F}) = 1 - P(E \cap F) = 1 - \frac{2}{15} = \frac{15}{15} - \frac{2}{15} = \frac{13}{15}. \] 

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