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Given $P(\overline{E}) = 0.6$, we can find $P(E)$ using the complement rule: $P(E) = 1 - P(\overline{E}) = 1 - 0.6 = 0.4$
We are given $P(E \cup F) = 0.8$. We also know that for any two events $E$ and $F$: $P(E \cup F) = P(E) + P(F) - P(E \cap F)$
Since $E$ and $F$ are independent events, $P(E \cap F) = P(E)P(F)$. Substituting this into the previous equation: $P(E \cup F) = P(E) + P(F) - P(E)P(F)$
Plugging in the known values: $0.8 = 0.4 + P(F) - 0.4P(F)$ $0.8 - 0.4 = P(F)(1 - 0.4)$ $0.4 = 0.6P(F)$ $P(F) = \frac{0.4}{0.6} = \frac{2}{3}$
Using De Morgan's Law: $P(\overline{E} \cup \overline{F}) = P(\overline{E \cap F}) = 1 - P(E \cap F)$ Since $E$ and $F$ are independent, $P(E \cap F) = P(E)P(F) = 0.4 \times \frac{2}{3} = \frac{0.8}{3} = \frac{4}{15}$ Therefore, $P(\overline{E} \cup \overline{F}) = 1 - \frac{4}{15} = \frac{15 - 4}{15} = \frac{11}{15}$
Final Answer: $P(F) = \frac{2}{3}$ and $P(\overline{E}\cup\overline{F}) = \frac{11}{15}$