$\displaystyle \mathrm{E}$ and $\displaystyle \mathrm{F}$ are two independent events such that $\displaystyle \mathrm{P}(\bar{\mathrm{E}})=0.6$ and $\displaystyle \mathrm{P}(\mathrm{E} \cup \mathrm{F})=0 \cdot 6$. Find $\displaystyle \mathrm{P}(\mathrm{F})$ and $\displaystyle \mathrm{P}(\overline{\mathrm{E}} \cup \overline{\mathrm{F}})$.

$\displaystyle \mathrm{P}(\overline{\boldsymbol{\mathrm{E}}})=0.6 \Rightarrow \boldsymbol{\mathrm{P}}(\boldsymbol{\mathrm{E}})=0.4$ $\displaystyle \mathrm{P}(\mathrm{E} \cup \mathrm{F})=\mathrm{P}(\mathrm{E})+\mathrm{P}(\mathrm{F})-\mathrm{P}(\mathrm{E} \cap \mathrm{F})$ $\displaystyle \Rightarrow 0.6=0.4+\mathrm{P}(\mathrm{F})-0.4 \mathrm{P}(\mathrm{F}) \Rightarrow \mathrm{P}(\mathrm{F})=\frac{1}{3}$ $\displaystyle \mathrm{P}(\overline{\boldsymbol{\mathrm{E}}} \cup \overline{\boldsymbol{\mathrm{F}}})=\mathbf{1}-\boldsymbol{\mathrm{P}}(\boldsymbol{\mathrm{E}} \cap \boldsymbol{\mathrm{F}})$=1-0.4 \times \frac{1}{3}=\frac{13}{15} $$