$E_1, E_2$ are events of a sample space such that $P\left(E_1\right)=\frac{1}{4},P\left(\frac{E_2}{E_1}\right)=\frac{1}{2},P\left(\frac{E_1}{E_2}\right)=\frac{1}{4}$ then $P\left(\frac{E_1}{E_2}\right)+P\left(\frac{E_1}{\overline{E_2}}\right)=$

$\begin{array}{rcl}\mathrm{P}\left({\mathrm{E}}_1\right)& =& \frac{1}{4}, \mathrm{P}\left(\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1}\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \frac{\mathrm{P}\left({\mathrm{E}}_1\cap {\mathrm{E}}_2\right)}{\mathrm{P}\left({\mathrm{E}}_1\right)}& =& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ & \Rightarrow & \mathrm{P}\left({\mathrm{E}}_1\cap {\mathrm{E}}_2\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\\ \mathrm{P}\left(\raisebox{1ex}{${\mathrm{E}}_1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{E}}_2$}\right.\right)& =& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \Rightarrow \frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}{\mathrm{P}\left({\mathrm{E}}_2\right)}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \Rightarrow \mathrm{P}\left({\mathrm{E}}_2\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ & & \mathrm{P}\left(\raisebox{1ex}{${\mathrm{E}}_1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{E}}_2$}\right.\right)+\mathrm{P}\left(\raisebox{1ex}{$\overline{{\mathrm{E}}_1}$}\!\left/ \!\raisebox{-1ex}{$\overline{{\mathrm{E}}_2}$}\right.\right)\\ & =& \frac{1}{4}+\frac{\mathrm{P}\left({\mathrm{E}}_1\cap \overline{{\mathrm{E}}_2}\right)}{\mathrm{P}\left(\overline{{\mathrm{E}}_2}\right)}=\frac{1}{4}+\frac{\mathrm{P}\left({\mathrm{E}}_1\right)-\mathrm{P}({\mathrm{E}}_1\cap {\mathrm{E}}_2)}{1-\mathrm{P}\left({\mathrm{E}}_2\right)}\\ & =& \frac{1}{4}+\frac{{\displaystyle \frac{1}{4}}-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{1-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{1}{4}+\frac{1}{4}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\end{array}$