$E_{1} \rightarrow A$ shows up 4

$\mathrm{E}_{2} \rightarrow \mathrm{B}$ shows up 2

$E_{3} \rightarrow \operatorname{Sum}$ is odd (i.e. even $+$ odd or odd $+$ even)

$\mathrm{P}\left(\mathrm{E}_{1}\right)= \frac{6}{6.6}=\frac{1}{6}$

$\mathrm{P}\left(\mathrm{E}_{2}\right)= \frac{6}{6.6}=\frac{1}{6}$

$\mathrm{P}\left(\mathrm{E}_{3}\right)= \frac{3 \times 3 \times 2}{6.6}=\frac{1}{2} $

${\rm{P}}\left( {{{\rm{E}}_1} \cap {{\rm{E}}_2}} \right) = \frac{1}{{6.6}} = {\rm{P}}\left( {{{\rm{E}}_1}} \right) \cdot {\rm{P}}\left( {{{\rm{E}}_2}} \right)$

$ \Rightarrow \mathrm{E}_{1} \, and \,  \mathrm{E}_{2} \text { are independent } $

${\rm{P}}\left( {{{\rm{E}}_1} \cap {{\rm{E}}_3}} \right) = \frac{{1.3}}{{6.6}} = {\rm{P}}\left( {{{\rm{E}}_1}} \right) \cdot {\rm{P}}\left( {{{\rm{E}}_3}} \right)$

$ \Rightarrow {{\rm{E}}_{1\,}}{\rm{and}}\,{{\rm{E}}_3}{\rm{ are independent  }}$

${\rm{P}}\left( {{{\rm{E}}_2} \cap {{\rm{E}}_3}} \right) = \frac{{1.3}}{{6.6}} = \frac{1}{{12}} = {\rm{P}}\left( {{{\rm{E}}_2}} \right) \cdot {\rm{P}}\left( {{{\rm{E}}_3}} \right)$

$ \Rightarrow \mathrm{E}_{2}\,and \, \mathrm{E}_{3} \text { are independent }$

$\mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2} \cap \mathrm{E}_{3}\right)=0$ ie imposible event.