$\mathrm{E}_{2} \rightarrow \mathrm{B}$ 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.
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Statement $p$ : The value of $sin\,120^o$ can be divided by taking $\theta\, = 240^o$ in the equation $2\,\sin \frac{\theta }{2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta } $
Statement $q$ : The angles $A, B, C$ and $D$ of any quadrilateral $ABCD$ satisfy the equation $\cos \left( {\frac{1}{2}\left( {A + C} \right)} \right) + \cos \left( {\frac{1}{2}\left( {B + D} \right)} \right) = 0$
Then the truth values of $p$ and $q$ are respectively.