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Let two fair six-faced dice $A$ and $B$ be thrown simultaneously. If $E_1$ is the event that die $A$ shows up four, $E_2 $ is the event that die $B$ shows up two and $E_3$ is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true $?$
$E_1 $ and $E_3 $ are independent.
$E_1 , E_2$ and $E_3 $ are independent.
$E_1$and $E_2$ are independent.
$E_2 $ and $E_3 $ are independent.
Solution
$E_{1} \rightarrow A$ 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.
