Let two fair six-faced dice A and B be thrown | vedclass

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Question

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

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

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

Vedclass · Accepted Answer

$E_1 , E_2$  and $E_3 $ are independent.

Vedclass · Answer

$E_{1} ightarrow A$ shows up 4

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

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

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

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

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

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

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

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

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

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

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

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

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 $?$

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