A) \[{{E}_{1}},{{E}_{2}}\]and \[{{E}_{3}}\]are independent.
B) \[{{E}_{1}}\]and \[{{E}_{2}}\]are independent.
C) \[{{E}_{2}}\]and \[{{E}_{3}}\]are independent.
D) \[{{E}_{1}}\]and\[{{E}_{3}}\] are independent.
Correct Answer: A
Solution :
\[{{E}_{1}}\to A\] shows up 4 \[{{E}_{2}}\to B\]shows up 2 \[{{E}_{3}}\to \]Sum is odd (i.e. even + odd or odd + even) \[P({{E}_{1}})=\frac{6}{6.6}=\frac{1}{6}\] \[P({{E}_{2}})=\frac{6}{6.6}=\frac{1}{6}\] \[P({{E}_{3}})=\frac{3\times 3\times 2}{6.6}=\frac{1}{2}\] \[P({{E}_{1}}\cap {{E}_{2}})=\frac{1}{6.6}=P({{E}_{1}}).P({{E}_{2}})\] \[\Rightarrow {{E}_{1}}\And {{E}_{2}}\]are independent \[P({{E}_{1}}\cap {{E}_{3}})=\frac{1.3}{6.6}=P({{E}_{1}}).P({{E}_{3}})\] \[\Rightarrow {{E}_{1}}\And {{E}_{3}}\]are independent \[P({{E}_{2}}\cap {{E}_{3}})=\frac{1.3}{6.6}=\frac{1}{12}=P({{E}_{2}}).P({{E}_{3}})\] \[\Rightarrow {{E}_{2}}\And {{E}_{3}}\]are independent \[P({{E}_{1}}\cap {{E}_{2}}\cap {{E}_{3}})=0\] ie impossible event.
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