A) \[{{E}_{1}}and{{E}_{2}}\] are independent
B) \[{{E}_{1}}\] and \[{{E}_{2}}\] are exhaustive
C) \[{{E}_{2}}\] is twice as likely to occur as \[{{E}_{1}}\]
D) Probabilities of the events\[{{E}_{1}}\cap {{E}_{2}}\], \[{{E}_{1}}\] and \[{{E}_{2}}\] are in GP.
Correct Answer: B
Solution :
[b] \[P({{E}_{2}}/{{E}_{1}})=\frac{P({{E}_{1}}\cap {{E}_{2}})}{P({{E}_{1}})}\] \[\Rightarrow \frac{1}{2}=\frac{P({{E}_{1}}\cap {{E}_{2}})}{1/4}\Rightarrow P({{E}_{1}}\cap {{E}_{2}})=\frac{1}{8}\] \[=P({{E}_{2}}).P({{E}_{1}}/{{E}_{2}})=P({{E}_{2}}).\frac{1}{4}\] \[\Rightarrow P({{E}_{2}})=\frac{1}{2}\sin ceP({{E}_{1}}\cap {{E}_{2}})\] \[=\frac{1}{8}=P({{E}_{1}}).P({{E}_{2}})\] \[\Rightarrow \] Events are independent Also \[P({{E}_{1}}\cup {{E}_{2}})=\frac{1}{2}+\frac{1}{4}-\frac{1}{8}=\frac{5}{8}\] \[\Rightarrow {{E}_{1}}\And {{E}_{2}}\] are non-exhaustive
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