A) E and \[{{F}^{c}}\](the complement of the event F) are independent
B) \[{{E}^{c}}\]and \[{{F}^{c}}\]are independent
C) \[P\,\left( \frac{E}{F} \right)+P\,\left( \frac{{{E}^{c}}}{{{F}^{c}}} \right)=1\]
D) All of the above
Correct Answer: D
Solution :
\[P(E\cap F)=P(E)\,.\,P(F)\] Now,\[P(E\cap {{F}^{c}})=P(E)-P(E\cap F)=P(E)[1-P(F)]=P(E)\,.P({{F}^{c}})\] and \[P({{E}^{c}}\cap {{F}^{c}})=1-P(E\cup F)=1-[P(E)+P(F)-P(E\cap F)\] \[=[1-P(E)][1-P(F)]=P({{E}^{c}})\,P({{F}^{c}})\] Also \[P(E/F)=P(E)\] and \[P({{E}^{c}}/{{F}^{c}})=P({{E}^{c}})\] \[\Rightarrow P(E/F)+P({{E}^{c}}/{{F}^{c}})=1.\]You need to login to perform this action.
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