Correct Answer - B::C::D
Since, E and F are independent events. Therefore, `P(E nn F) = P(E) * P(F) ne 0`, so E and F are not mutually exclusive events.
Now, `P(E nn bar(F)) = P(E) - P(E nn F) = P(E) - P(E) * P(F)`
`=P(E) [1-P(F)] = P(E) * P(bar(F))`
and `P(bar(E) nn bar(F)) = P(bar(E nn F)) = 1-P(E uu F)`
` = 1-[1-P(bar(E) * P(bar(F))] [because E " and F are independent"]`
` =P(bar(E)) * P(bar(F))`
So, E and `bar(F)` as well as `bar(E)` and `bar(F)` are independent events.
Now, `P(E//F) + P(bar(E) //F) = (P(E nn F) + P(bar(E) nn F))/(P(F))`
`= (P(F))/(P(FF)) = 1`