If E and Fare independent events such that `0 < P(E) <1` and `0 < P(F) <1`, then

Options

A. E and `F^(c )` are independent
B. `E^(c )` and `F^(c )` are independent
C. `P((E )/(F))+P ((E^(c ))/(F^(c )))=1`
D. All of the above

Answer

D

Explanation

`because P(E nn F)=P(E ).P(F)`
Now, `P(E nn F^(c ))=P(E )-P(E nn F)`
`=P(E )[1-P(F)]`
`=P(E ).P(F^(c ))`
and `P(E^(c )nn F^(c ))=1-P(E uu F)`
`=1 -[P(E )+P(F)-P(E nn 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 ))`
`rArr P(E//F)+P(E^(c )//F^(c ))=1`