Independence Complement Theorem
If sets E and F are independent, then so are E and F^', where F^' is the complement of F (i.e., the set of all possible outcomes not contained in F). Let union denote "or" and intersection denote "and." Then
P(E) = P(EF union EF^')
(1)
= P(EF)+P(EF^')-P(EF intersection EF^'),
(2)
where AB is an abbreviation for A intersection B. But E and F are independent, so
| P(EF)=P(E)P(F). |
(3)
|
Also, since F and F^' are complements, they contain no common elements, which means that
| P(EF intersection EF^')=0 |
(4)
|
for any E. Plugging (4) and (3) into (2) then gives
| P(E)=P(E)P(F)+P(EF^'). |
(5)
|
Rearranging,
| P(EF^')=P(E)[1-P(F)]=P(E)P(F^'), |
(6)
|
Q.E.D.
See also
Independent SetExplore with Wolfram|Alpha
WolframAlpha
More things to try:
Cite this as:
Weisstein, Eric W. "Independence Complement Theorem." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/IndependenceComplementTheorem.html