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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 $\cup$ denote ``or'' and $\cap$ denote ``and.'' Then

$\displaystyle P(E)$ $\textstyle =$ $\displaystyle P(EF\cup EF')$ (1)
  $\textstyle =$ $\displaystyle P(EF)+P(EF')-P(EF\cap EF'),$ (2)

where $AB$ is an abbreviation for $A\cap B$. But $E$ and $F$ are independent, so
\end{displaymath} (3)

Also, since $F$ and $F'$ are complements, they contain no common elements, which means that
P(EF\cap EF')=0
\end{displaymath} (4)

for any $E$. Plugging (4) and (3) into (2) then gives
P(E)= P(E)P(F)+P(EF').
\end{displaymath} (5)

P(EF') = P(E)[1-P(F)] = P(E)P(F'),
\end{displaymath} (6)


See also Independent Statistics

© 1996-9 Eric W. Weisstein