Two Events Being Independent Proves that Their Complements Are Also Independent
Theorem
The following are equivalent. $$ P(A \cap B) = P(A)P(B) \\ P(A \cap B^c)=P(A)P(B^c) \\ P(A^c \cap B)=P(A^c)P(B) \\ P(A^c \cap B^c)=P(A^c)P(B^c) $$
Explanation
Not only is this fact highly beneficial to know, but it is also useful as a formula.
Proof
Let’s assume $P(A \cap B) = P(A)P(B)$. In other words, events $A$ and $B$ are independent. According to the property of the complement, $$ P(A)=1-P(A^{ c }) \\ P(B)=1-P(B^{ c }) $$ therefore, the right-hand side of $P(A \cap B) = P(A)P(B)$ is $$ \begin{align*} P(A)P(B)&=(1-P(A^{ c }))(1-P(B^{ c })) \\ =& 1-P(A^{ c })-P(B^{ c })+P(A^{ c })P(B^{ c }) \end{align*} $$ and the left-hand side, by De Morgan’s theorem, is $$ \begin{align*} P(A\cap B) =& 1-P((A\cap B)^{ c }) \\ =& 1-P(A^{ c }\cup B^{ c }) \end{align*} $$ Substituting both sides into $P(A \cap B) = P(A)P(B)$ and simplifying gives us $$ P(A^{ c }\cup B^{ c })=P(A^{ c })+P(B^{ c })-P(A^{ c })P(B^{ c }) $$ Since events $A$ and $B$ are independent, by the addition theorem of probability, $$ P(A^{ c }\cup B^{ c })=P(A^{ c })+P(B^{ c })-P(A^{ c }\cap B^{ c }) $$ Therefore, $$ P(A^{ c })P(B^{ c })=P(A^{ c }\cap B^{ c }) $$ This means, if $A$ and $B$ are independent, then $A^c$ and $B^c$ are also independent. On the other hand, $$ \begin{align*} P(A)P(B^{ c })&=P(A){1-P(B)} \\ &=P(A)-P(A)P(B) \\ &=P(A)-P(A)-P(B)+P(A\cup B) \\ &=P(A\cup B)-P(B) \\ &=P(A\cap B^{ c }) \end{align*} $$ and this is the same for $A^c$ and $B$.
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