> [!NOTE] Lemma > Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a [[Probability Space|probability space]]. Let $A,B\in \mathcal{F}.$ The events $A$ and $B$ are [[Independence of Two Events|independent]] iff $\Omega \setminus A$ and $B$ are independent, where $\Omega \setminus A$ denotes a [[Set Difference|set difference]]. **Proof**: Note that $\Omega\setminus A \cap B = B \setminus A$since $\begin{align} \Omega\setminus A \cap B &= \{ x\in \Omega : x\not \in A \} \cap B \\ &= \{ x\in \Omega : x \not \in A \land x\in B \} \\ &= \{ x\in \Omega \land x\in B : x \not \in A \} \\ &= \{ x\in B : x \not \in A \} \tag{*} \\ & = B \setminus A \end{align}$using $B \subset A$ to gain $(*).$ Thus by [[Probability of Set Difference of Events]], $\begin{align} \mathbb{P}(\Omega \setminus A \cap B) &= \mathbb{P}(B\setminus A) \\ &= \mathbb{P}(B) - \mathbb{P}(A \cap B) \\ &= \mathbb{P}(B) - \mathbb{P}(A) \cdot \mathbb{P}(B) \\ &=\mathbb{P}(B)[1-\mathbb{P}(A)] \\ &= \mathbb{P}(B) \mathbb{P}(\Omega\setminus A) \end{align}$using [[Probability of Complement of Event]] to give the last line. $\square$ # Applications **Consequences**: By [[Independent Events iff Independent Complements]].