Pairwise Independence (For 2 events) Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $\mathcal{G}_1,\mathcal{G}_2\subset\mathcal{F}$. $\mathcal{G}_{1},\mathcal{G}_{2}$ are independent if $$\forall A_{1}\in\mathcal{G}_{1},\forall A_{2}\in\mathcal{G}_{2}: \ P(A_1\cap A_2)=P(A_1)P(A_2)$$Joint Independence (For n events) Let $A_1,...,A_n\in\mathcal{F}$. $A_1,...,A_n$ are independent if for any $2\le k\le n$ and $1\le i_1 \lt\ldots\lt i_k\le n$, $$P(A_{i_1}\cap\ldots\cap A_{i_k})=P(A_{i_1})\ldots P(A_{i_k})$$ Independence for countable events Let $A_{1},A_{2},\dots$ be a countable sequence of events. We say they’re independent if $$\mathbb{P}(A_{i_{1}}\cap\dots \cap A_{i_{n}})=\mathbb{P}(A_{i_{1}})\dots \mathbb{P}(A_{i_{n}}),\quad\forall \{ i_{1},\dots,i_{n} \}\subset \mathbb{N}$$

Pairwise Independence (For 2 rvs) For $X,Y$ rvs, $X\perp\!\!\!\perp Y$ means $$\mathbb{P}(X\in S_{1},Y\in S_{1})=\mathbb{P}(X\in S_{1})\mathbb{P}(Y\in S_{2}),\quad\forall S_{1},S_{2}\in \mathcal{B}(\mathbb{R})$$ Joint Independence (For $n$ rvs) Let $X_1,\cdots,X_n$ be RVs. They are independent if for any $A_1,\cdots,A_n\in \mathcal{B}(\mathbb{R})$ $$P(X_1\in A_1,\cdots,X_n\in A_n)=P(X_1\in A_1)\cdots P(X_n\in A_n)$$