By Cameron Bell 1. Given (Ω, F, P) be a probability space and A1, A2, A3 ∈ F mutually independent events. WTS $A_1$ will also be independent of $A_2 ∩ A_3$ and $A_2 ∪ A_3. $\mathbb{p} (A_1 \cap (A_2 \cap A_3))=\mathbb{p}(A_1 \cap A_2 \cap A_3)=^{by\, mutual\,independence}\mathbb{P}(A_1)\mathbb{P}(A_2)\mathbb{P}(A_3)=\mathbb{P}(A_1)\mathbb{P}(A_2 \cap A_3)$which shows that $A_1, A_2 \cap A_3$ are independent. $\mathbb{P}(A_1 \cap (A_2 \cup A_3)) = \mathbb{P}((A_1 \cap A_2)\cup(A_1 \cap A_3)) $ 2. 1. $X: \Omega \to \mathbb{R}$ Let $E=\{W_1, \ldots, W_8, B_1, \ldots, B_4, G_1, G_2\}$ $\Omega=\{\{w_1,w_2\}:w_1,w_2 \in E, w_1 \ne w_2\}$ $F = P(\Omega)$ 3. Let $X=$ \#Empty Baskets $\quad \quad= \sum_{i=1}^{n}(\mathbb{I}(basket: empty)$ Note: $\mathbb{I} = \begin{cases} 1 & \text{if A happens} \\ 0 & \text{otherwise} \end{cases}$ So $E(X)=\sum_{i=1}^n E(\mathbb{I}(basket: empty))$ by linearity of expectation $\quad \quad=\sum_{i=1}^n \mathbb{P}(basket:empty)$