1. If two die are rolled, the sample space is given by $\Omega = \{ 1,2,3,4,5,6 \}^{2}$. Let $\begin{align} A_{1} &= \{ \text{sum of the die is 7} \} \\ A_{2} &= \{ \text{the first dice is even} \} \\ A_{3} &= \{ \text{the second dice is odd} \} \end{align}$These events are pairwise independent since $\begin{gathered} \mathbb{P}(A_1\cap A_2) =\frac{1}{12}=\mathbb{P}(A_1)\mathbb{P}(A_2) \\\mathbb{P}(A_1\cap A_3) =\frac1{12}=\mathbb{P}(A_1)\mathbb{P}(A_3) \\\mathbb{P}(A_2\cap A_3) =\frac1{4}=\mathbb{P}(A_2)\mathbb{P}(A_3). \end{gathered}$Now $\begin{align}\mathbb{P}(A_{1} \cap (A_{2} \cap A_{3})) &= \mathbb{P}(A_{1} \cap A_{2} \cap A_{3} ) = \frac{1}{12} \neq \frac{1}{24} = \mathbb{P}(A_{1}) \cdot \mathbb{P}(A_{2} \cap A_{3} ) \end{align}$which shows that $A_{1}$ is not independent of $A_{2} \cap A_{3}$. Now if 3 die are rolled, the 2. (a) Let $a_{i}, \, i =1,2,3,\dots,60$ and $b_{i}, i =1,2,3,\dots, 20$ denote the 60 untagged fish and 20 tagged fish respectively. Define $F=\{ a_{i}, b_{j}, \, i=1,2,3,\dots, 60, \, j = 1,2,3,\dots, 20 \}$ The sample space is given by $\begin{align}\Omega &= \{ (f_{1},f_{2},f_{3},\dots,f_{10}): f_{i} \in F \text{ and } f_{i} \neq f_{j} \text{ for all } i,j =1,2,3\dots 10 \text{ s.t. } i \neq j\} \\ &= O_{80,10}(F)\end{align}$ So $|\Omega| = {80 \choose 10}$ Define $T=\{b_{i}, \, i= 1,\dots,20\}$ $\begin{align}\{ X=x \} &= \{ \omega \in \Omega: X(\omega) = x \} \\ &= \{(f_{1}, f_{2}, f_{3}, \dots f_{x}): f_{i} \in T \} \end{align}$ So $p_{X}(x) = \mathbb{P}(\{X=x\})= \frac{|\{X=x\}|}{|\Omega|} = \frac{{20 \choose x}}{80 \choose 10 }$