Why does |A1|=18 and |A2|=6 in this rolling a die random variables question? Rolling a die.
Let $X_i$ be the score on the $i$-th roll of the die, then $X_1, X_2, \dots$ form an independent sequence of random variables.
When rolling the die twice, we have sample space $\Omega = \{1,2,3,4,5,6\}^2$, and random variables $X_1$ and $X_2$: $\Omega \to \mathbb R$.
$X_1(w)=i$, $X_2(w) = j$, where $w(i,j)$ belongs to the sample space $\Omega$.
Let $A_1$ be the event that the score on the first roll is even, then $A_1 = \{w \in \Omega : X_1(w) \in \{2,4,6\}\}$. So $|A_1|=18$ and $P(A_1)=1/2$. Similarly $A_2$ is the event that the second roll results in the number 3, then $A_2 = \{w \in \Omega : X_2(w) = 3\}$. So $|A_2|=6$ and $P(A_2)=1/6$.
I understand how to get the probabilities but where are the $|A_1|=18$ and $|A_2|=6$ coming from? What does $|\textrm{An event}|=$ something mean exactly, and do you work it out the same way everytime?
 A: $A_1$, while an event, is also a set (the subset of the sample space containing each instance where the first roll is even). $|A_1|$ refers to the cardinality, or size, of this set. By saying that $|A_1| = 18$, you are saying, "Of the 36 total possibilities for die rolls, 18 of them fall into this set."
A: $ (1,1);(1,2);(1,3);(1,4);(1,5);(1,5) \text{ all the first entries on this row here are odd  }          \\ (2,1);(2,2);(2,3);(2,4);(2,5);(2,6) \text{ all the first entries on this row are even }    \\ (3,1);(3,2);(3,3);(3,4);(3,5);(3,6) \text{ all the first entries on this row here are odd } \\ (4,1);(4,2);(4,3);(4,4);(4,5);(4,6) \text{ all the first entries on this row are even  } \\   (5,1);(5,2);(5,3);(5;4);(5,5);(5,6) \text{ all the first entries on this row here are odd } \\ (6,1);(6,2);(6,3);(6,4);(6,5);(6,6) \text{ all the first entries on this row are even} \\ \text{  So of the 6(6)=36 possibilities there are } 3(6)=18 \text{  that have the first entry as even.} \\ \text{ Now looking at these pairs, I see 6 of them that have the second entry as 3. }$
