Probability of Union of Events in a Probability Product Space By Counting Event Size This question is about probability of union of events in a probability product space. Let’s say a fair die is thrown twice and we’re interested to find out probability of getting face value one in 1st throw or two in  2nd throw. There are many ways to solve this as few of them are listed below.
Option 1) 
We could think of both the throws as one experiment with sample space $S$. In such case, each sample point will be a double valued point e.g. $S=${{1,1}, {1,2}…{6,6}}. We could define an event $C$ that either 1st throw has face value one or 2nd throw has face value 2, implying $C=$ {{1,1}, {1,2}…,{1,6}, {2,2}, {3,2}…{6,2}}
Finally applying $P(C)$ =  size of {$C$} / size of sample space, we could get $P(C) =$ 11/36
Option 2) 
Alternately, we could decompose the previous single experiment into two experiments say $E1$ the 1st throw and $E2$ the 2nd throw. We could define two events $A$ and $B$ that a face value of one and of two respectively. In such case, the event $A$ is scoped within $E1$ with its own sample space $S1=${1,2,..6}. Similarly $B$ is scoped within $E2$ and has a different sample space $S2=${1,2,..6}. We could apply sum-rule of probability augmented by independence formula to derive the probability as follows.
$P(A\cup B)=P(A)+P(B)-P(A \cap B) = P(A)+P(B)-P(A).P(B) = 1/6+1/6 – 1/6.1/6= 11/36$
Option 3) 
Here I would like to use the decomposed view of two experiments $E1$ and $E2$ like option-2, but would like to apply the size of event set like option-1 e.g. 
$P(A \cup B)= $size of {$A\cup B$}/size of product space {$S1 \times S2$}
The question is how I find the size of {$A \cup B$} in terms of size of individual events given that each event is restricted within its sample space or experiment. More precisely, can I apply the set theoretic formula of {$A \cup B$}={$A$}$+${$B$}$-${$A \cap B$} here? It seems I need to use rather something like {$A \cup B$}$=${$A \times S2$}$+${$B \times S1$}$-${$A \cap B$}. Wondering is this something that is mathematically addressed somewhere!
Further, most of the books proves addition rule of probability using Venn diagram using two events, say $A$ and $B$, of same experiment with same sample space, say $S$. How can we prove that the same rule is applicable even when events $A$ and $B$ belong to different experiments with own sample space.
Will appreciate any help on this! I have not taken up Measure Theory or $\sigma$-algebra yet. So exposition of relevant topic without reference to Measure Theory would be highly appreciated.
 A: The correct approach is always to consider the common sample space. If the events given are independent then the common sample space can be constructed. If they are not independent then either the dependence structure is given on a common sample space or the common sample space is not defined.
In the case of independent dice rolling the sample space is always
$$\Omega=\{(i,j): 1\le i\le 6, 1\le i\le 6\}$$
with
$$P(\{(i,j)\})=\frac1{36}.$$
Option 1) follows this route by counting the elements of event $C$.
Option 2) is also following this approach if we interpret $A$ and $B$ as
$$A=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\}$$
and
$$B=\{(1,2),(2,2),(3,2),(4,2),(5,2),(6,2)\}.$$
With this 
$$A\cup B=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(3,2),(4,2),(5,2),(6,2)\}$$
and 
$$A\cap B=\{(1,2)\}.$$
The result will be the same because in $P(A)+P(B)$ the probability $P(\{(1,2)\})$ is enumerated twice so $$P(A\cap B)=P(\{(1,2)\})$$
has to be subtracted.
Option 3) This approach is wrong at the beginning because the $A$ and $B$ belong to different universes. Then the argumentation sneaks back to the right track.
