Calculating probability on sets I was reading about calculating the support and confidence in regard to "associate rule mining" and found the following definitions:
An association rule is defined as: $A \rightarrow B$ where $A\subset T$, $B\subset T$, and $A \cap B = \emptyset$.
Support: $c(A \rightarrow B) = P(A \cup B) $. In the other words, Support should be the ratio of the transactions that contains both $\{A\}$ and $\{B\}$ divided by total number of the transactions in the database. 
For example, consider the following transactions stored in the Database:
\begin{array}{|c|c|}
\hline
{\bf ID} & {\bf Transaction} \\ \hline
1 & \{Beer, Dipper, Milk\} \\ \hline
2 & \{Beer,  Milk\} \\ \hline
3 & \{Beer,  Potato Chips\} \\ \hline
4 & \{Dipper,  Cheese, Butter \} \\ \hline
\end{array}
So based on the above definitions and description I want to calculate the support for $c(\{Beer\} \rightarrow \{Milk\})$.
Therefore, I have to compute the $P(\{Beer\} \cup \{Milk\})$ (the probablity that a given transaction contains Beer or Milk). What is confusing to me is, given that $\{Beer\}$ and $\{Milk\}$ are sets, should I compute the union by constructing the $\{Beer, Milk\}$ set and then compute the probability of $P(\{Beer, Milk\})$ ? 
Case 1) If we don't give precedence to union operation before computing the probability:
$P(\{Beer\} \cup \{Milk\}) = P(\{Beer\} ) + P(\{Milk\}) - P(\{Beer\} \cap  \{Milk\}) $
$P(\{Beer\} \cup \{Milk\}) = \frac{3}{4} + \frac{2}{4} - \frac{2}{4} = \frac{3}{4} = 0.75$
Case 2) But if we assume that sets are not events, and we have to compute the union of two sets and then compute the probability:
$P(\{Beer\} \cup \{Milk\}) = P(\{Beer, Milk\}) = \frac{2}{4} = 0.5$
My Question) To me, case-1 is mathematically correct with the information provided, but case-2 is the right answer. Which one is mathematically correct in terms of writing? Is valid to say $P(\{Beer\} \cup \{Milk\})$ = P({Beer, Milk}) since they are sets and not variables? 

The first answer is the PIE formula for beer or milk, but the question is asking for beer and milk.
The second answer is correct - if you think about it, picking a random transaction, 2 of the 4 transactions have beer and milk, and so the probability is $\frac24$.
As for your notation, $P(\{\text{Beer, Milk}\}=P(\{Beer\}\cap\{Milk\})$, and not $P(\{Beer\}\cup\{Milk\})$ (the difference is the direction the cap is pointing).
