The addition rule (inclusion–exclusion for two events)

Step 1 (partition the union into disjoint pieces): split $A \cup B$ into three pairwise disjoint pieces: $A \setminus B$ (outcomes only in $A$), $A \cap B$ (outcomes in both), and $A \cap B$$B \setminus A$ (outcomes only in $B$). Every outcome of $A \cup B$ falls into exactly one of these three pieces.

Step 2 (count the union by the addition rule for disjoint sets): since the three pieces are pairwise disjoint, the number of outcomes satisfies $|A \cup B| = |A \setminus B| + |A \cap B| + |B \setminus A|$.

Step 3 (express $|A|$ and $|B|$ using the same three pieces): similarly, $|A| = |A \setminus B| + |A \cap B|$ (splitting $A$ by whether it overlaps $B$) and $|B| = |B \setminus A| + |A \cap B|$ (splitting $B$ the same way). Adding these two equations gives $|A| + |B| = |A \setminus B| + |B \setminus A| + 2|A \cap B|$.

Step 4 (combine and divide by $|\Omega|$): comparing Step 2 and Step 3, $|A| + |B| - |A \cap B| = |A \setminus B| + |B \setminus A| + |A \cap B| = |A \cup B|$. Dividing both sides by $|\Omega|$ and applying the classical probability formula $P(A) = \frac{|A|}{|\Omega|}$ to each term gives exactly $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.