Let A and B be two events defined on a sample space $\Omega$. Suppose $A^C$ denotes the complement of A relative to the sample space $\Omega$. Then the probability $P\Bigg{(}(A\cap{B}^C)\cup({A}^C\cap B)\Bigg{)}$ equals

Given: Two events \( A \) and \( B \) defined on sample space \( \Omega \). We are to find the probability:

$$ P\left((A \cap B^c) \cup (A^c \cap B)\right) $$

Step 1: This is the probability of events that are in exactly one of A or B (but not both), i.e., symmetric difference of A and B:

$$ (A \cap B^c) \cup (A^c \cap B) = A \Delta B $$

Step 2: So, we use:

$$ P(A \Delta B) = P(A) + P(B) - 2P(A \cap B) $$

Final Answer:

$$ \boxed{P(A) + P(B) - 2P(A \cap B)} $$