Q.If $P(A) = 0.35$ for an event $A$, find $P(\text{not } A)$.

For any event $A$, the complementary event $A'$ ("not $A$") satisfies $A \cup A' = S$ and $A \cap A' = \phi$, so $A$ and $A'$ are mutually exclusive and together exhaust $S$. Applying Axiom 3 to their union gives $P(A) + P(A') = P(S) = 1$ by Axiom 2, so

This rule is derived from the axioms, not assumed separately. It is especially useful for events described with "at least one" or "at most", which are often far easier to compute as $1$ minus the probability of the complementary, simpler-to-describe event — e.g. "no heads at all" instead of directly counting every "at least one head" case.