Two Problem Header

Here are two simple probability problems for the sake of adjusting the, padding, text size, etc.

By additivity of a probability measure we know that \(P(A\cup A^c)=P(A)+P(A^c)\) since \(A \cap A^c= \emptyset\). Since \(P(A\cup A^c) = P(\Omega) = 1\), then \(1 = P(A) + P(A^c)\), hence \(P(A^c) = 1 - P(A)\).

First note that \[A = A \cap \Omega =A\cap (B\cup B^c)=(A\cap B) \cup (A\cap B^c).\] Note that \((A\cap B^c) \cap (A\cap B) = \emptyset\), so by additivity of probability measure \(P\) we have that \[P(A)=P((A\cap B) \cup (A \cap B^c))=P(A \cap B) + P( A\cap B^c).\] Thus, \[P(A \cap B^c) = P(A)-P(A \cap B).\]