How to Solve Probability Foundations — General probability — complements, De Morgan, conditional Questions on Exam P

We are given \(\Pr(A\cup B) = \frac{11}{20}\) and \(\Pr(A\cup B^{\mathrm c}) = 0.88\). Taking complements and applying De Morgan's laws gives \(\Pr((A\cup B)^{\mathrm c}) = \Pr(A^{\mathrm c} \cap B^{\mathrm c}) = 1 - \frac{11}{20}\) and \(\Pr((A\cup B^{\mathrm c})^{\mathrm c}) = \Pr(A^{\mathrm c} \cap B) = 1 - 0.88 = 0.12\). The events \((A^{\mathrm c} \cap B^{\mathrm c})\) and \((A^{\mathrm c} \cap B)\) are disjoint and partition \(A^{\mathrm c}\). Thus \(\Pr(A^{\mathrm c}) = \Pr(A^{\mathrm c} \cap B^{\mathrm c}) + \Pr(A^{\mathrm c} \cap B) = (1 - \frac{11}{20}) + (1 - 0.88) = 2 - \frac{11}{20} - 0.88\). Taking the complement gives \(\Pr(A) = 1 - \Pr(A^{\mathrm c}) = \frac{11}{20} + 0.88 - 1 = \frac{43}{100}\). Therefore \(\Pr(A) = \frac{43}{100}\). The evaluated result is \(\frac{43}{100}\), which matches the correct option.