How to Solve Probability Foundations — General probability — set operations / inclusion-exclusion Questions on Exam P

Let \(x = \Pr(A\cap B\cap C)\) be the triple-intersection probability. The event \(A\cap B\) consists of the exactly-two region \((A\cap B\cap C^{\mathrm c})\), with probability \(0.09\), and the triple intersection, with probability \(x\), so \(\Pr(A\cap B) = x + 0.09\). The given conditional probability is \(\Pr(A\cap B\cap C\mid A\cap B) = \frac{x}{x+0.09} = \frac{2}{5}\). Solving \(x = \frac{2}{5}(x+0.09)\) yields \(x(1-\frac{2}{5}) = \frac{2}{5}(0.09)\), so \(x = \frac{3}{50}\). The universe consists of seven mutually disjoint regions plus the none region. Summing all seven feature regions gives \(\Pr(A\cup B\cup C) = 3(0.07) + 3(0.09) + x = \frac{27}{50}\). Thus the none region has probability \(\Pr((A\cup B\cup C)^{\mathrm c}) = 1 - \frac{27}{50} = \frac{23}{50}\). Feature \(A\) contains its single region (\(0.07\)), two pair regions (\(2\times0.09\)), and the triple intersection (\(x\)), so \(\Pr(A) = 0.07 + 2(0.09) + x = \frac{31}{100}\). Its complement is \(\Pr(A^{\mathrm c}) = 1 - \frac{31}{100} = \frac{69}{100}\). The conditional probability of selecting none given \(A^{\mathrm c}\) is \(\frac{\Pr(\text{none})}{\Pr(A^{\mathrm c})} = \frac{\frac{23}{50}}{\frac{69}{100}} = \frac{2}{3}\). The evaluated result is \(\frac{2}{3}\), which matches the correct option.