How to Solve Probability Foundations — General probability — two-stage experiments / total probability Questions on Exam P
Sample Practice Problem
ID: #AU91A
Let \(X_A\) and \(X_B\) be the monthly incident costs at sites \(A\) and \(B\). They have zero monthly incident cost with probabilities 60% and 72%. Given a positive cost, their independent totals are normal with means 100 and 92 and common standard deviation 20. Calculate the probability that \(B\)'s total exceeds \(A\)'s.
📖 Worked Solution & Strategy
The sample space breaks into four mutually exclusive cost states: (1) Both zero (probability \(0.60 \times \frac{18}{25}\)), where \(X_A = X_B = 0\); (2) Site \(A\) positive, site \(B\) zero (probability \(0.40 \times \frac{18}{25}\)), where \(X_A > X_B\); (3) Site \(A\) zero, site \(B\) positive (probability \(0.60 \times \frac{7}{25}\)), where \(X_B > X_A\); (4) Both positive (probability \(0.40 \times \frac{7}{25}\)), where \(X_B - X_A \sim \operatorname{Normal}(92-100, 20^2+20^2) = \operatorname{Normal}(-8, 800)\). In State 4, \(\Pr(X_B > X_A \mid \text{both positive}) = \Pr(X_B - X_A > 0) = 1 - \Phi\left(\frac{0 - (-8)}{\sqrt{800}}\right) = 1 - \Phi(8/\sqrt{800}) = 0.3886\). Site \(B\)'s total exceeds \(A\)'s in State 3 and State 4: \(\Pr(X_B > X_A) = 0.60(\frac{7}{25}) + 0.40(\frac{7}{25})(0.3886) = 0.2115\). The evaluated result is \(0.2115\), which matches the correct option.