How to Solve Continuous Random Variables — Univariate RV — moments of a density on (-1, 1) Questions on Exam P
Sample Practice Problem
ID: #Y0NF2
Two service warranties each charge a one-time fee of 180 and pay 5000 if the covered technician leaves within five years. The probabilities that only technician 1 remains, only technician 2 remains, and both remain are 0.03, 0.02, and 0.92. Given that technician 1 remains, calculate the expected excess of the two collected fees over warranty payouts.
📖 Worked Solution & Strategy
Let \(R_1, R_2\) be the events that technician 1 and technician 2 remain. We are given \(\Pr(R_1 \cap R_2^{\mathrm c}) = 0.03\), \(\Pr(R_1^{\mathrm c} \cap R_2) = 0.02\), and \(\Pr(R_1 \cap R_2) = 0.92\). Conditioning on technician 1 remaining (\(R_1\)), the conditioning event has total probability \(\Pr(R_1) = 0.92 + 0.03 = 0.95\). Under this condition, technician 2 leaves with probability \(\Pr(R_2^{\mathrm c} \mid R_1) = \frac{\Pr(R_1 \cap R_2^{\mathrm c})}{\Pr(R_1)} = \frac{0.03}{0.95} = \frac{3}{95}\). A warranty payout of \(5000\) occurs if technician 2 leaves, so expected payout is \(5000\left(\frac{3}{95}\right) = \frac{5000}{47}\). Collecting two fees of \(180\) yields gross revenue \(2(180) = 360\). Net credit is \(2(180) - 5000\left(\frac{3}{95}\right) = \frac{11920}{47}\). The evaluated result is \(\frac{11920}{47}\), which matches the correct option.