How to Solve Probability Foundations — General probability — Bayes' theorem / law of total probability Questions on Exam P
Sample Practice Problem
ID: #86NQN
Let \(T\in\{A,B,C\}\) denote a server's reliability tier, with probabilities \(0.45,0.35,0.20\), and let \(H\) denote a one-month alert. The conditional alert probabilities for tiers \(A,B,C\) are \(0.002\), \(0.006\), and \(0.002\), respectively. Calculate \(\Pr(T=A\mid H)\).
📖 Worked Solution & Strategy
Let \(T\in\{A,B,C\}\) denote the server's tier and \(H\) denote the occurrence of an alert. The prior tier probabilities are \(\Pr(T=A)=0.45\), \(\Pr(T=B)=0.35\), and \(\Pr(T=C)=0.20\), and the conditional alert rates are \(\Pr(H\mid T=A)=\frac{1}{500}\), \(\Pr(H\mid T=B)=0.006\), and \(\Pr(H\mid T=C)=0.002\). By the Law of Total Probability, the unconditional probability of an alert is \(\Pr(H) = \sum_{j\in\{A,B,C\}} \Pr(H\mid T=j)\Pr(T=j) = 0.45(\frac{1}{500}) + 0.35(0.006) + 0.20(0.002) = 0.0034\). Applying Bayes' Theorem gives \(\Pr(T=A\mid H) = \frac{\Pr(H\mid T=A)\Pr(T=A)}{\Pr(H)} = \frac{\frac{9}{20}(\frac{1}{500})}{0.0034} = \frac{9}{34}\). The evaluated result is \(\frac{9}{34}\), which matches the correct option.