How to Solve Probability Foundations — General probability — multiplication rule for chained conditionals Questions on Exam P
Sample Practice Problem
ID: #OUN2C
Let \(T_A\) and \(T_I\) be independent exponential waiting times to a network alert and a safety inspection, with means \(5\) and \(3\) hours. Their respective limits are \(2\) and \(2\) hours. Calculate the probability that both occur within their limits.
📖 Worked Solution & Strategy
For exponential waiting time \(T\) with mean \(\mu\), \(\Pr(T \le t) = 1 - e^{-t/\mu}\). For alert time \(T_A\) (mean 5, limit \(2\)), \(p_A = \Pr(T_A \le 2) = 1 - e^{-2/5} = 0.3297\). For inspection time \(T_I\) (mean 3, limit 2), \(p_I = \Pr(T_I \le 2) = 1 - e^{-2/3} = 0.4866\). Independence allows multiplying event probabilities. The probability that both events occur within their limits is \(p_A p_I = (0.3297)(0.4866) = 0.1604\). The evaluated result is \(0.1604\), which matches the correct option.