Travelers face long waits at some US airports amid DHS shutdown

Let’s break down the problem step by step. --- ## Step 1: Understanding the problem We have a large number of people (1,000,000) and a small number of people who have a rare disease (1 in 1,000). The test for the disease has: - Sensitivity = 99% (probability of a positive test given the person has the disease) - Specificity = 99% (probability of a negative test given the person does not have the disease) We are told that a randomly selected person tests positive. We need to find the probability that this person actually has the disease. --- ## Step 2: Define events Let: - \( D \) = event that the person has the disease - \( T^+ \) = event that the test is positive We are given: - \( P(D) = \frac{1}{1000} = 0.001 \) - \( P(T^+ \mid D) = 0.99 \) (sensitivity) - \( P(T^- \mid D^c) = 0.99 \) → \( P(T^+ \mid D^c) = 1 - 0.99 = 0.01 \) (false positive rate) We want \( P(D \mid T^+) \). --- ## Step 3: Apply Bayes’ theorem Bayes’ theorem says: \[ P(D \mid T^+) = \frac{P(T^+ \mid D) P(D)}{P(T^+ \mid D) P(D) + P(T^+ \mid D^c) P(D^c)} \] --- ## Step 4: Plug in numbers \[ P(D) = 0.001, \quad P(D^c) = 0.999 \] \[ P(T^+ \mid D) = 0.99, \quad P(T^+ \mid D^c) = 0.01 \] Numerator: \[ P(T^+ \mid D) P(D) = 0.99 \times 0.001 = 0.00099 \] Denominator: \[ P(T^+ \mid D) P(D) + P(T^+ \mid D^c) P(D^c) = 0.00099 + (0.01 \times 0.999) \] \[ = 0.00099 + 0.00999 = 0.01098 \] --- ## Step 5: Compute the conditional probability \[ P(D \mid T^+) = \frac{0.00099}{0.01098} \] Multiply numerator and denominator by 100000 to avoid decimals: \[ \frac{99}{1098} \] Simplify: \[ \frac{99}{1098} = \frac{33}{366} = \frac{11}{122} \] Numerically: \[ \frac{0.00099}{0.01098} \approx \frac{0.00099}{0.011} \approx 0.09 \] So the probability is about \( \frac{11}{122} \approx 0.0901639344262295 \). --- ## Step 6: Match to the given options The problem likely expects the answer as a fraction in simplest form. \[ \frac{11}{122} \] Checking the options: - A) \(\frac{1}{11}\) ≈ 0.0909 → close but not exact - B) \(\frac{1}{10} = 0.1\) → no - C) \(\frac{1}{101}\) ≈ 0.0099 → no - D) \(\frac{1}{2}\) = 0.5 → no None match exactly \(\frac{11}{122}\), but \(\frac{1}{11}\) is very close (relative error ~1%). Given typical multiple-choice setups, they might approximate \( \frac{11}{122} \approx \frac{1}{11} \). --- \[ \boxed{A} \] --- **Final note**: The exact answer is \(\frac{11}{122}\), but since that’s not an option, A is the best approximation.