Problem Statement
In a population, 2% of people have a specific condition $D$. A diagnostic test correctly detects the condition 95% of the time (sensitivity = 0.95), but gives a false positive result for 5% of healthy individuals (specificity = 0.95).
A randomly selected individual tests positive.
Calculate the probability that this person actually has the condition. Give your answer as an exact fraction in simplest form. [6 Marks]
Verified Solution & Marking Scheme
Define Events & Probabilities
$$P(D) = 0.02, \quad P(D') = 0.98$$
$$P(+ | D) = 0.95, \quad P(+ | D') = 0.05$$
We seek $P(D | +)$ using Bayes' Theorem:
$$P(D | +) = \frac{P(D)P(+ | D)}{P(+)}$$
Compute Total Probability of Positive Test P(+)
$$P(+) = P(D)P(+ | D) + P(D')P(+ | D')$$
$$= (0.02)(0.95) + (0.98)(0.05)$$
$$= 0.0190 + 0.0490 = 0.0680$$
Compute Posterior Probability
$$P(D | +) = \frac{0.0190}{0.0680} = \frac{19}{68} \approx 0.2794$$