Conditional Probability: Real-World Scenarios (6)

Question

In a city, 1% of people have a rare disease. A test detects the disease correctly in 99% of sick people, but also gives a false positive in 5% of healthy people. If a randomly chosen person tests positive, what is the probability they actually have the disease?

This is the classic Bayes’ theorem problem — counter-intuitive and a JEE Advanced favourite.

Solution — Step by Step

Final answer: $\approx 16.7\%$.

Why This Works

The result feels wrong because the test is “99% accurate” — but that statistic refers to true-positive rate, not the predictive value of a positive result. Most healthy people test positive too, simply because there are vastly more healthy people than sick people.

Bayes’ theorem balances the prior probability ($P(D) = 1\%$, very small) against the likelihood ratio ($0.99/0.05 \approx 20$, fairly informative). The product gives the posterior — moving from $1\%$ to $\sim 17\%$ is a 17-fold update, not a slam-dunk diagnosis.

Alternative Method

Use a population of $10000$ people. Expected: $100$ sick, $9900$ healthy. True positives: $99$ ($99\%$ of $100$). False positives: $495$ ($5\%$ of $9900$). Total positives: $594$. Probability of disease given positive = $99/594 = 1/6 \approx 16.7\%$.

This “natural frequencies” approach is what physicians and statisticians prefer — no formulas to remember.

Common Mistake

Reading “99% accurate” and concluding $P(D|+) = 0.99$. The conditional flips: $P(+|D) = 0.99$ does NOT imply $P(D|+) = 0.99$. The base rate $P(D)$ matters enormously.