Does a 99% accurate test mean a 99% chance of disease?

Here is a question that tests this misconception head-on, in the spirit of JEE Advanced 2024 Paper 1 Q2:

The misconception answer reads the accuracy straight off: "99% accurate, positive result, so about a 99% chance." The correct move is to count real people in a town of $100{,}000$, where the base rate $\tfrac{1}{1000}$ can do its work:

The $99$ are the true positives, the $999$ are the false alarms from the huge healthy group, and the answer is about 9%, not 99% — the base rate decides, not the accuracy.

The accuracy $P(\text{positive} \mid \text{disease})$ is the number on offer, and it feels like the answer to "am I ill?". So the gut reads it straight across as $P(\text{disease} \mid \text{positive})$, discarding the base rate it never thought to ask for. The swap is invisible because both quantities are "the chance of disease given a test" in ordinary English — only the direction of conditioning differs, and language hides direction.

Gigerenzer and Hoffrage (1995) showed this error is far less likely when the same information is given as natural frequencies — counts of real people — rather than conditional probabilities. Being shown Bayes' theorem rarely dislodges the intuition; students plug numbers in and revert to "99%" once the symbols are gone. What dislodges it is counting the town by hand and watching the $99{,}900$ healthy people produce more positives than the $100$ ill ones.

$P(A \mid B)$ and $P(B \mid A)$ are different quantities. The one you are given (the test's accuracy) is not the one you want (your chance of illness). Connect them with Bayes' theorem, which forces the base rate back into view:

The two conditionals coincide only when $P(A) = P(B)$. For a rare event $P(A)$ is tiny, so $P(\text{disease} \mid \text{positive})$ sits far below $P(\text{positive} \mid \text{disease})$. The reliable habit is to count people: build the population, tally true positives against false positives, and read the answer as a fraction — the count is Bayes' theorem, written out.

Count a town of $100{,}000$ people for a disease affecting $1$ in $1000$, tested by a 99% accurate test.

1. The genuinely ill. $100{,}000 \times \tfrac{1}{1000} = 100$ people have the disease.
2. True positives. The test catches 99% of them: $100 \times 0.99 = 99$ ill people test positive.
3. The healthy group. $100{,}000 - 100 = 99{,}900$ are healthy, and the test wrongly flags 1% of them: $99{,}900 \times 0.01 = 999$ false positives.
4. Put the groups together. Total positives $= 99 + 999 = 1098$, of which only the $99$ are truly ill:
   $$P(\text{disease} \mid \text{positive}) = \frac{99}{1098} \approx 0.090 = 9\%.$$
5. Conclude. The gut said 99%; the count says about 9%. The accuracy never changed — the base rate did all the work. That count is exactly Bayes' theorem: $\tfrac{99}{99 + 999}$ is $\tfrac{P(\text{pos}\mid\text{dis})P(\text{dis})}{P(\text{pos})}$.

The same framework settles JEE Advanced 2024 Paper 1 Q2: a student answers a quiz question correctly, and you are asked $P(\text{knew} \mid \text{correct})$. The trap is to read off the "knew" rate; the fix is Bayes, weighting against the chance a guesser also lands the correct option, so the base rate of guessing pulls the posterior down — never the transposed read.