Base rate fallacy

Example

A disease affects $1\%$ of a population. A test is positive for $90\%$ of the sick people and for $9\%$ of the healthy people. Let $S$ be the event "sick" and $T$ the event "positive test". Then $P(S) = 0.01$, $P(T \mid S) = 0.9$ and $P(T \mid S^c) = 0.09$. By the law of total probability, $P(T) = 0.9 \cdot 0.01 + 0.09 \cdot 0.99 = 0.0981$. By Bayes' theorem, $P(S \mid T) = 0.009 / 0.0981 \approx 0.092$. So only about $9\%$ of the people with a positive test are sick. This is true although the test is right for $90\%$ of the sick and for $91\%$ of the healthy.

Remarks

The base rate fallacy is the error to ignore the base rate $P(S)$ and to take $P(S \mid T)$ to be close to $P(T \mid S)$. When the condition is rare, most positive results are false positives, even for a good test.

Explanations

L1 human

Among $10000$ people, $100$ are sick and $90$ of them test positive. Of the $9900$ healthy people, $891$ also test positive. So of the $981$ positive tests, only $90$ come from sick people.