Bayes' Theorem

Inverting Conditional Probabilities

Bayes' theorem lets you flip a conditional probability — turning $P(E \mid H)$ into $P(H \mid E)$:

$P(H \mid E) = \frac{P(E \mid H) \cdot P(H)}{P(E)}$

The denominator $P(E)$ comes from the law of total probability:

$P(E) = P(E \mid H) \cdot P(H) + P(E \mid H^c) \cdot P(H^c)$

The Medical Test Example

A disease affects 1% of the population. A test has 99% sensitivity (true positive rate) and 5% false positive rate.

You test positive. What is the probability you have the disease?

$P(H) = 0.01, \quad P(E \mid H) = 0.99, \quad P(E \mid H^c) = 0.05$

$P(E) = 0.99 \times 0.01 + 0.05 \times 0.99 = 0.0594$

$P(H \mid E) = \frac{0.99 \times 0.01}{0.0594} \approx 0.1667$

Only 17% — the low base rate dominates. This is the base-rate fallacy.

Terminology

• $P(H)$ — prior: your belief before seeing evidence
• $P(H \mid E)$ — posterior: updated belief after evidence
• $P(E \mid H) / P(E)$ — likelihood ratio: how much the evidence updates you

Your Task

Implement bayes(p_h, p_e_given_h, p_e_given_not_h) that returns $P(H \mid E)$, rounded to 4 decimal places.