In Bayes' rule you reversed a conditional probability with a tree. With several possible causes the idea is the same: the law of total probability supplies the denominator, and each cause gets its share. Bayes' theorem is how evidence updates a belief.
Idea clave
For a partition $A_1, \ldots, A_n$ and evidence $B$ with $P(B) > 0$: $$P(A_j \mid B) = \frac{P(B \mid A_j)P(A_j)}{\sum_i P(B \mid A_i)P(A_i)}.$$ The posterior is proportional to prior × likelihood, rescaled so the posteriors add to 1.
Por qué funciona
By definition, $P(A_j \mid B) = \frac{P(A_j \cap B)}{P(B)}$. The numerator is $P(B \mid A_j)P(A_j)$ (the multiplication rule) and the denominator is $\sum_i P(B \mid A_i)P(A_i)$ (total probability). In tree language: among all the paths that end in B, the posterior of cause $A_j$ is the share of probability flowing through $A_j$.
Worked examples
Ejemplo resuelto: Which machine?
Machines A, B, C make 50%, 30%, 20% of the parts with defect rates 1%, 2%, 4%. A random part is defective. Find the probability that each machine made it.
- Joint probabilities: $A$: $0.005$, $B$: $0.006$, $C$: $0.008$; total $P(D) = 0.019$.
- $P(A \mid D) = \frac{0.005}{0.019} = \frac{5}{19}$, $P(B \mid D) = \frac{6}{19}$, $P(C \mid D) = \frac{8}{19}$.
- They add to 1. C makes the fewest parts but is the most likely source of a defect.
Respuesta: $\frac{5}{19}$, $\frac{6}{19}$, $\frac{8}{19}$
Ejemplo resuelto: Two positive tests
A condition affects 1% of people. A test is positive for 95% of people with it and for 5% without it. A person tests positive twice, and the two results are independent given their condition. Find the probability they have the condition.
- After one positive: $$\frac{0.95(0.01)}{0.95(0.01) + 0.05(0.99)} = \frac{0.0095}{0.059} = \frac{19}{118} \approx 0.161$$
- Use $0.161$ as the new prior. Or in one step, multiply the likelihoods: $$\frac{0.95^{2}(0.01)}{0.95^{2}(0.01) + 0.05^{2}(0.99)} = \frac{0.009025}{0.0115} = \frac{361}{460}$$
- $\approx 0.785$. One positive is weak evidence for a rare condition; two independent positives are much stronger. (A negative second result would instead multiply by $0.05$ on the "condition" side and $0.95$ on the other, pushing the probability back down.)
Respuesta: $\frac{361}{460} \approx 0.785$
The same update in odds form: posterior odds = prior odds × likelihood ratio. Here the prior odds are $\frac{1}{99}$, and each positive multiplies them by $\frac{0.95}{0.05} = 19$: after two positives the odds are $\frac{361}{99}$, a probability of $\frac{361}{460}$.
Error común
Reading the test's accuracy as the answer: $P(+ \mid \text{condition}) = 0.95$ is not $P(\text{condition} \mid +)$, which is only $0.161$ after one test. The other common slip is forgetting to divide by $P(B)$, so the "posteriors" do not add to 1. Check: list every cause and make sure your posteriors add to 1.