Probability and Bayes’ rule
A probability depends on which outcomes you are counting. Combine groups to find an overall rate, then restrict the same counts to see what new evidence tells you.
A flag changes which group we count
A detector flags a component for inspection. How likely is a flagged component to be defective? A high detection rate alone cannot answer this: flags may also come from good components. Use the following chosen batch of 1000 components, with hypothetical detector outcomes.
| Actual condition | Flagged | Not flagged | Total |
|---|---|---|---|
| Defective | 18 | 2 | 20 |
| Good | 49 | 931 | 980 |
| Total | 67 | 933 | 1000 |
Imagine selecting one of these 1000 components uniformly. An event is a set of possible selected components. Write H for “defective” and E for “flagged.” Then $P(H)=20/1000$, the fraction in H. The intersection $H\cap E$ consists of components that are both defective and flagged; its probability is 18/1000.
The conditional probability $P(H\mid E)$ means the probability of a defect given that the component is flagged. The condition keeps only the 67 flagged components. Of those, 18 are defective, so the answer is 18/67. Reversing the condition asks a different question: among the 20 defective components, how many are flagged? That is $P(E\mid H)=18/20=9/10$.
$$P(H\mid E)=\frac{P(H\cap E)}{P(E)},\qquad P(E)>0$$
Conditioning divides the overlap by the probability of the group being kept. It is not a claim that H caused E.
We can recover the overlap from the other conditional: $P(H\cap E)=P(E\mid H)P(H)$. Here (18/20)(20/1000)=18/1000. Substitute this into the preceding fraction to obtain Bayes' theorem.
$$P(H\mid E)=\frac{P(E\mid H)P(H)}{P(E)}$$
The displayed likelihood P(E|H) requires P(H)>0, and the posterior requires P(E)>0. A zero-prior hypothesis can instead be handled by its zero joint probability.
The prior $P(H)$ is our probability before using this new evidence; it can already reflect other knowledge. The likelihood $P(E\mid H)$ describes how often the evidence occurs under the hypothesis. The posterior $P(H\mid E)$ is the probability after restricting to the evidence. The denominator P(E) counts every route to that evidence, including flags from good components.
Build the denominator from both kinds of flag
- Let $H^c$ mean the complement of H: the component is good. The prior probabilities are $P(H)=1/50$ and $P(H^c)=49/50$. They add to one because every component is in exactly one of these two groups.
- The detector's true-positive rate is $P(E\mid H)=9/10$. Its false-positive rate is $P(E\mid H^c)=49/980=1/20$. Multiply each rate by the share of components to which it applies: true flags have probability 9/500 and false flags 49/1000.
- These are disjoint routes to a flag, so $P(E)=9/500+49/1000=67/1000$. Therefore $P(H\mid E)=(18/1000)/(67/1000)=18/67\approx0.269$.
- For a non-flagged component, use the other column. There are 2 defective components among 933 non-flags, giving $P(H\mid E^c)=2/933$. Its numerator can also be computed as $(1-9/10)(1/50)=2/1000$.
The flag raises the defect probability from 1/50 to about 0.269, but it does not raise it to 0.9. Good components are much more numerous, and their 49 false flags outnumber the 18 true flags. The probability of a false flag among good components, 1/20, is also different from the probability of being good among flags, 49/67.
If the overall evidence probability is already supplied, no two-case calculation is needed. For example, chosen probabilities $P(E\mid H)=3/5$, $P(H)=1/4$ and $P(E)=1/2$ give $P(H\mid E)=(3/5)(1/4)/(1/2)=3/10$. The overlap 3/20 is divided by the evidence probability, not by the prior again.
Odds compare a probability with its complement: prior odds are $P(H)/P(H^c)$. In the detector table they are 20/980=1/49. Among flags the posterior odds are 18/49. Dividing Bayes' formula for H by its formula for Hᶜ cancels the common evidence denominator.
$$\frac{P(H\mid E)}{P(H^c\mid E)}=\frac{P(E\mid H)}{P(E\mid H^c)}\frac{P(H)}{P(H^c)}$$
Posterior odds equal likelihood ratio times prior odds. This finite-ratio form requires both prior groups and the false-positive likelihood to have positive probability.
Here the likelihood ratio is (9/10)/(1/20)=18, so the odds multiply by 18: (1/49)·18=18/49. Odds of 18 to 49 correspond to probability 18/(18+49)=18/67; odds and probability have different denominators.
Evidence need not change a prior. If a different flag occurs with probability 2/5 in both defective and good components, its overall probability is also 2/5. Bayes gives $P(H\mid E)=(2/5)P(H)/(2/5)=P(H)$. Equal likelihoods provide no distinction between the hypotheses. If P(E)=0 instead, elementary conditional probability is undefined: its formula cannot divide by zero.