Bayesian inference
A detector catches almost every fault. Yet most of its alerts can still be false. Explore the missing piece: how common the fault was before the alert.
In a hypothetical fleet, 1% of machines have a fault. The detector alerts on 90% of faulty machines and 5% of healthy machines. A randomly chosen machine has an alert.
What is the chance it actually has a fault?
All numbers in this lesson are invented teaching examples. No fleet or detector was measured.
You will learn to distinguish a detector's sensitivity from the probability of a fault after an alert, explain the denominator, and recognize when the model's assumptions fail. Percentages are enough to begin.
The base rate describes the machines before checking alerts. Sensitivity is the fraction of faulty machines that alert. The false-positive rate is the fraction of healthy machines that alert.
Start with your prediction above, or open the experiment now.
Among all machines, how many have a fault?
Among faulty machines, how many alert?
Among healthy machines, how many alert? Specificity is 100% minus this rate.
Among machines with an alert
90 true alerts out of 585 total alerts.
The alert raises the probability from 1% to 15.38%.
This bar has a new denominator: all alerts. The table below is the numeric text alternative to both graphics.
| Actual state | Alert | No alert | Total |
|---|---|---|---|
| Faulty | 90 | 10 | 100 |
| Healthy | 495 | 9,405 | 9,900 |
| Total | 585 | 9,415 | 10,000 |
With no alert, the probability of a fault is 0.11%.
These are expected counts, not a random sample. Fractional counts are valid expectations. Displayed values are rounded to at most two decimals; calculations use full precision.
Predict what happens at 20%, then try it. Why does an unchanged detector now produce mostly true alerts?
Start at 1%, 90%, 5%. Compare raising sensitivity to 100% with lowering false positives to 0.5%. Which change makes an alert more trustworthy here?
Set sensitivity equal to the false-positive rate, with a rate above 0%. Does an alert change the prior? Explain why.
At 20% base rate, 1,800 of 2,200 alerts are true: 81.82%. The healthy group is smaller and the faulty group larger.
At 1% base rate, raising sensitivity to 100% gives 100 / 595 = 16.81%. Lowering the false-positive rate to 0.5% with sensitivity at 90% gives 90 / 139.5 = 64.52%. Here, the large healthy group makes false alarms the dominant issue. This does not imply detector thresholds can improve sensitivity and false positives independently in practice.
If sensitivity equals the false-positive rate, an alert is equally likely in either state. For an alert with nonzero probability, the posterior equals the prior.
An alert has two possible routes: a fault triggered it, or a healthy machine triggered it. Bayesian inference compares the probability mass arriving by each route.
The key move: after seeing an alert, restrict attention to machines with an alert. The fraction that are faulty is the posterior probability.
The alert is still useful evidence: it raises the probability from 1% to about 15%. But 90% sensitivity answers a different question. Its denominator is faulty machines; the posterior's denominator is alerted machines.
This uses conditional probability and the law of total probability. See MIT 18.05, reading 3.
The vertical bar means “given.” In formulas, use probabilities from 0 to 1: 5% is 0.05.
The numerator is P(H and +) = P(+ | H)P(H). The denominator adds both mutually exclusive ways an alert can occur: P(+) = sp + f(1 − p). Dividing their intersection by P(+) is the definition of conditioning.
For a population of size N, expected true alerts are Nsp and false alerts are Nf(1 − p). N cancels in their ratio. A larger fleet changes counts, but not this probability if the rates stay fixed.
For 0 < p < 1 and f > 0:
In the first example, prior odds are 1 / 99 and the likelihood ratio is 0.9 / 0.05 = 18. Posterior odds are 18 / 99 = 2 / 11. Converting odds to probability gives 2 / (2 + 11) = 2 / 13.
An alert supports the fault hypothesis when s > f, leaves it unchanged when s = f, and counts against it when s < f, provided the prior is not certain and the observation is possible.
At the starting settings this is 10 / 9,415 ≈ 0.1062%. The chance of a healthy machine after no alert is its complement, about 99.89%.
If sp + f(1 − p) = 0, the model predicts no alerts at all. Conditioning on an alert is undefined, not 0%. The same applies to “no alert” when its probability is zero. Try these boundaries in the experiment. A supposedly impossible observation is a reason to revisit the model.
The two states are exhaustive and mutually exclusive. The base rate applies to the population being examined, and the sensitivity and false-positive rate apply there too. We treat the rates as known and the true fault state as well defined.
Real rates are estimated with uncertainty and can change with operating conditions, detector thresholds, or how machines are selected. A single binary observation needs no independence assumption. Multiplying likelihood ratios across observations requires the right conditional model. Repeated evidence is not automatically independent.
The arithmetic transfers when you can define a hypothesis, evidence, and credible probabilities. The domain determines whether those inputs are defensible.
Classification
H is “this message is spam.” Evidence could be a flagged phrase. Its prevalence in spam and legitimate mail matters, along with the prior spam rate.
Boundary: correlated words cannot be treated as independent evidence without justification. A changed inbox can invalidate old rates.
Scientific evidence
H is “a source contains the target phenomenon.” An instrument signature updates the prior through how likely that signature is with and without the phenomenon.
Boundary: choosing candidates after scanning many signals changes the relevant model. A posterior alone does not establish a causal explanation.
Health statistics
Sensitivity and specificity describe conditional test behavior. Predictive value also depends on prevalence in the tested population.
Boundary: population figures alone do not determine an individual's diagnosis. Clinical context and applicability of test estimates matter. See the diagnostic-performance reference.
Belief is not a decision rule. Whether to inspect a machine depends on the costs of inspection, missed faults, and unnecessary shutdowns. Bayes' rule updates a probability; it does not choose those costs.
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Recognition & explanation
They reversed P(alert | fault) and P(fault | alert). Sensitivity uses faulty machines as its denominator. The posterior uses alerted machines. You also need the relevant base rate and false-positive rate.
Self-check: name both denominators and explain why healthy machines contribute to the alert pool. Repeating “base-rate neglect” without the mechanism is recognition, not yet explanation.
Near transfer · new quantities
In 10,000 fragments, 200 come from the kiln and 9,800 do not. Expect 160 marked kiln fragments and 392 marked other fragments. The posterior is 160 / 552 = 20 / 69 ≈ 28.99%.
Self-check: both routes to the marker must appear in the denominator. The marker raises the prior from 2% substantially, yet the kiln is still less likely than all alternatives combined. The sampling and marker-rate assumptions are essential.
Far transfer · dependent evidence
A copied display contains no additional evidence once the original reading is known. Squaring the ratio double-counts the same observation. The second update needs P(second evidence | H, first evidence) divided by P(second evidence | ¬H, first evidence).
For an exact copy after the first alert, both probabilities are 1, giving a ratio of 1. Multiplying the original ratio again is justified only if a new observation has those same rates and is conditionally independent of the first under both H and ¬H.
Self-check: explain the dependency and name the conditional assumption. Two devices or two reports do not by themselves establish it.
Far transfer · model limits
The relevant prior is conditional on what is already known, including unusual vibration. You need P(fault | unusual vibration), or a joint model that updates on vibration and detector evidence without double-counting. Detector performance may also differ in that selected group.
A shutdown decision needs consequences and alternatives: damage risk, inspection cost, downtime, and what additional information can be gathered. Probability alone does not set the action threshold.
Self-check: identify the selected population and separate inference from decision-making. Do not invent a replacement rate from the information given.
In the private curriculum, record the date and whether each answer was unaided, with support, or needs a revisit. Save one sentence about the misconception or assumption you noticed.
Keep recognition, explanation, and transfer separate. This page cannot assess free-text answers or certify mastery. A later successful attempt with a fresh problem is stronger evidence than an answer revealed today.
The mathematical model is a binary Bayesian update with fixed rates. It does not estimate those rates, simulate a random sample, or model a changing machine over time.
Version 1 · 2026-10-10. References need a connection; the lesson does not. Author-reviewed calculations and browser checks are recorded in the private learning registry. Learner understanding has not been assessed.