Question bank — Bayes' theorem and applications

False. These are different questions ("A given B" vs "B given A") and are only equal in the special case $P(A)=P(B)$; the whole point of Bayes is that they usually differ.

False. Accuracy is a likelihood $P(T|D)$; the posterior $P(D|T)$ also depends on the prior (base rate). A rare disease can still give a low posterior after a positive test.

False. Evidence against $A$ (evidence more likely under $\neg A$) pushes the posterior below the prior; Bayes updates in both directions.

True. When $B$ is equally likely whether or not $A$ holds, it fails to distinguish the hypotheses, so the posterior equals the prior — the evidence is uninformative.

Mostly true in spirit. When the prior is tiny, even a small false-positive rate produces far more false positives than true positives, so a single positive test barely raises the posterior — you need a second independent test.

False. It is pure algebra: rearranging the two definitions of $P(A\cap B)$ from conditional probability. It introduces no new axiom.

True. $P(B)$ does not depend on which $A_i$ you ask about, so it is a shared normalizing constant — this is why it cancels in ratios of posteriors.

False. "Naive" flags the assumption as usually wrong; it is a tractability choice. The classifier often still ranks classes correctly despite the false assumption.

The prior $P(D)$ is missing from the numerator. The correct form is $P(D|T)=\frac{P(T|D)P(D)}{P(T)}$ — dropping $P(D)$ silently assumes $P(D)=1$.

This only counts the world where $A$ is true. The law of total probability needs both branches: $P(B)=P(B|A)P(A)+P(B|\neg A)P(\neg A)$.

A probability above 1 is impossible, signalling a wrong $P(T)$. Since $P(T)\ge P(T|D)P(D)$ always, a numerator of $0.8$ forces $P(T)\ge 0.8$; the arithmetic exposes a mis-computed denominator.

They need not. $P(T|D)$ and $P(\neg T|D)$ sum to 1 (same condition), but $P(T|D)$ and $P(T|\neg D)$ are probabilities under different conditions and are unrelated.

The $84\%$ is the probability a particular email is spam given it says "FREE", not an error rate over a population. Confusing a single-event posterior with a long-run frequency is a category error.

The predicted class comes from the largest posterior, which depends only on rank order. Poor absolute calibration can coexist with correct ranking and good accuracy.

A prior of exactly 0 kills the numerator $P(B|A)P(A)=0$, so the posterior stays 0. Bayes can never resurrect a hypothesis you assigned zero prior — assign a tiny nonzero prior instead.

The many healthy people (large $P(\neg D)$) generate a flood of false positives that outnumber the few true positives, so most positives come from healthy people.

The numerator $P(B|A)P(A)$ is an unnormalized weight; dividing by $P(B)$ rescales all hypotheses so their posteriors sum to 1 and stay $\le 1$.

$P(B)$ is identical for both classes, so it cancels in the ratio $\frac{P(C_1|\mathbf{x})}{P(C_2|\mathbf{x})}$; only the numerators decide the winner.

Both $P(A|B)$ and $P(B|A)$ are built from the same joint $P(A\cap B)$. Bayes substitutes one expression for that joint into the other, swapping the direction of conditioning.

A posterior is just your updated belief; feeding it back as the prior for the next piece of evidence is exactly what "updating incrementally" means, and repeated application is consistent as long as the evidences are conditioned correctly.

The bigger the gap between $P(B|A)$ and $P(B|\neg A)$, the more one hypothesis explains the data than the other, so the ratio of numerators — and hence the shift from prior to posterior — is larger.

MLE maximizes only the likelihood $P(B|A)$ and ignores the prior $P(A)$; Bayes weights the likelihood by the prior, so with a strong prior the two can pick different hypotheses.

Undefined — Bayes requires $P(B)\ne 0$ because you cannot condition on an event that never happens (division by zero).

The posterior stays 1 regardless of evidence: you were already certain, so no observation can change a probability of exactly 1 (or exactly 0).

It becomes 0. If $A$ makes $B$ impossible yet $B$ occurred, $A$ is ruled out completely — this is how evidence can eliminate a hypothesis.

The whole product $\prod P(x_i|C)$ collapses to 0, wrongly forbidding the class from ever winning. This is why smoothing (e.g. Laplace) adds a small count so no likelihood is exactly zero.

They are equal — the evidence cannot distinguish them, so you leave with a 50/50 split (for two hypotheses) exactly as you started.

Nothing about the diagnosis. With $P(T|D)=P(T|\neg D)$ the posterior equals the prior — the test is diagnostically worthless despite producing "positive" results.

It equals the prior for each hypothesis. Uniformly likely evidence is uninformative, so Bayes leaves your beliefs untouched — a useful sanity check.