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The numerator $P(B\mid A)P(A)$ equals $P(A\cap B)$, while $P(A\mid B)=P(A\cap B)/P(B)$. Thus, wherever the original quotient is defined,In particular a zero numerator and denominator cannot be cancelled; the original expression requires $P(A\mid B)>0$. This is the rearrangement of Bayes' theorem.

By Bayes' theorem,The denominator sums red outcomes from both possible jar choices, as calculated in (i).

The lazy choice weights are $1/w$ and $1/(3w)$, in ratio three to one. Normalizing gives $\boxed{l=3/4}$. Let $Q$ denote selection of the first question and let $q=P(\text{Lazy}\mid Q)$. Its conditional expected value, divided by $m$, isso $q=6/7$. If the overall lazy fraction is $p$, the law of total probability gives $P(Q)=(3/4)p+(1/2)(1-p)=(p+2)/4$. Bayes' theorem then givesSolving $6(p+2)=21p$ yields $\boxed{p=4/5}$. The observed mean is conditioned on the chosen question, so it cannot be equated directly to the unconditional mean over all students.

Insert the values from (a) into Bayes' theorem:Thus the identity is satisfied. Equivalently, both $P(A\mid B)P(B)$ and $P(B\mid A)P(A)$ equal the joint probability $4/9$.

The law of total probability givesFor $\alpha >0$, Bayes' theorem therefore yieldsAt the allowed endpoint $\alpha =0$, no bulb is defective: $\mathbb{P}(D)=0$. This conditional probability is then undefined, rather than $2/5$. The value $2/5$ is its limit as $\alpha \to 0^{+}$, but does not define conditioning on the impossible event within the stated discrete model.

Let $A,B,C$ denote the mutually exclusive supplier events and $D$ the defective-bulb event. Their probabilities and conditional defect rates areThese supplier events partition the possible bulbs. The following calculations use the law of total probability and Bayes' theorem with that partition.

Let $C$ mean control-group membership and $E$ the adverse event. Equal allocation gives $P(C)=P(C^c)=1/2$, so the law of total probability gives $P(E)=(p_0+p_1)/2$. Bayes' theorem then givesThis is defined when $p_0+p_1>0$. If both are zero, the conditioning event has zero probability and the stated conditional probability is undefined.

The sole green ball comes from the first bag exactly on the event $(X,Y)=(1,0)$. Divide this joint probability by the conditioning event's probability:This is the conditional probability form of Bayes' theorem.

By Bayes' theorem, provided $p+q>0$,If $p=q=0$, singing has probability zero and the stated observation cannot occur in this model; that conditional probability is undefined.

Divide the relevant joint probability by the marginal stopping probability. By Bayes' theorem,The denominator is positive under the stated assumptions.

The event includes stopping positions zero through $k$. Applying Bayes' theorem to their joint and marginal probabilities givesThis conditional probability is defined whenever its conditioning event has positive probability. At $k=n$, it reduces to $p$.

By the definition of conditional probability, or Bayes' theorem,Its denominator is positive throughout the specified range. Observing two blue balls raises the probability of replacement for $0<r<1$, because replacement makes a repeated colour more likely.

Let $C={\overline{B}}_1\cap B_2$. Under replacement its joint probability iswhereas without replacement it isIn the latter case the first removed ball is non-blue, so all $N_B$ blue balls remain. Applying Bayes' theorem and cancelling the positive common factors yields the posterior replacement probability after different-colour draws:It does not depend on $N_B$. To prove the requested bound directly,Equality holds at $r=0$ and $r=1$, and the inequality is strict between them. Its sketch joins $(0,0)$ to $(1,1)$, stays below the diagonal, and is increasing and convex becauseThe interpretation is that removing a non-blue ball increases the chance of blue on the second draw; seeing the two different colours therefore favours the no-replacement explanation.

Each position has apple probability $1/2$. Applying Bayes' theorem to the preceding conditional probability givesThis also agrees with exchangeability of the second and third positions.

Of the ten outcomes in $F$, all except $(1,1,1)$ have exactly one repeated pair. Hence $P(T\mid F)=9/10$. Applying Bayes' theorem in the requested direction,Equivalently there are nine favorable outcomes among the ninety outcomes in $T$. This direct count confirms the conditional probability calculation.

Let $0<p(x)<\infty$, take $q_{\lambda }$ supported where $p(z,x)>0$, and assume the following expectations exist. By the definition of Kullback-Leibler divergence and Bayes' theorem,Therefore the evidence lower bound isNonnegativity of Kullback-Leibler divergence proves $\mathcal{L}(\lambda )\le \log p(x)$. The marginal likelihood $p(x)$ is constant with respect to the variational parameter $\lambda$, so maximizing $\mathcal{L}$ minimizes the stated Kullback-Leibler divergence.

The practical advantage is that evaluating logarithms of joint probabilities and $\log q_{\lambda }$ requires no evaluation of the usually intractable marginal likelihood. One can estimate $\mathcal{L}$ and its gradient with Monte Carlo method. Equality with the log marginal likelihood occurs only when $q_{\lambda }=p(\cdot \mid x)$ almost everywhere. Merely maximizing the evidence lower bound over a restricted family neither computes the exact posterior distribution nor proves that the maximum attains this equality.

For any chosen proper prior density $\rho (\mu ,\sigma )$ on $\mathbb{R}\times (0,\infty )$ with positive evidence, Bayes' theorem givesThe prior density comes from the first two random draws. The likelihood $L$ is the probability of satisfying all observed-label constraints after integrating the past duration noises. The factor $\Phi ((\mu -30)/\sigma {)}^5$ is the probability of the five-day outcome conditional on the shared parameters. The denominator is the probability of acceptance, which converts the numerator's joint probability into the required conditional probability.

For the independent uniform priors in part (a), the constant $\rho =1/(60\cdot 19)$ cancels and both integrals reduce to $0<\mu <60$, $1<\sigma <20$. A prior on the variance $v={\sigma }^2$ would require transforming the density: this chosen prior has density $1/(38\sqrt{v})$ for $1<v<400$ conditional on $\mu$. One must not silently replace it by a uniform prior on $v$.

There is no unique numerical probability until the prior distribution is chosen. As a check for the stated uniform priors, independent Gaussian quadrature calculations with 256 and 512 nodes per coordinate agree at the shown precision:The prior-predictive probability of all observed labels is approximately $0.00030422536$, explaining the rejection cost. These numbers are illustrations of the explicit model, not additional observed data.

A subtle point is that the five future days are independent only conditional on $(\mu ,\sigma )$. Integrating shared parameter uncertainty gives $\mathbb{E}[{\pi }^5\mid A]$, not $(\mathbb{E}[\pi \mid A]{)}^5$. The latter is about $0.08631110$ for these priors and understates the five-day probability. Jensen's inequality gives $(\mathbb{E}[\pi \mid A]{)}^5\le \mathbb{E}[{\pi }^5\mid A]$, while $0\le \pi \le 1$ gives $\mathbb{E}[{\pi }^5\mid A]\le \mathbb{E}[\pi \mid A]$. For two future late-arrival indicator functions, their posterior covariance is $\mathrm{Var}(\pi \mid A)\ge 0$. This is the posterior predictive repeated-success identity.

Bayes' theorem solves a precise part of the scientific confirmation problem: it gives a coherent relation between prior commitments, evidential likelihoods and updated probabilities. It does not select those commitments from no premises, and it should not be identified with every meaning of confirmation.

For background information $K$, a hypothesis $H$ and newly considered evidence $E$, let $p=P(H\mid K)$, $a=P(E\mid H,K)$ and $b=P(E\mid \neg H,K)$. With a positive denominator,The first formula requires a positive denominator; the odds formula additionally assumes $0<p<1$ and $b>0$. The second equation separates prior odds from the Bayes factor. For $0<p<1$, the evidence raises the probability of $H$ exactly when $a>b$. It is not enough that $E$ is likely under $H$; it must be more likely than under the relevant alternative mixture. This guards against an inverse-probability error in evaluating scientific tests.

For example, let the prior be 0.01 and the two likelihoods be 0.8 and 0.08. The Bayes factor is ten, but the posterior probability is only $10/109\simeq 0.0917$. The evidence strongly favours $H$ relative to its previous odds without making $H$ probable in an absolute sense. Evidence can therefore confirm a hypothesis while leaving a rational investigator doubtful. The comparison also depends on which alternatives are included in $\neg H$ and how their probabilities are distributed.

The raven paradox shows another benefit. Logically equivalent claims receive identical probabilities in the same probabilistic model, but this does not force every investigation of a white object to provide the same evidence about ravens. If an object is selected because it is already known to be a shoe, observing that it is not a raven may have equal likelihood under the rival hypotheses and provide a Bayes factor of one. Random selection from a specified population can give different likelihoods. The sampling procedure and background information must be represented, rather than letting a grammatical positive-instance rule determine evidential support.^[ref][ref]

The problem of old evidence is a genuine limit of the simplest updating account. If $E$ already belongs to $K$, conditioning on it again produces no increase. Yet an old measurement can be important support for a newly formulated theory. One may assess support relative to a specified earlier or counterfactual background, or model learning a previously unrecognized connection between theory and evidence. These are substantive additions. Simply applying the same conditioning rule to an already certain fact does not represent the new intellectual achievement.^[ref]

Other limitations concern model construction and priors. A hypothesis assigned zero prior probability remains at zero after ordinary conditioning on evidence of positive total probability. A badly specified likelihood can generate confident but unreliable results. Different reasonable priors need not disappear under finite or non-discriminating evidence. Nor does the theorem supply a noncircular solution to the problem of induction: the model's connection to future physical events remains an empirical commitment.

Thus the Bayesian framework gives a rigorous account of probability change, comparison of alternatives and some apparent paradoxes. These are real achievements. It is incomplete as a general recipe for scientific judgment unless the hypotheses, backgrounds, likelihoods and forms of learning are also defended. A mathematical rule of coherent updating can organize evidential reasoning without establishing that every coherent input represents nature correctly.

Scientific underdetermination occurs when the relevant evidence does not select a unique theoretical account. Its force depends on the scope of “the evidence”. Several models can agree with a finite data set while disagreeing about new measurements; a suitable experiment can then discriminate them. The stronger claim of empirical equivalence says that rivals agree on every possible observation within a stated domain, so further evidence of that same kind cannot separate them.

The Duhem problem gives a related but distinct source of ambiguity. If a target theory $T$ and auxiliary hypotheses $A$ jointly imply an observation $E$, then observing $\neg E$ shows $\neg (T\wedge A)$, not uniquely $\neg T$. Independent evidence about the apparatus or background assumptions can redistribute support. It does not make the original conjunction logically true; it provides additional information relevant to which revision is warranted.

A finite-data example makes the limits clear. If $p_0$ interpolates observations at distinct $x_1,\dots ,x_n$, then every polynomial $p_c(x)=p_0(x)+c{\prod }_{i=1}^n(x-x_i)$ has the same sampled values. Thus polynomial interpolation does not determine a unique prediction outside the sampled points when the degree is unrestricted. A degree bound of $n-1$ would instead give the familiar uniqueness result, so the allowed model class matters. That does not show that every interpolant is equally credible for extrapolation. Smoothness assumptions, independently motivated mechanisms and further measurements can matter. But those extra commitments must be stated rather than smuggled into the claim that the original data uniquely determined a model.

In a probabilistic comparison, Bayes' theorem giveswhen the quantities are defined. Equal likelihoods leave the prior odds unchanged by this evidence. Neither equality nor unequal priors follows merely from both theories being logically compatible with $E$. Thus compatibility, equal fit in a particular measure and full empirical equivalence should not be conflated.

The philosophical significance is a constraint on inference, not a universal proof of irrationality. An empiricist may use it to motivate restraint about hidden structure; a scientific realist may defend further explanatory or intervention-based reasons for belief. A convincing sceptical argument must identify genuine rivals and the relevant equivalence, while a convincing realist reply must explain why its additional virtues track the claims at issue. Scientific knowledge can be selective and fallible even when evidence does not deductively choose a unique complete theory.

A Bayesian confirmation account can dissolve the apparent conflict between logical equivalence and ordinary judgements about relevant evidence. It does so by rejecting an unrestricted positive-instance rule and specifying how the evidence was sampled. Bayes' theorem alone does not guarantee that every white shoe is irrelevant, or that every black raven provides strong support.

Let $H$ assert that every raven is black. Its contrapositive is logically equivalent to $H$, so the two descriptions must receive the same conditional probability. The raven paradox arises if every positive instance of a universal statement automatically confirms it: a white nonraven then confirms the contrapositive and hence $H$. The disputed step is that unrestricted confirmation rule, not the logical equivalence.^[ref]

For background $K$, set $p=P(H\mid K)$, $a=P(E\mid H,K)$ and $b=P(E\mid \neg H,K)$. For $0<p<1$ and positive evidence probability,Thus it is the comparison of likelihoods, including the sampling protocol and alternatives, that determines support. Merely rewriting $H$ cannot change it.

The finite-population model of raven confirmation makes the intuitive difference quantitative. Fix $r$ ravens and $M$ nonblack objects. Compare an all-black-ravens hypothesis with the alternative that exactly one raven is nonblack. A uniform draw among nonblack objects that finds a nonraven has likelihood ratio $1/(1-1/M)$. A uniform draw among ravens that finds a black one has likelihood ratio $1/(1-1/r)$. If $M\gg r>1$, the former is much nearer one. With equal prior probabilities, $M=10^6$ and $r=100$ give posterior probabilities approximately $0.50000025$ and $0.50251256$. A tiny amount of confirmation need have no practical significance.

This example is not a theorem about every possible background. Inspecting an object already known to be a white shoe adds no relevant information about its being a nonraven. Other alternatives or selection mechanisms change the likelihoods and can remove or reverse the support. Conditioning only on the observed colour while silently changing how the object was selected would compare different experiments.^[ref]

The resolution is therefore qualified. A coherent probability model preserves equivalence and explains why superficially parallel observations need carry radically different evidence. It does not select a unique prior or sampling model without scientific judgement. Those remaining modelling questions are limitations of an application, rather than a contradiction in Bayesian inference.