The finale keeps the exact count table visible while naming the modeling boundary. The arithmetic is exact; the independence assumption and any broader performance claim stay outside the computation.

highlighted = computed this step

What is exact

The priors, likelihoods, scores, and posteriors are exact rational values from the shown integer counts.

countsratiosproductsposteriors\text{counts}\rightarrow\text{ratios}\rightarrow\text{products}\rightarrow\text{posteriors}
Exact Naive Bayes surfaceThe table keeps counts, products, posteriors, and the assumption boundary visible.Naive Bayes from countstotal=6; counts=A:4, B:2feature counts: F1=1: A=3, B=1; F2=1: A=2, B=1query: F1=1, F2=1; smoothing=noneclassP(c)likelihood count/countscoreposteriorsumargmaxAP(A)=4/6 (=2/3)P(F1=1|A)=3/4P(F2=1|A)=2/4 (=1/2)1/43/41ABP(B)=2/6 (=1/3)P(F1=1|B)=1/2P(F2=1|B)=1/21/121/41posteriors sum = 1argmax=Anaive independence assumed; no smoothing; no logs or decimals

What is assumed

The independence statement is a modeling assumption, not a transcendental boundary. The smoothing policy here is none, so zero likelihoods are rejected.

smoothing=none\text{smoothing}=\text{none}
Exact Naive Bayes surfaceThe table keeps counts, products, posteriors, and the assumption boundary visible.Naive Bayes from countstotal=6; counts=A:4, B:2feature counts: F1=1: A=3, B=1; F2=1: A=2, B=1query: F1=1, F2=1; smoothing=noneclassP(c)likelihood count/countscoreposteriorsumargmaxAP(A)=4/6 (=2/3)P(F1=1|A)=3/4P(F2=1|A)=2/4 (=1/2)1/43/41ABP(B)=2/6 (=1/3)P(F1=1|B)=1/2P(F2=1|B)=1/21/121/41posteriors sum = 1argmax=Anaive independence assumed; no smoothing; no logs or decimals

Summary

The priors, likelihoods, products, and posteriors are exact rational values from the shown counts. The independence assumption is a stated modeling choice; this pins counting mechanics, not a claim about accuracy, calibration, or generalization.

exact counting mechanics only\text{exact counting mechanics only}
Exact Naive Bayes surfaceThe table keeps counts, products, posteriors, and the assumption boundary visible.Naive Bayes from countstotal=6; counts=A:4, B:2feature counts: F1=1: A=3, B=1; F2=1: A=2, B=1query: F1=1, F2=1; smoothing=noneclassP(c)likelihood count/countscoreposteriorsumargmaxAP(A)=4/6 (=2/3)P(F1=1|A)=3/4P(F2=1|A)=2/4 (=1/2)1/43/41ABP(B)=2/6 (=1/3)P(F1=1|B)=1/2P(F2=1|B)=1/21/121/41posteriors sum = 1argmax=Anaive independence assumed; no smoothing; no logs or decimals