Conditional probability: Difference between revisions

Revision as of 16:43, 25 August 2025

Revision as of 16:44, 25 August 2025

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{{Main|Confusion of the inverse}}

[[File:Bayes theorem visualisation.svg|thumb|450x450px|A geometric visualization of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that <math>P(A\mid B) P(B) = P(B\mid A) P(A)</math> i.e. <math>P(A\mid B) = \frac{P(B\mid A)} {P(A)\cdot P(B)}</math> . Similar reasoning can be used to show that <math>P(\bar A\mid B) = \frac{P(B\mid\bar A) P(\bar A)}{P(B)}</math> etc.]]

In general, it cannot be assumed that ''P''(''A''|''B'')&nbsp;≈&nbsp;''P''(''B''|''A''). This can be an insidious error, even for those who are highly conversant with statistics.<ref>{{cite book |last=Paulos, |first=J. A. (|year=1988) ''|title=Innumeracy: Mathematical Illiteracy and its Consequences'', |publisher=Hill and Wang. {{ISBN|isbn=0-8090-7447-8}} (|at=p. 63 ''et seq.'') }}</ref> The relationship between ''P''(''A''|''B'') and ''P''(''B''|''A'') is given by [[Bayes' theorem]]:

:<math>\begin{align}

P(B\mid A) &= \frac{P(A\mid B) P(B)}{P(A)}\\