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Bayes' Theorem .
Pr
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{\displaystyle \Pr(A|B)={\frac {\Pr(B|A)\,\Pr(A)}{\Pr(B|A)\Pr(A)+\Pr(B|A^{'})\Pr(A^{'})}}\!}
Assume that
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{\displaystyle \Pr(A^{'})=1-\Pr(A)}
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{\displaystyle \Pr(A|B)={\frac {\Pr(B|A)\,\Pr(A)}{\Pr(B|A)\Pr(A)+\Pr(B|A^{'})-\Pr(B|A^{'})\Pr(A)}}\!}
↓
{\displaystyle \downarrow }
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{\displaystyle \Pr(A|B)={\frac {\Pr(B|A)\,\Pr(A)}{(\Pr(B|A)-\Pr(B|A^{'}))\Pr(A)+\Pr(B|A^{'})}}\!}
↓
{\displaystyle \downarrow }
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{\displaystyle {\frac {\Pr(A|B)}{\Pr(B|A)\,\Pr(A)}}={\frac {1}{(\Pr(B|A)-\Pr(B|A^{'}))\Pr(A)+\Pr(B|A^{'})}}\!}
↓
{\displaystyle \downarrow }
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{\displaystyle {\frac {\Pr(B|A)\,\Pr(A)}{\Pr(A|B)}}=(\Pr(B|A)-\Pr(B|A^{'}))\Pr(A)+\Pr(B|A^{'})\!}
↓
{\displaystyle \downarrow }
Pr
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[
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=
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{\displaystyle \Pr(A)\left[{\frac {\Pr(B|A)}{\Pr(A|B)}}-(\Pr(B|A)-\Pr(B|A^{'}))\right]=\Pr(B|A^{'})\!}
↓
{\displaystyle \downarrow }
Pr
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{\displaystyle \Pr(A)={\frac {\Pr(B|A^{'})}{{\frac {\Pr(B|A)}{\Pr(A|B)}}-(\Pr(B|A)-\Pr(B|A^{'}))}}\!}
Now assume that a positive integer number X (between 1 and 1 million) is picked at random.
let
A
{\displaystyle A\,}
and
let
B
{\displaystyle B\,}
We have
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{\displaystyle \Pr(B|A^{'})=0}
Thus
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{\displaystyle \Pr(A)={\frac {0}{{\frac {\Pr(B|A)}{\Pr(A|B)}}-(\Pr(B|A)-\Pr(B|A^{'}))}}\!}
↓
{\displaystyle \downarrow }
Pr
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A
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=
0
{\displaystyle \Pr(A)=0\!}
Therefore if we pick a positive integer between 1 and 1 million at random, the number we pick will not be an even number.
This is of course WRONG! But I can't see where the mistake is.
You can have fun with this:
let
A
{\displaystyle A\,}
and let
B
{\displaystyle B\,}
202.168.50.40 22:59, 28 November 2006 (UTC) [ reply ]
ab = c → a = c /b . Then you conclude that c = 0 implies a = 0. But if you substitute c := 0 in the equation ab = c , it becomes ab = 0. You cannot conclude from there to a = 0. --Lambiam Talk 23:32, 28 November 2006 (UTC) [ reply ] 
![{\displaystyle \Pr(A)\left[{\frac {\Pr(B|A)}{\Pr(A|B)}}-(\Pr(B|A)-\Pr(B|A^{'}))\right]=\Pr(B|A^{'})\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bba2413574be0ad52793f6b71d3d5b8f10b1a92e)
