Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.
Bayes' Theorem .
Pr
(
A
|
B
)
=
Pr
(
B
|
A
)
Pr
(
A
)
Pr
(
B
|
A
)
Pr
(
A
)
+
Pr
(
B
|
A
′
)
Pr
(
A
′
)
{\displaystyle \Pr(A|B)={\frac {\Pr(B|A)\,\Pr(A)}{\Pr(B|A)\Pr(A)+\Pr(B|A^{'})\Pr(A^{'})}}\!}
Assume that
Pr
(
A
′
)
=
1
−
Pr
(
A
)
{\displaystyle \Pr(A^{'})=1-\Pr(A)}
Pr
(
A
|
B
)
=
Pr
(
B
|
A
)
Pr
(
A
)
Pr
(
B
|
A
)
Pr
(
A
)
+
Pr
(
B
|
A
′
)
−
Pr
(
B
|
A
′
)
Pr
(
A
)
{\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 }
Pr
(
A
|
B
)
=
Pr
(
B
|
A
)
Pr
(
A
)
(
Pr
(
B
|
A
)
−
Pr
(
B
|
A
′
)
)
Pr
(
A
)
+
Pr
(
B
|
A
′
)
{\displaystyle \Pr(A|B)={\frac {\Pr(B|A)\,\Pr(A)}{(\Pr(B|A)-\Pr(B|A^{'}))\Pr(A)+\Pr(B|A^{'})}}\!}
↓
{\displaystyle \downarrow }
Pr
(
A
|
B
)
Pr
(
B
|
A
)
Pr
(
A
)
=
1
(
Pr
(
B
|
A
)
−
Pr
(
B
|
A
′
)
)
Pr
(
A
)
+
Pr
(
B
|
A
′
)
{\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 }
Pr
(
B
|
A
)
Pr
(
A
)
Pr
(
A
|
B
)
=
(
Pr
(
B
|
A
)
−
Pr
(
B
|
A
′
)
)
Pr
(
A
)
+
Pr
(
B
|
A
′
)
{\displaystyle {\frac {\Pr(B|A)\,\Pr(A)}{\Pr(A|B)}}=(\Pr(B|A)-\Pr(B|A^{'}))\Pr(A)+\Pr(B|A^{'})\!}
↓
{\displaystyle \downarrow }
Pr
(
A
)
[
Pr
(
B
|
A
)
Pr
(
A
|
B
)
−
(
Pr
(
B
|
A
)
−
Pr
(
B
|
A
′
)
)
]
=
Pr
(
B
|
A
′
)
{\displaystyle \Pr(A)\left[{\frac {\Pr(B|A)}{\Pr(A|B)}}-(\Pr(B|A)-\Pr(B|A^{'}))\right]=\Pr(B|A^{'})\!}
↓
{\displaystyle \downarrow }
Pr
(
A
)
=
Pr
(
B
|
A
′
)
Pr
(
B
|
A
)
Pr
(
A
|
B
)
−
(
Pr
(
B
|
A
)
−
Pr
(
B
|
A
′
)
)
{\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
Pr
(
B
|
A
′
)
=
0
{\displaystyle \Pr(B|A^{'})=0}
Thus
Pr
(
A
)
=
0
Pr
(
B
|
A
)
Pr
(
A
|
B
)
−
(
Pr
(
B
|
A
)
−
Pr
(
B
|
A
′
)
)
{\displaystyle \Pr(A)={\frac {0}{{\frac {\Pr(B|A)}{\Pr(A|B)}}-(\Pr(B|A)-\Pr(B|A^{'}))}}\!}
↓
{\displaystyle \downarrow }
Pr
(
A
)
=
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 [ 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)
