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[edit] November 28

[edit] Bayes Theorem Equation

Looking at the equation for Bayes' Theorem.

\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)

\Pr(A|B) = \frac{\Pr(B | A)\, \Pr(A)}{\Pr(B|A)\Pr(A) + \Pr(B|A^') - \Pr(B|A^')\Pr(A)} \!
\downarrow
\Pr(A|B) = \frac{\Pr(B | A)\, \Pr(A)}{(\Pr(B|A) - \Pr(B|A^')) \Pr(A) + \Pr(B|A^') } \!
\downarrow
\frac{\Pr(A|B)}{\Pr(B | A)\, \Pr(A)} = \frac{1}{(\Pr(B|A) - \Pr(B|A^')) \Pr(A) + \Pr(B|A^') } \!
\downarrow
\frac{\Pr(B | A)\, \Pr(A)}{\Pr(A|B)} = (\Pr(B|A) - \Pr(B|A^')) \Pr(A) + \Pr(B|A^') \!
\downarrow
\Pr(A) \left [ \frac{\Pr(B | A)}{\Pr(A|B)} - (\Pr(B|A) - \Pr(B|A^')) \right ] = \Pr(B|A^') \!
\downarrow
\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 \, be "X is divisible by 2"

and

let B \, be "X is divisible by 4"

We have

\Pr(B|A^') = 0

Thus

\Pr(A) = \frac{0}{\frac{\Pr(B | A)}{\Pr(A|B)} - (\Pr(B|A) - \Pr(B|A^'))}\!
\downarrow
\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 \, be "George Bush is an American"
and
let B \, be "George Bush is the president of the United States"

202.168.50.40 22:59, 28 November 2006 (UTC)

Hint. At some point, rather late in your derivation, you take a step of the form 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.  --LambiamTalk 23:32, 28 November 2006 (UTC)