Wikipedia:Reference desk/Archives/Mathematics/2006 November 28

= November 28 =

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. --Lambiam Talk  23:32, 28 November 2006 (UTC)