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Incorrect comments on Help me teach Bayes' - Less Wrong Discussion

2 Post author: tomme 11 April 2012 05:56PM

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Comment author: Incorrect 11 April 2012 06:27:10PM 3 points [-] 
P(A and B) = P(B and A)
P(A and B) = P(A) * P(B|A)

=>

P(A and B) = P(A) * P(B|A)
P(B and A) = P(A) * P(B|A)
P(B) * P(A|B) = P(A) * P(B|A)
P(A|B) = P(A) * P(B|A) / P(B)
Comment author: XiXiDu 11 April 2012 07:13:37PM *  0 points [-]

P(B and A) = P(A) * P(B|A)

To make it clearer, shouldn't this step be = P(B) * P(A|B)?

Also, are middle school pupil in the U.S. familiar with the notation? Maybe one should state it in English instead?

The probability of A if B is known to occur is equal to the probability of A times the probability of B if A is known to occur divided by the probability of B.

ETA

P(A and B) = P(A) * P(B|A)

The probability of A and B to occur at the same time is the same as the probability of A to occur alone times the probability of B to occur under the condition that A is known to occur.

Comment author: army1987 11 April 2012 07:06:40PM -1 points [-]

Spot the typo.

Comment author: Incorrect 11 April 2012 07:28:51PM 0 points [-]

Don't see one. Could you please tell me where?

Comment author: army1987 11 April 2012 07:34:05PM 0 points [-]

I think the second line after the => was supposed to be P(B and A) = P(B) * P(A|B).

Comment author: Incorrect 11 April 2012 07:52:55PM 1 point [-]

No, it's a substitution on the left hand side. I substituted P(A and B) for P(B and A)

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