Comment on Open Thread: April 2010

Fri, 30 Apr 2010 20:48:04 GMT

[Jordan] said: P(O) = P(B) * P(O|B)

Bayes's theorem: P(O) P(B|O) = P(B) P(O|B)

I agree that Jordan's equation needs to be adjusted (corrected), but I humbly suggest that in this context, it is better to adjust it to the product rule:

P(O and B) = P(B) * P(O|B).

ADDED. Yeah, minor point.

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