[Jordan] said: P(O) = P(B) * P(O|B) Bayes’s theorem: P(O) P(B|O) = P(B) P(O|B)
[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.
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.