[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.