How can the same equality be rearranged into both forms of Bayes' theorem?

Starting from the equality

P(A)P

(B|A) =

P(B)P

(A|B), you can solve for either conditional probability by dividing by the corresponding marginal probability. Dividing both sides by

P(B)

isolates P(A|B), giving P(A|B) =

P(A)P

(B|A)/P(B). Dividing both sides by

P(A)

isolates P(B|A), giving P(B|A) =

P(B)P

(A|B)/P(A).

Conditions

Both A and B have positive probability for the ordinary conditionals used here.
The algebraic rearrangement requires the denominators $P(B)$ and $P(A)$ to be nonzero.

Write down the equality of the two joint-probability decompositions: $P(A)P$ (B|A) = $P(B)P$ (A|B).
To find P(A|B), divide both sides of the equation by $P(B)$ .
Simplify to get P(A|B) = $P(A)P$ (B|A)/P(B).
To find P(B|A), divide both sides of the original equation by $P(A)$ .
Simplify to get P(B|A) = $P(B)P$ (A|B)/P(A).

Overlooking the need for nonzero denominators in the rearranged formulas; the clip displays the divisions but does not explicitly state this condition aloud.