Why can the probability that the second draw is defective given the first draw was defective be directly written as $3/19$?

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).

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.

The event 'A and B' is logically identical to 'B and A', so any valid decomposition of its probability must agree. This symmetry forces the equality of the two product formulas: $P(A)P$(B|A) = $P(B)P$(A|B).

Conditions: The argument uses commutativity of logical conjunction for events.; Both A and B have positive probability for the ordinary conditionals used here.

In the square diagrams, the braces represent proportions of areas. $P(A)$ is the fraction of the total sample space where event A occurs (a vertical strip).

Conditions: The visualization assumes probabilities can be represented by relative areas.; Both A and B have positive probability for the ordinary conditionals used here.

Substitute the given numerical values into the rearranged formula P(B|A) = $P(B)P$(A|B)/P(A). For example, if $P(B) = 1/21$, P(A|B) = $4/10$, and $P(A) = 24/210$, the calculation is ($1/21 * 4/10$) / ($24/210$).

Conditions: The formula used is the rearranged Bayes identity for P(B|A).; The given probabilities must be consistent with the formula's assumptions (e.g., $P(A) > 0$).

The book and magnifying glass icons represent specific events in a probability space, serving as labels for the variables in Bayes' theorem. The formula shown is P(Book|Magnifier) = P(Book)P(Magnifier|Book)/P(Magnifier).

Conditions: The icons are used as event labels in the displayed formula.; No fixed letter-to-icon assignment is imposed across displays in the video.