Why are the values in the classroom example described as illustrative inputs?

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 model illustrates the difference by showing that $P(+\mid D)$ (sensitivity) is fixed at $95\%$, while $P(D\mid +)$ (positive predictive value) is calculated to be only $~16.1\%$. This demonstrates that the probability of a positive test given disease is not the same as the probability of disease given a positive test, especially when prevalence is low.

Conditions: Sensitivity is $95\%$.; Calculated $P(D\mid +)$ is $~16.1\%$.; Prevalence is $1\%$.

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.

Using the assumed values from the classroom example where $A$ denotes alcohol dependence and $B$ denotes being a man, you divide the probability of the intersection $P(A\cap B)$ by the probability of the conditioning event $P(B)$. With $P(A\cap B)=0.0225$ and $P(B)=0.5$, the calculation is $0.0225/0.5=0.045$, or $4.5\%$.

Conditions: $P(A\cap B) = 0.0225$.; $P(B) = 0.5$.; These are illustrative inputs, not verified population statistics.

False positives are more numerous because the healthy population is much larger than the diseased population. Even though the false positive rate is low ($5\%$), applying it to a large base (1,980 people) yields a higher absolute count (99) than applying the high sensitivity rate ($95\%$) to a small base (20 people), which yields 19.

Conditions: Prevalence is $1\%$ (20 diseased, 1,980 healthy).; False positive rate is $5\%$.; Sensitivity is $95\%$.