Bayes' Theorem of Probability With Tree Diagrams & Venn Diagrams

The denominator $P(B)$ in Bayes' theorem represents the total probability of the evidence (event B). In the context of a screening test, a positive result can arise from two mutually exclusive scenarios: a true positive (person has the disease and tests positive) or a false positive (person does not have the disease but tests positive).

Conditions: The evidence event B can occur via multiple mutually exclusive paths.; The law of total probability applies.

Bayes' theorem is derived by expressing the same intersection $P(A \cap B)$ in two different ways using the definition of conditional probability. First, $P(A \mid B) = P(A \cap B) / P(B)$ implies $P(A \cap B) = P(A \mid B)P(B)$.

Conditions: Events A and B have positive probability ($P(A) > 0, P(B) > 0$).; The definition of conditional probability applies.

The positive predictive value, or the probability of having the disease given a positive test result ($P(D \mid +)$), is calculated by dividing the probability of a true positive by the total probability of a positive result. The total probability of a positive result includes both true positives and false positives.

Conditions: Prevalence $P(D) = 0.12$.; Sensitivity $P(+ \mid D) = 0.95$.; False positive rate $P(+ \mid D^c) = 0.06$.; The events are mutually exclusive and exhaustive for the disease status.

In a sample space of nine equally likely outcomes, let event A contain 5 elements and event B contain 6 elements, with their intersection containing 2 elements. The conditional probability $P(A \mid B)$ is calculated directly as the size of the intersection divided by the size of B: $2/6 = 1/3$.

Conditions: The sample space has nine equally likely outcomes.; Event A has 5 elements, Event B has 6 elements, and $A \cap B$ has 2 elements.

The denominator $P(B)$ is the marginal probability of the evidence. It acts as a normalizing constant that ensures the posterior probability $P(A \mid B)$ is a valid probability (i.e., between 0 and 1).

Conditions: $P(B) > 0$.; The events A and B are defined on the same sample space.

The result $P(A \mid B) = 1/3$ means that if event B has occurred, there is a one-in-three chance that event A has also occurred. In the specific example with nine equally likely outcomes, B contains 6 outcomes.

Conditions: The sample space consists of nine equally likely outcomes.; Event B has occurred.; We are evaluating the likelihood of A within the context of B.