Total probability theorem

Once the tree is drawn, the marginal probability of the event at the second level (for example $B$) is obtained in a geometrically transparent way: you add up the probabilities of the paths that end in $B$. No extra formula: it is all already on the drawing.

Theorem — Total probability, classical formulation

Let $\{A_1, A_2, \ldots, A_n\}$ be a partition of the sample space (pairwise mutually exclusive events whose union covers everything). Then for every event $B$: $\boxed{P(B) = \sum_{i=1}^n P(A_i)\cdot P(B\mid A_i).}$

Proof — Visual, via the paths of the tree

We build the tree with first level $\{A_1,\ldots,A_n\}$ and with $B,\overline{B}$ at the second level under each $A_i$. The leaves "$\ldots\cap B$" are pairwise disjoint (different outcomes of the first experiment). By additivity over mutually exclusive events: $P(B) = P\!\Bigl(\bigcup_{i=1}^n (A_i\cap B)\Bigr) = \sum_{i=1}^n P(A_i\cap B) = \sum_{i=1}^n P(A_i)\cdot P(B\mid A_i),$ where in the last step the product formula is applied to each path. ∎

Example — Diagnostic test, $P(T^+)$ via the paths

We take up again the tree of the rare disease. The paths that end in $T^+$ are two, highlighted in red:

The two red paths both end in $T^+$: the total probability of testing positive is their sum.

Adding the two red paths: $P(T^+) = \underbrace{0,001\cdot 0,99}_{\text{percorso }M\to T^+} + \underbrace{0,999\cdot 0,02}_{\text{percorso }\overline{M}\to T^+} \approx 0,021.$

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Topics: Probability
Concepts: Tree diagram · Partition · Conditional probability · Total probability
Methods: Total probability theorem
Skills: Calculating probability · Proving