Theorem of total probability

Let {E₁, E₂, …, Eₙ} be a partition of S with every P(Eⱼ) non-zero, and let A be any event. Then P(A) = Σ P(Eⱼ)P(A|Eⱼ), summed over j = 1 to n: one product P(case) × P(A | case) for each part.

Why: the partition cuts A into the pieces A ∩ E₁, …, A ∩ Eₙ. The pieces are disjoint, so their probabilities add, and the multiplication rule gives each one as P(Eᵢ)P(A|Eᵢ).

On a tree diagram: multiply along each branch that ends in A, then add the branch products.

P(A) is a weighted average of the conditional probabilities P(A|Eᵢ), with the P(Eᵢ) as weights. It therefore lies between the smallest and the largest of them.

A construction job: P(strike) = 0.65, P(on time | no strike) = 0.80, P(on time | strike) = 0.32. With the partition {strike, no strike}, P(on time) = 0.65 × 0.32 + 0.35 × 0.80 = 0.208 + 0.28 = 0.488.

The cases must form a partition. If they overlap or leave something out, the sum is not P(A).