ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Three pairwise-independent events that are not mutually independent

Example

On the uniform space $Ω = {00, 01, 10, 11}$ , let $A$ be the event that the first bit is $0$ , let $B$ be the event that the second bit is $0$ , and let $C$ be the event that the two bits agree. Then $A, B, C$ are pairwise independent but not mutually independent.

Facts & Assumptions

Given: The uniform four-point space and events $A, B, C$ in the Example.

[L1]

In a uniform finite space, probability is cardinality divided by the outcome count (The uniform probability space on a nonempty finite set).

[L2]

Pairwise independence requires each pairwise intersection to have product probability, while mutual independence also requires the triple identity (Independent events, pairwise independence, and mutual independence of a finite family).

Verification

technique · direct

1.1

Each of $A, B, C$ has two outcomes and probability $1 / 2$ . Moreover $A \cap B = A \cap C = B \cap C = {00}$ , so every pairwise intersection has probability $1 / 4$ .

2.1

Therefore every pair satisfies $P (E \cap F) = 1 / 4 = (1 / 2) (1 / 2)$ .

step 1.1L2algebra

2.2

The triple intersection is ${00}$ and has probability $1 / 4$ , whereas the product of the three probabilities is $1 / 8$ .

step 1.1L1algebra

3.1

Thus the events are pairwise independent but not mutually independent.

step 2.1step 2.2L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 20 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Matousek and J. Vondrak, The Probabilistic Method, Section 1.1 (standard reference, not scraped)