CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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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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Unions of overlapping independent events need not remain independent

Statement refuted

Whenever $A$ and $B$ are independent events, and $C$ overlaps both of them, the unions $A \cup C$ and $B \cup C$ are also independent.

Facts & Assumptions

Given: The uniform four-point space $Ω = {00, 01, 10, 11}$ and the events $A : = {00, 01}, B : = {00, 10}, C : = {00, 11} .$

[L1]

In a uniform finite space, event probability is cardinality divided by the total number of outcomes. (The uniform probability space on a nonempty finite set)

[L2]

Two events are independent exactly when $P (E \cap F) = P (E) P (F)$ . (Independent events, pairwise independence, and mutual independence of a finite family)

Counterexample

technique · constructive

1.1

By [L1], the events $A$ and $B$ have probability $1 / 2$ , and $A \cap B = {00}$ has probability $1 / 4$ . Hence $A$ and $B$ are independent by [L2]. Also $C$ overlaps both of them because $00 \in A \cap C$ and $00 \in B \cap C$ .

L1L2construct

2.1

The unions are $A \cup C = {00, 01, 11}, B \cup C = {00, 10, 11},$ so each has probability $3 / 4$ , while $(A \cup C) \cap (B \cup C) = {00, 11}$ has probability $1 / 2$ .

step 1.1L1

3.1

Since $P ((A \cup C) \cap (B \cup C)) = \frac{1}{2} \neq \frac{9}{16} = \frac{3}{4} \cdot \frac{3}{4},$ [L2] shows that $A \cup C$ and $B \cup C$ are not independent. This refutes the statement.

step 2.1L2discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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