Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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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.
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A loaded die as a nonuniform finite probability space

Example

On Ω={1,2,3,4,5,6} assign weights w(1)=0,w(2)=110,w(3)=w(4)=15,w(5)=w(6)=14. This is a loaded die with a possible zero-weight outcome. Let A={4,5,6} and B={2,4,6}.

Facts & Assumptions

Given: The weights and events in the Example.

[L1]

Finite probability spaces allow nonnegative outcome weights summing to 1 (Finite probability spaces, outcome weights, events, and event probabilities).

[L2]
[L3]

P(AB)=P(AB)/P(B) when P(B)>0 (Conditional probability P(AB) for P(B)>0).

[L4]

The multiplication rule is P(AB)=P(AB)P(B) (The multiplication rule and finite chain rule for conditional probability).

Verification

technique · direct
1.1

The six weights sum to 1, so [L1] gives a finite probability space; {1} is nonempty but has probability zero.

L1algebra
1.2

Direct summation gives P(A)=7/10, P(B)=11/20, and P(AB)=9/20.

L1L2algebra
2.1

Since P(B)>0, [L3] gives P(AB)=(9/20)/(11/20)=9/11.

step 1.2L3algebra
3.1

Finally (9/11)(11/20)=9/20=P(AB), verifying [L4] in this nonuniform space.

step 1.2step 2.1L4algebra

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 39 results over 12 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