Independence Proof | Probability Visualizer

Problem 3.1 • If A and B are independent, then A and Bᶜ are also independent

A and B are independent events.

P(A \cap B) = P(A) \cdot P(B)

Any event A can be split using B and Bᶜ:

A = (A \cap B) \cup (A \cap Bᶜ)

These are disjoint, so probabilities add.

Using the addition rule for disjoint events:

P(A) = P(A \cap B) + P(A \cap Bᶜ)

Replace P(A ∩ B) using the independence condition:

P(A) = P(A)\cdotP(B) + P(A \cap Bᶜ)

Rearranging the equation:

P(A \cap Bᶜ) = P(A) - P(A)\cdotP(B)

Factor P(A) from the right side:

P(A \cap Bᶜ) = P(A)\cdot(1 - P(B))

Since P(Bᶜ) = 1 − P(B):

P(A \cap Bᶜ) = P(A)\cdotP(Bᶜ) ✓

This is the definition of independence! Q.E.D.

The Venn diagram shows how A is partitioned into (A∩B) and (A∩Bᶜ).

Independence is symmetric across complements. If knowing B doesn't help predict A, then knowing "not B" also doesn't help predict A.

• A ⊥ B → Aᶜ ⊥ B
• A ⊥ B → Aᶜ ⊥ Bᶜ

Independence Proof: A⊥B → A⊥Bᶜ