Introduction to Probability · Problem 4.1.33

Conditional Probability — Grinstead & Snell

Mathematics — Probability · mutual-independence

Problem

Let A1A_1, A2A_2, and A3A_3 be events, and let BiB_i represent either AiA_i or its complement A~i\tilde{A}_i. Then there are eight possible choices for the triple (B1,B2,B3)(B_1, B_2, B_3). Prove that the events A1A_1, A2A_2, A3A_3 are independent if and only if P(B1∩B2∩B3)=P(B1)P(B2)P(B3),P(B_1 \cap B_2 \cap B_3) = P(B_1)P(B_2)P(B_3) , for all eight of the possible choices for the triple (B1,B2,B3)(B_1, B_2, B_3).
Intuitive explanation

Independence of three events means the probability of any intersection factors into a product of individual probabilities—but this must hold not just for A1,A2,A3A_1, A_2, A_3 themselves, but for every combination of the events and their complements. The key insight is that if factorization holds for all eight cases (each BiB_i being either AiA_i or A~i\tilde{A}_i), you've verified the multiplicative property in every possible "corner" of the probability space, which is exactly what independence demands. Conversely, true independence guarantees this factorization everywhere by definition.

Intuitive explanation

Independence of three events means the probability of any intersection factors into a product of individual probabilities—but this must hold not just for A1,A2,A3A_1, A_2, A_3 themselves, but for every combination of the events and their complements. The key insight is that if factorization holds for all eight cases (each BiB_i being either AiA_i or A~i\tilde{A}_i), you've verified the multiplicative property in every possible "corner" of the probability space, which is exactly what independence demands. Conversely, true independence guarantees this factorization everywhere by definition.

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