Introduction to Probability · Problem 4.1.5

Conditional Probability — Grinstead & Snell

Mathematics — Probability · conditional-probability-independence-events

Problem

A coin is tossed three times. Consider the following events AA: Heads on the first toss. BB: Tails on the second. CC: Heads on the third toss. DD: All three outcomes the same (HHH or TTT). EE: Exactly one head turns up. (a) Which of the following pairs of these events are independent? ((1)) A,BA, B ((2)) A,DA, D ((3)) A,EA, E ((4)) D,ED, E (b) Which of the following triples of these events are independent? ((1)) A,B,CA, B, C ((2)) A,B,DA, B, D ((3)) C,D,EC, D, E
Intuitive explanation

Two events are independent when knowing one happened tells you nothing about the other—mathematically, P(X∩Y)=P(X)⋅P(Y)P(X \cap Y) = P(X) \cdot P(Y). The key is recognizing which events are defined by different tosses (automatically independent) versus which share information (potentially dependent). For triples, all three pairs must be independent and the triple condition P(X∩Y∩Z)=P(X)⋅P(Y)⋅P(Z)P(X \cap Y \cap Z) = P(X) \cdot P(Y) \cdot P(Z) must hold—a stricter requirement than just pairwise independence.

Intuitive explanation

Two events are independent when knowing one happened tells you nothing about the other—mathematically, P(X∩Y)=P(X)⋅P(Y)P(X \cap Y) = P(X) \cdot P(Y). The key is recognizing which events are defined by different tosses (automatically independent) versus which share information (potentially dependent). For triples, all three pairs must be independent and the triple condition P(X∩Y∩Z)=P(X)⋅P(Y)⋅P(Z)P(X \cap Y \cap Z) = P(X) \cdot P(Y) \cdot P(Z) must hold—a stricter requirement than just pairwise independence.

Related theory

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