6.9. Chain Rules and Total Probability

If \(\{ A_i \}\) is a partition of \(S\) and \(B \subset S\) is an event, then \(A_i \cap A_j = \emptyset\) also implies that

\[ B_i \cap B_j = (B \cap A_i) \cap (B \cap A_j) = B \cap (A_i \cap A_j) = B \cap \emptyset =\emptyset, \]

and

\[ \bigcup_i B_i = \bigcup_i \left(A_i \cap B\right ) = B \cap \left( \bigcup_i A_i \right) = B \cap S = B. \]

These two properties imply that \(B_0, B_1, \ldots, B_{n-1}\) are a partition for \(B\).

Quiz 1: Using Chain Rules

Quiz 2: Chain Rules and the Magician’s Coin

Proofs Omitted in Text

Proof that all events \(B_j, ~B_k,~j \ne k\) are mutually exclusive $\( B_j \cap B_k = (B \cap A_j) \cap (B \cap A_k) = B \cap (A_j \cap A_k) = B \cap \emptyset =\emptyset. \)$

Proof that the union of all events \(B_j\) is equal to \(B\) $\( \bigcup_j B_j = \bigcup_j \left(A_j \cap B\right ) = B \cap \left( \bigcup_j A_j \right) = B \cap S = B. \)$

Quiz 3: Total Probability and the Magician’s Coin

Self-Assessment

\(~~~~\mbox{ }\)

Terminology Review