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7 suppose ph2h125 and pt2t15 where we are either

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Unformatted text preview: union law (for non ­disjoint sets) and the fact that P(X)<=1 for any set X that P(AB)>= P(A)+P(B)  ­1. 6. Prove our theorem (from definitions of conditional probability) that if P(AB) = P(A)P(B) then P(A|B)= P(A|Bc) where P(B)and P(Bc) are not 0. 7. Suppose P(H2|H1...
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