chapter 1

Chapter 2: Probability24. Since A is contained in B, then B can be written as the union of A and(B ∩ A′), two mutually exclusive events. (See diagram).From Axiom 3, P[A ∪ (B ∩ A′)] = P(A) + P(B ∩ A′). Substituting P(B),P(B) = P(A) + P(B ∩ A′) or P(B) - P(A) = P(B ∩ A′) . From Axiom 1,P(B ∩ A′) ≥ 0, so P(B) ≥ P(A) or P(A) ≤ P(B). For general events A and B, P(A ∩ B) ≤ P(A),and P(A ∪ B) ≥ P(A).25. P(A ∩ B) = P(A) + P(B) - P(A∪B) = .65P(A ∩ C) = .55, P(B ∩ C) = .60P(A ∩ B ∩ C) = P(A ∪ B ∪ C) – P(A) – P(B) – P(C)+ P(A ∩ B) + P(A ∩ C) + P(B ∩ C)= .98 - .7 - .8 - .75 + .65 + .55 + .60= .53a. P(A ∪ B ∪ C) = .98, as given.b. P(none selected) = 1 - P(A ∪ B ∪ C) = 1 - .98 = .02c. P(only automatic transmission selected) = .03 from the Venn Diagramd. P(exactly one of the three) = .03 + .08 + .13 = .2459