BU255 Practice Questions

4.2 Given X = {1, 3, 5, 7, 8, 9}, Y = {2, 4, 7, 9}, and Z = {1, 2, 3, 4, 7}, solve the following. a. X ∪ Z = (1,2,3,4,5,7,8,9) b. X ∩ Y = (7,9) c. X ∩ Z = (1,3,7) d. X ∪ Y ∪ Z = (1,2,3,4,5,7,8,9) e. X ∩ Y ∩ Z = (7) f. (X ∪ Y) ∩ Z = (1,2,3,4,7) g. (Y ∩ Z) ∪ (X ∩ Y) = (2,4,7,9) h. X or Y = (1,2,3,4,5,7,8,9) i. Y and Z = (2,4,7) 4.3 If a population consists of the positive even numbers through 30 and if A = {2, 6, 12, 24}, what is Aʹ? A’ = 4,8,10,14,16,18,20,22,26,28,30 4.10 Use the values in the matrix to solve the equations given E F A 0.10 0.03 0.13 B 0.04 0.12 0.16 C 0.27 0.06 0.33 D 0.31 0.07 0.38 0.72 0.28 a. P(A ∪ F) = P(A) + P(F) - P(A∩F) = ( 0.13 + 0.28) - 0.03 = 0.38 b. P(E ∪ B) = P(E) + P(B) - P(E∩B) = (0.72 + 0.16) - 0.04 = 0.84 c. P(B ∪ C) = P(B) + P(C) = 0.16 + 0.33 = 0.49 d. P(E ∪ F) = P(E) + P (F) = 0.72 + 0.28 = 1.00 4.13 A survey estimated that 67% of all households with television subscribe to a digital streaming service. Seventy-four percent of all households with television have two or more TV sets. Suppose 55% of all households with television subscribe to a digital streaming service and have two or more TV sets. A household with television is randomly selected. P(A) = 0.67,,,, P(B) = 0.74,,,,, P(A∩B) = 0.55,,,,, a. What is the probability that the household subscribes to a streaming service or has two or more TV sets? P(A ∪ B) = P(A) + P(B) - P(A∩B) = (0.67 + 0.74) - 0.55 = 0.86 b. What is the probability that the household subscribes to a streaming service or has two or more TV sets but not both?