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AP Statistics – Chapter 7 Exercises Solutions: 7.11: Rolling Two Dice a. The 36 possibilities of “up faces” are: (1, 1), (1, 2), … , (1, 6) (2, 1), (2, 2), … , (2, 6) * * * * * * * * * (6, 1), (6, 2), … , (6,6) b. Each pair has a probability of 1/36 c. Let X = sum of “up faces” [ex: X = 2 for the pair (1, 1) and X = 12 for the pair (6, 6)] X P(X) 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 8 5/36 9 4/36 10 3/36 11 2/36 12 1/36 d. P( X = 7 or X = 11) = 6/36 + 2/36 = 8/36 = 2/9 e. P(X ≠ 7) = 1 – P(X = 7) = 1 – 6/36 = 30/36 = 5/6 7.14: Car Ownership a. All probabilities are between 0 and 1 and the sum of all probabilities equals 1. b. The event {X ≥ 1} means that the household owns at least one car. P(X ≥ 1) = 1 – P(X = 0) = 0.91 c. P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) = 0.20 20% of households own more cars than a two-car garage can hold. 7.16: Student Fees a. Let S = {student supports funding} Let O = {student opposes funding} P(SSO) = .6 * .6 * .4 = 0.144 b. 8 possibilities: SSS, SSO, SOS, SOO, OOO, OOS, OSO, OSS P(SSS) = .216 P(SSO) = P(SOS) = P(OSS) = .144 P(SOO) = P(OSO) = P(OOS) = .096 P(OOO) = .064 c. Probability Distribution for X = number of students opposed X 0 1 P(X) .216 .432 d. {X ≥ 2} or {X > 1} P(X > 1) = .352 2 .288 3 .064