AP Statistics Answers to Chapter 6 Homework

AP Statistics Answers to Chapter 6 Homework 4(a) 0 (b) 1 (c) 0.01 (d) 0.6 6 a) We expect probability 1/2 (for the first flip, or any flip of the coin). b) The theoretical probability that the first head occurs on an odd-numbered toss of a fair coin is 1 ⎛ 1⎞ 3 ⎛ 1⎞ 5 2 + ⎜ ⎟ + ⎜ ⎟ + ... = . Most answers should be between about 0.47 and 0.87. ⎝ ⎠ ⎝ ⎠ 2 2 2 3 10 a) With n=20, nearly all answers will be 0.4 or greater. With n=80, nearly all answers will be between 0.58 and 0.88. With n=320, nearly all answers will be between 0.66 and 0.80. 14(a) If two coins are tossed, then by the multiplication principle, there are (2)(2)=4 possible outcomes. The outcomes are illustrated in the tree diagram: H H HH HT TH TT T H T T The sample space is {HH, HT, TH, TT} (b) If three coins are tossed, then there are (2)(2)(2)=8 possible outcomes. The outcomes are illustrated in the tree diagram: H T H T H T H T H T H T H T The sample space is {HHH, HTT, HTH, HTT, THH, THT, TTH, TTT} (c) If four coins are tossed, then by the multiplication principle, there are (2)(2)(2)(2)=16 possible outcomes. The sample space is {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}. 20(a) The sum of the given probabilities is 0.9, so P(blue) =0.1. (b) The sum of the given probabilities is 0.7, so P(blue) =0.3. (c) P(plain M&M is red, yellow, or orange)= 0.2 + 0.2 + 0.1 = 0.5. P(peanut M&M is red, yellow, or orange)= 0.1 + 0.2 + 0.1 = 0.4. 26 a) P(D) = P(1,2,or3) = .301+.176+.125 = .602 b) P(B∪D) = P(B)+P(D) = .602+.222 = .824 HHH HTT HTH HTT THH THT TTH TTT c c) P(D ) = 1-P(D) = 1-.602 = .398 d) P(C∩D) = P(1 or 3) = .301+.125=.426 e) P(B∩) = P(7 or 9) = .058+.046 = .104 28. (1 − 0.05)12 = (0.95)12 = 0.5404 36 a) Sum of given probabilities = .705, so P(car has some other color)=1-.705=.295. b) P(silver or white) = P(silver)+P(white) = .176+.172= .348. 2 c) Assuming the vehicle choices are independent, P(both silver or white) = (.348) = .121 1 1 1 ⋅ = . 2 2 4 1 1 1 P(both of two children are albino) = ⋅ = . 4 4 16 44. P(first child is albino) = ⎛ 1⎞ 2 9 P(neither child is albino) = ⎜1 - ⎟ = . ⎝ 4⎠ 16 46. P(A or B) = P(A) + P(B) – P(A and B) = .125+.237-.077=.285. 47. Ac & Bc .692 S A & Bc .054 54 a) 18669/103870=.18 b) 8270/18669=.443 c) 8270/103870=.08. 56. 0.1472 60. a) 0.25 b) .2 64a) A&B .080 Ac & B .174 b) P(test positive) = (.01)(.9985)+(.99)(.006)= .016 c) P(antibody present | test positive) = (.01)(.9985)/(.016)= .624 70 a) P(C) = .20, P(A) = .1, P(A|C) = .05 b) P(A and C) = P(C)×P(A|C) = (.20)(.05) = .01 73. tree diagram only 76. 1/4 82. 0.4 87. tree diagram only