Chapter 4 Solutions: Section 4.1: 1.) 2.) 3.) 481 » 0.184 = 18.4% 2620 371 P(brown) = » 0.142 = 14.2% 2620 483 P(green) = » 0.184 = 18.4% 2620 544 P(orange) = » 0.208 = 20.8% 2620 372 P(red) = » 0.142 = 14.2% 2620 369 P(yellow) = » 0.141 = 14.1% 2620 Probability of choosing a particular car: (total number of people = 300) P(blue) = P(reason) Safety 84 84 = 300 0.28 = 28% Reliability Cost 62 46 62 46 » » 300 300 0.21 0.15 = 21% = 15% Performance 34 34 » 300 0.11 = 11% Comfort 47 47 » 300 0.16 = 16% Section 4.2: 483 + 372 855 = » 0.326 = 32.6% 2620 2620 481+ 372 + 369 1222 b.) P(blue, red, or yellow) = = » 0.466 = 46.6% 2620 2620 371 2249 c.) P(not brown) = 1- P ( brown ) = 1= » 0.858 = 85.8% 2620 2620 483 2137 d.) P(not green) = 1= » 0.816 = 81.6% 2620 2620 1.) a.) P(green or red) = 2.) 3.) a.) SS = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} 3 b.) P(2 heads) = P({HHT, HTH, THH}) = = 0.375 = 37.5% 8 4 c.) P(at least 2 heads) = P({HHH, HHT, HTH, THH}) = = 0.50 = 50% 8 Looks 27 27 = 300 0.09 = 9% d.) P(odd number of heads) = P({HHT, HTH, THH, TTT}) = 4 = 0.50 = 50% 8 4.) 2 = 0.25 = 25% 8 f.) P(two heads or two tails) = P({HHT, HTH, THH, TTH, THT, HTT}) = 6 = 0.75 = 75% 8 g.) P(not an odd number of heads) = 4 4 1- P ( odd number of heads ) = 1- = = 0.50 = 50% 8 8 - 5.) a.) e.) P(all heads or all tails) = P({HHH, TTT}) = SS = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 2 b.) P(sum of 3) = P({(1,2), (2,1)}) = » 0.056 = 5.6% 36 c.) P(1st die a 4) = P({(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)}) = 6 » 0.167 = 16.7% 36 5 » 0.139 = 13.9% 36 e.) P(sum of 3 or sum of 8) = P({(1,2), (2,1), (6,2), (5,3), (4,4), (3,5), (2,6)}) 7 = » 0.194 = 19.4% 36 or P ( sum of 3 or sum of 8 ) = P ( sum of 3) + P ( sum or 8 ) d.) P(sum of 8) = P({(6,2), (5,3), (4,4), (3,5), (2,6)}) = 2 5 7 + = » 0.194 = 19.4% 36 36 36 f.) P(sum of 3 or 1st die a 4) = P({(1,2), (2,1), (4,1), (4,2), (4,3), (4,4), (4,5), 8 (4,6)}) = » 0.222 = 22.2% 36 or P ( sum of 3 or 1st die a 4 ) = P ( sum of 3) + P (1st die a 4 ) = 2 6 8 + = » 0.222 = 22.2% 36 36 36 g.) P(sum of 8 or 1st die a 4) = P({(6,2), (5,3), (4,4), (3,5), (2,6), (4,1), (4,2), (4,3), 10 (4,5), (4,6)}) = » 0.278 = 27.8% 36 or = P ( sum of 8 or 1st die a 4 ) = P ( sum of 8 ) + P (1st die a 4 ) - P ( sum of 8 or 1st die a 4 ) = 5 6 1 10 + = » 0.278 = 27.8% 36 36 36 36 h.) P(not getting a sum of 8) = 1- P ( sum of 8 ) = 16.) - 7.) a.) P(red ball) = 8.) 5 31 = » 0.861 = 86.1% 36 36 5 = 0.625 = 62.5% 8 3 b.) P(blue ball) = = 0.375 = 37.5% 8 58 5 8 5 c) odds of red ball = = * = = 5 to 3 38 8 3 3 38 3 8 3 d.) odds of blue ball = = * = = 3 to 5 58 8 5 5 - Section 4.3: 1.) 2.) 3.) 4.) 5.) 1.) 2.) 3.) 4.) 5.) Owning a refrigerator and owning a car are independent events, since knowing that a person owns one of these items does not change the probability of the person owning the other item. (One could argue that world-wide that these are dependent events. This is because in some parts of the world if a person can afford one, then there is a different probability that they could afford to own the other too.) Passing your statistics class and passing your biology class are dependent events, since if you put more work into one class to pass it then you have less time for the other. So the probability of passing could in fact change the probability of the other. (One could argue that these events are independent, since passing one class does not change the probability that you will pass the other in content knowledge.) SS = {2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AS, 2C, 3C, 4C, 5C, 6C, 7C, 8C, 9C, 10C, JC, QC, KC, AC, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, 10D, JD, QD, KD, AD, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH, AH} a.) - Number of meals = 2*3*5 = 30 meals Number of phone numbers = 8*10*10*10*10*10*10 = 8,000,000 phone numbers Number of CA license plates = 10*26*26*26*10*10*10 = 175,760,000 plates 10 P6 = 151,200 6.) 7.) 8.) 9.) 10.) 11.) 20 C4 = 4845 Permutation since order matters. 25 P9 = 7.41´1011 Combination since order doesn’t matter. 20 C7 = 77,520