13.2 Option (i) is better, because only one thing has to happen. In option (ii) both events must occur. 13.3 Option (ii) is better. Option (ii) includes the possibility of option (i), but there are lots of other ways to win (e.g., diamond, spade, club, heart; spade, club, diamond, heart; etc.). 13.4 The chances of the first card being an ace is 4/52. The second card being an ace, given that the first is an ace is 3/51. The third card being another ace is 2/50, and the fourth being the final ace is 1/49. That the fifth card is a king is 4/48. 4/52 * 3/51 * 2/50 * 1/49 * 4/48 = about 1/ 3,000,000 13.5 Color and number are independent. Knowing the color of a ticket tells you nothing about its number; likewise, knowing the number of the ticket doesn't tell you anything about its color. 13.6 a) True. b) True. See example 2 on p. 226. c) False. The two events are not independent (knowing the first card tells you something about the second card; e.g., that the first card is the ace of clubs tells you that the second card can't be the ace of clubs). The chance of c) is 1/52 * 1/51. 13.7 Both sequences are equally likely, as each flip of the coin is independent of the previous flips. (c). 13.8 a) On each roll, the chance of getting 3 or more spots on a six sided die is 4 in 6, or 4/6. So, the chance of getting 3 or more spots on four rolls in a row is (4/6)^4, or about 20%. b) There is a 2 in 6 chance that a roll of a die will show less than three spots, so the chance of having four consecutive rolls all with less than three spots is (2/6)^4, or about 1.2%. c) This question asks you for the opposite of a). So, the chance would be given by 100%-20% = 80%. 13.10 The second box (ii) is better, since you will get the 1 on 50% of the trials and the 5 on 50% of the trials. Over 100 trials, that would average out to around $300. 13.12 Your chances of having written down the first ball that comes out are 3 in 100, since you have three numbers out of the 100 possible. Your chances for getting the second ball as well are only 2 in 99, while you have only a 1 in 98 chance of getting the third ball (after also picking the first two). 3/100 * 2/99 * 1/98 = about 6/1,000,000 14.1 a) Both show three spots: 1/6 * 1/6 = 1/36 b) The chance is 6/36 = 1/6, see figure 1 in Ch. 14. Or, imagine it this way: roll one of them, then the other; no matter how the first one lands, the second has a one in six chance of landing the same way. 14.3 a) False. The events are not mutually exclusive, so you can not add the probabilities (see Sec. 14.4). b) False. Same reason. 14.4 Option (i) is better, since you have two chances to win the queen, rather than one. 14.5 a) False. A and B could happen together with chance of 1/3 * 1/10. b) True. If A happens, then B cannot happen, so they are not independent. 14.7 The chance can be found by looking at its opposite, that "no 2 is drawn". There is a 3/5 chance that no two is drawn on any one try, so the chance that none is drawn on four tries is (3/5)^4, or about 13%. So the answer is: 100%-13%= about 87%. 14.9 Draw the list of possible outcomes: 11 12 13 14 21 22 23 24 31 32 33 34 The first number represents the draw from box 1, the second represents the draw from box 2. All are equally likely (1/12 chance). a) The first number is larger than the second in three events (3/12). b) The numbers are equal on three events (3/12). c) The second is larger than the first on six events (6/12). 14.11 a) 13/52 * 12/51 * 11/50 = about 1% b) 39/52 * 38/51 * 37/50 = about 41% c) That not all are diamonds is equal to the opposite of a) or about 99%. 14.13 That you will not get a red marble on each draw is given by 98/100. So, the question asks you to solve this equation: (98/100)^X = 50%. That is, find "X" where .98 to the X is (slightly greater than) 50 percent. You can do this using logarithms, or you can just use trial and error. By trial and error, after 35 draws you have a 50.7% chance of having drawn a red marble. 15.2 The chance of not getting a sex in one roll is 5/6. option (iii) calculates the chance that a die rolled 10 times never lands on a six by the multiplication rule (or also, by the binomial formula). 15.3 The chance of getting 4 girls is (1/2)^4=1/16. The chance of getting 3 girls is 4/16 by the binomial formula. The total chance of getting more girls than boys in a 4-children family is 5/16, by the addition rule. 15.5 True. The first person gets 8!/(2! * 6!) = 28 committees. (Write the 8 names out in a row; then put down 2 C's and 6 N's under the names, where C means ``in the committee'' and N means ``not in the committee''; the binomial coefficient tells you how many ways there are to do that.) By similiar reasoning, the second person gets 8!/(5! * 3!)= 56 committees. 15.8 The chance of getting 2 heads among the first 5 tosses is: 5!/(2! * 3!) * (1/2)^5 = 10/32 The chance of getting 4 heads among the last 5 tosses is: 5!/(4! * 1!) * (1/2)^5 = 5/32 The first 5 tosses are independent of the last 5. So the chance that there will be exactly 2 heads in the first 5 tosses and exactly 4 heads among the last 5 tosses is simply the product: 10/32 * 5/32 = 50/1024 = .0488 = 5% 15.9 a. (i) Mulitplication rule. b. (viii) The chance is 0. c. (iv) Addition rule. d. (viii) The events are not mutually exclusive. (comment: the chance is 2/52 - 1/52 * 1/51 ) e. (vii) 15.10 Without, then it's bound to happen.