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ch.13&amp;14

# ch.13&amp;14 - Homework Ch13-14 Solutions Chapter 13 2...

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Unformatted text preview: Homework Ch13-14 Solutions Chapter 13 2. is better, because the event in (ii) is harder to get. 3. (ii) is better. The event in is harder to get because of the restriction in the order of the suites. 4. 4/52 x 3/51 x 2/50 x 1/49 x 4/48 2 3 x10”. 6. (a) True, (b) True, (c) False. The chance of getting the ace of clubs and then the ace of diamonds is 1/52 x 1/51. 7. (a) and (b) are False. (c) is True, the chance of any sequence being 1/2 x1/2 x . . . x 1/2 : 2 6 1/ 8. (a) (4/6)4 = (2/3)4 2 0.197. (b) That means all the rolls must show two or less spots, so the chance is (2/6)4 : 0.012. (c) the opposite event of “not all the rolls show 3 or more spots” is “all of the rolls show 3 or more spots”. The chance of the latter event was obtained in (a), 0.197. So the chance of the event “not all the rolls show 3 or more spots” is 100% — 0.197 = 0.803. 9. (a) (1/6)10 =1/60,466,176. (b) 1M 1/60,466,176 z 1. (c) (5/6)”) a 0.16 = 16%. 11. (ii) is better, you can’t lose. 12. 3/100 x 2/99 x 1/98 a: 6/1,000,000. Chapter 14. 1. (a) 1/6 ><1/6 =1/36. (b) Using the addition rule, it is the chance that they are both aces, or both two, or both three etc..., the chance is 1/36 +1/36 + + 1/36 = 6/36 = 1/6. 2. There are two ways of getting a sum of 11, (5,6) or (6,5) (see ﬁgure 1 in chapter 14). So the chance is 2/36 2 1/18. 3. Both (a) and (b) are false. In both cases, the events are not mutually exclusive, so the addition rule can not be used. To answer these questions, ﬁnd the Opposite event (like The paradox of the Chevalier De Mere). 4. is better because you have two tries. 6. If you want to ﬁnd the chance that at least one of two events will happen, check to see if they are mutually exclusive. If so, you can add the chances. If you want to ﬁnd the chance that both events will happen, check to see if they are independent. If so, you can multiply the chances. 7. Use the opposite event. The opposite of “2 is drawn at least one” is “None of the draws is a two”. The chance of the latter event is (3/ 5)4. So the chance of the event “2 is drawn at least one” is 100% — (3/5)4 x 87%. 8. 100%. 9. List all the possibilities: 11121314 21 22 23 24 31 32 33 34 For example, 12 means that you got a 1 from box A and a 2 from box B. Each outcome has chance 1/12. (a) The outcomes are 21 31 and 32, so the chance is 3/12 2 25%. (b) 3/12 2 25%. (c) One way is to use the opposite event: take 100% and subtract the answers from (a) and (b): 100% — (25% + 25%) : 50%. 10. (ii) is better: on each roll the chance to win a dollar is 2/6 but on each draw, it is 3/6. 11. (a) 13/52 x 12/51 x 11/50 x 1%. (b) 39/52 x 38/51 X 37/50 :3 41%. (c) This chance is 100% minus the chance that all are diamonds, i.e., 1%, the chance is 99%. ...
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ch.13&amp;14 - Homework Ch13-14 Solutions Chapter 13 2...

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