Chapter 7 Homework Answers

118 CHAPTER 7: PROBABILITY: LIVING WITH THE ODDS 9. a. The sum of all the probabilities for a probability distribution is always 1. UNIT 7A 10. a. If Triple Treat’s probability of winning is 1/4, then the probability he loses is 3/4. This implies TIME OUT TO THINK the odds that he wins is (1/4) ÷ (3/4), or 1 to 3, Pg. 417. Birth orders of BBG, BGB, and GBB are the which, in turn, means the odds against are 3 to 1. outcomes that produce the event of two boys in a family of 3. We can represent the probability of DOES IT MAKE SENSE? two boys as P(2 boys). This problem is covered 7. Makes sense. The outcomes are HTTT, THTT, later in Example 3. TTHT, TTTH. Pg. 419. The only change from the Example 2 8. Does not make sense. The probability of any event solution is that the event of a July 4 birthday is is always between 0 and 1. only 1 of the 365 possible outcomes, so the 9. Makes sense. This is a subjective probability, but probability is 1/365. it’s as good as any other guess one might make. Pg. 421. (Should be covered only if you covered Unit 10. Does not make sense. The two outcomes in 6D). The experiment is essentially a hypothesis question are not equally likely. test. The null hypothesis is that the coins are fair, 11. Does not make sense. The sum of P(A) and P(not in which case we should observe the theoretical A) is always 1. probabilities. The alternative hypothesis is that the 12. Makes sense. This is the method we use to coins are unfair. The results in this case do not compute empirical probabilities. differ significantly from the results expected under the null hypothesis. Therefore we do NOT reject BASIC SKILLS & CONCEPTS the null hypothesis, so we lack sufficient evidence 13. There are 15 × 3 = 45 choices. for claiming that the coins are unfair. 14. There are 9 × 8 × 12 = 864 packages. 4 Pg. 423. Four coins have 2 = 16 possible outcomes, 15. There are 3 × 6 × 5 = 90 three-course meals. but only five possible events: 0 heads, 1 head, 2 16. Assuming each district elects exactly one heads, 3 heads, and 4 heads. representative, there are 2 × 3 × 4 = 24 possible commissions. QUICK QUIZ 17. There are 28 × 4 = 112 combinations. 1. b. HHT is a different outcome than HTH due to 18. There are 4 × 3 × 5 × 2 × 3 = 360 sets. the order in which the heads appear, but both are 19. There are 7 × 9 × 11 = 693 systems. the same event because each has two heads. 20. There are 2 × 2 × 2 × 8 = 64 versions. 2. b. This is a probability based on observations 21. There are 4 queens in a standard deck, so the about the number of free throws Shawna probability is 4/52 = 1/13. We are assuming that attempted and made, and so it is empirical. each card is equally likely to be chosen. 3. a. This is a theoretical probability because we are 22. There are 26 black cards, so the probability is assuming each outcome is equally likely, and no 26/52 = 1/2. We are assuming each card is equally experiment is being carried out to determine the likely to be chosen. probability. 23. The probability is 2/6 = 1/3. We are assuming 4. c. If the probability of an event is P(A), the equally likely outcomes. probability it does not occur is 1 – P(A). 24. There are 6 × 6 = 36 possible outcomes, and 3 of 5. c. The most likely outcome when a coin is tossed these sum to 4 (see Table 7.3), so the probability is several times is that half of the time heads will 3/36 = 1/12. It is assumed that each outcome is appear. equally likely to occur. 6. b. Table 7.3 shows that the probability Mackenzie 25. There are 12 royalty cards, so the probability is wins is 5/36, whereas the probability that Serena 12/52 = 3/13. It is assumed that each card is wins is 4/36. equally likely to be chosen. 7. c. There are six possible outcomes for each of the 26. There are 3 even numbers on a die, so the four dice, and the multiplication principle says we probability is 3/6 = 1/2. We are assuming equally multiply these outcomes together to get the total likely outcomes. number of outcomes when all four dice are rolled. 27. The probability is 2/52 = 1/26 because there are 2 8. b. The lowest sum that could occur when four dice red aces. We are assuming equally likely are rolled is 4, and the highest is 24, and every outcomes. sum in between is possible. 28. There are 6 ways to get the same number on both dice, and 36 possible outcomes, so the probability UNIT 7A: FUNDAMENTALS OF PROBABILITY 119 is 6/36 = 1/6. It is assumed that all outcomes are 47. The probability is 5/6, because there are 5 equally likely. outcomes that aren’t 2s. It is assumed that each 29. There are ten possible outcomes for the last digit outcome is equally likely. in a phone number, and two of these is are 1 or 2, 48. There are 8 possible outcomes when 3 heads are so the probability is 2/10 = 1/5. It is assumed that tossed (HHH, HHT, HTH, THH, TTH, THT, each digit is equally likely to occur at the end of a HTT, TTT), and only one of them consists of all 3 phone number. tails. Thus the probability of tossing 3 tails is 1/8, 30. The probability is 1/7, assuming that births are and the probability of not tossing 3 tails is 1 – 1/8 uniformly distributed over the days of the week. = 7/8. Equally likely outcomes are assumed. 31. Assuming that people are equally likely to be born 49. The probability he will make the next free throw is during any hour of the day, the probability is 1/24. 75% = 0.75, so the probability he will miss is 1 – 32. There are 12 months in the year, so the probability 0.75 = 0.25, or 25%. The probabilities computed is 2/12 = 1/6, assuming births are uniformly here are based on his past performance. distributed across the months of the year. If using 50. The probability that the next person you meet was the number of days in the months, there are 30 + born on in July is 1/12 (assuming each month is 31 = 61 total days in April and October, so the equally likely), so the probability that the person probability is 61/365, assuming births are was not born In July is 1 – 1/12 = 11/12. uniformly distributed across the days of the year. 51. The probability of rolling a sum of 7 is 6/36 (see 33. There are 4 possible outcomes with a 2-child Table 7.3), so the probability of not rolling a sum family (BB, BG, GB, GG), and 2 of these have of 7 is 1 – 6/36 = 30/36 = 5/6. Equally likely one boy and one girl, so the probability is 2/4 = outcomes are assumed. 1/2. This assumes each outcome is equally likely 52. The probability is 100% – 30% = 70% = 0.7. This to occur. answer assumes that “not rain” means “sunny day” 34. There are 8 possible outcomes with a 3-child (rather than, say, a cloudy day with no rain), and it family (BBB, BBG, BGB, GBB, GGB, GBG, assumes the subjective/empirical probability of BGG, GGG), and 3 of these have exactly two rain is a reasonable prediction. girls, so the probability is 3/8. This assumes each outcome is equally likely to occur. 53. The probability of having a 100-year flood is 35. There are 8 possible outcomes with a 3-child 1/100, so the probability of not having one is 1 – family (BBB, BBG, BGB, GBB, GGB, GBG, 1/100 = 99/100 = 0.99. This assumes that each BGG, GGG), and 1 of these has exactly three year the probability of a 100-year flood is 1/100. girls, so the probability is 1/8. This assumes each 54. The probability of meeting someone born in May outcome is equally likely to occur. or June is 2/12, so the probability that this does not 36. There are 22 pairs of socks, and 8 are white, so the occur is 1 – 2/12 = 10/12 = 5/6. This assumes probability is 8/22 = 4/11, assuming that each pair births are uniformly distributed across the year. of socks is equally likely to be chosen. 55. 37. Your address ends in one of ten digits, and the Result Probability probability that a randomly selected person has the 0 H 1/16 same ending digit is 1/10. This assumes that each digit is as likely as the others to be at the end of an 1 H 4/16 address, and it assumes that addresses end in 2 H 6/16 digits. 3 H 4/16 38. You will wait longer than 15 minutes if you arrive 4 H 1/16 at any of the first 15 minutes of a given half hour (30 minutes), so the probability is 15/30 = 1/2. 56. This assumes your arrival times are uniformly Sum Probability distributed across any given half-hour.

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