Modeling Randomness

Section 11.1

11.1.1 C.

11.1.2 B.

11.1.3 C.

11.1.4 E.

11.1.5 A.

11.1.6 a. Yes b.  Yes c.  No d. No e.  Yes f.  Yes (but not too much longer using a computer)

11.1.7  Number the index cards 1, 2, and 3 (they each represent a phone). Shuffle them and place them on the table in spots labeled 1, 2, and 3. If all the numbers on the cards fail to match all the numbered spots on the table call that a success; anything else call a failure. Repeat this 1,000 times. The number of successes divided by 1,000 is an estimate for the probability none of them would get their own phones.

11.1.8 a.  ABC, ACB, BAC, BCA, CAB, CBA b.  BCA, CAB c.  2/6 = 1/3 d.  If the three executives dropped their phones and randomly picked them up many, many times, about 1/3 of the time nobody would get their own phone.

11.1.9 a. 1/2 b. 0 c. 1/6 d. 2/3

11.1.10  With 2 babies the probability is 0.5 that they both get returned to the correct mothers. As the number of babies increases, the probability they all get returned to the correct mothers goes down. For 4 mothers the probability is 0.0417. For 6 mothers the probability is 0.0014. For 8 mothers the probability is 0.0002.

11.1.11 With 2 babies the probability is 0.5 that at least on gets returned to the correct mothers. For 4, 6, and 8 babies, the probabilities that at least one gets returned to the correct mother are all very similar, around 0.63.

11.1.12  The average number of babies returned to the correct mother is always 1.

11.1.13 a.  As the number of dice increases, the probability of getting at least one 5 or 6 will also increase. (Imagine rolling 100 dice. It is almost a sure thing that you will get at least one 5 or 6.) Therefore the probability that the price will be 50 cents or less will decrease as the number of dice increases. Also, rolling additional dice will never drop a price from above 50 cents to below it, but it can move it from below 50 cents to above it. b.  Because getting a “large” number is more likely with the more dice, the price is more likely to be higher and so the mean price will also increase as the number of dice increases. c. If you can choose one die (and you don’t have to multiply the outcome by 10) then you will always be able to afford the ice cream cone! However, if you have to use more than one die, you should choose two dice because the probability of an outcome of 50 cents or less is the greatest.

11.1.14

= 10/30 c. The cards. This makes sense because while the numerator is the same, six pairs were eliminated from the denominator when we switched to the cards, therefore that probability increased in value.

11.1.15 a.  B1B2B3, G1B2B3, B1G2B3, B1B2G3, G1G2B3, B1G2G3, G1B2G3, G1G2G3 b. B1B2B3, G1B2B3, B1G2B3, B1B2G3 c. 2/8 = 1/4

11.1.16 a.  B1B2B3B4, G1B2B3B4, B1G2B3B4, B1B2G3B4, B1B2B3G4, G1G2B3B4, G1B2G3B4, G1B2B3G4, B1G2G3B4, B1G2B3G4, B1B2G3G4, B1G2G3G4, G1B2G3G4, G1G2B3G4, G1G2G3B4, G1G2G3G4 b.  B1B2B3B4, G1B2B3B4, B1G2B3B4, B1B2G3B4, B1B2B3G4 c.  2/16 = 1/8 d.  A 3 to 1 breakdown is more likely because this probability is 8/16 whereas two boys and two girls is 6/16.

11.1.17 a.  Write “Male” on four cards and “Female” on two cards. Shuffle the six cards and randomly choose two of them. If they are the two Female cards call this a success. Repeat this for a total of 1,000 times. The proportion of successes out of 1,000 is an estimate for the probability. b.  If you do it with dice, you can think each number represents a different person. However, you could roll doubles (e.g. (3,3)) and you can’t pick the same person twice, so this is not a good model for the simulation.

11.1.18 a.  AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF b. 1/15 c. If you repeatedly choose two people from the committee at random, about 1 in every 15 times you would choose both of the women. d.  9/15

11.1.19 a.  H6H7; H6T7H8; T6H7H8; T6T7H8H9; T6H7T8H9; H6T7T8H9; T6T7T8; H6T7T8T9; T6H7T8T9; T6T7H8T9 b.  Heather wins on the first 6 listed in part (a) and Tom wins on the last 4 listed. c.  No, as they are different length sequences the probabilities won’t all be the same.

11.1.20 a.  Flip four coins and count the number of heads. Repeat this until you have done it 1,000 times. Count how many times 2 or more heads occurred. Divide that number by 1,000 and that is an estimate for the probability that Heather would win. b. 0.687163 c.  0.750627

11.1.21 a.  I would expect that Heather has a smaller probability of winning. Heather’s one-point advantage is a bigger deal in the short term than in the long run. b. Because Heather’s one-point advantage is almost nothing if they have to go all the way to 100, her probability of winning should be very slightly above 1/2. a. HHH, HHT, HTH, THH, TTH, THT, HTT, TTT b.

11.2.12  The events are not mutually exclusive. To be mutually exclusive there would need to be no one that uses both Facebook and Instagram.

11.2.14 The events are not complements. To be complements there would need to be no one that uses both Twitter and Instagram and everyone uses one or the other. 11.2.15 a.

− (0.42 + 0.10 + 0.04) = 0.44 b.

11.2.16 a. P(Y) = 336/1,864 ≈ 0.1803 and it is the probability a randomly chosen student contracted the norovirus. b.  P(F) = 1,261/1,864 ≈ 0.6765 and it is the probability a randomly chosen student is a female. c.  P(F c) = 603/1,864 ≈ 0.3235 and it is the probability a randomly chosen student is a male. d. P(Y and F) = 212/1,864 ≈ 0.1137 and it is the probability a randomly chosen student is both female and has contracted the norovirus. e. P(Y or F) = (212 + 124 + 1,049)/ 1,864 = 1,385/1,864 ≈ 0.7430 and it is the probability a randomly chosen student has either contracted the norovirus or is female.

11.2.17 a.  (Y or F)c means the event is neither Yes nor Female, so it means that it is the males that did not get the norovirus so it is the same as (N and M). That probability is 479/1,864 = 0.2570. b. a) (Y c and F c) means the event is not Yes (so No) and not Female (so Male), so it means that it is the males that did not get the norovirus so it is the same as (N and M). That probability is 479/1,864 = 0.2570 as in part (a).

11.2.18 a.  1,955/3,222 ≈ 0.6068 b.  789/3,222 ≈ 0.2449 c.  457/3,222 ≈ 0.1418 d. (1,955 + 332)/3,222 = 2,287/3,222 ≈ 0.7098 e.  Because it is measuring those that are not female and not sophomores it is measuring those males that are freshmen, juniors, seniors or non-degree which is 935/3,222 ≈ 0.2902. f.  Because it is measuring those that are neither female nor sophomores it is measuring those males that are freshmen, juniors, seniors, or non-degree which is 935/3,222 ≈ 0.2902, the same as the previous question.

11.2.19 a. 1,267/3,222 ≈ 0.3932 b.  264/3,222 ≈ 0.0819 c.  (1,267 + 445)/3,222 = 1,712/3,222 ≈ 0.5313 d.  264/1,267 ≈ 0.2084 e.  264/709 ≈ 0.3724

11.2.20 a.  GGG, GGB, GBG, BGG, BBG, BGB, GBB, BBB b.  1/8 (the outcome is GGB) c.  3/8 (the outcomes are BGG or GBG or GGB)

Section 11.3

11.3.1 A.

11.3.2 C.

11.3.3 B.

11.3.4 A.

11.3.5 B, D, G.

11.3.6  Because P(A) ≠ 0 and P(B) ≠ 0 then P(A)P(B) ≠ 0 and because A and B are independent, P(A and B) = P(A)P(B). Therefore, P(A and B) ≠ 0 which implies that A and B are not disjoint.

11.3.7 a.  0.2/0.5 = 0.4 b. 0.2/0.4 = 0.5 c. 0.4 + 0.5 – 0.2 = 0.7 d. Yes because P(A and B) = P(A)P(B) = 0.2. e.

11.3.8 a. 0.4/0.5 = 0.8 b.  0.4/0.8 = 0.5 c. 0.8 + 0.5 – 0.4 = 0.9 d.  Yes because P(A and B) = P(A)P(B). e.

11.3.9 a. TTT, TTH, THT, HTT, HHT, HTH, THH, HHH b.  i.  7/8 ii.  3/4 iii. There are 7 outcomes with at least one tail and 3 of these start with H: 3/7

11.3.11 Sample space: BBB, BBG, BGB, GBB, GGB, GBG, BGG,

= 0.95 b.

= 0.29 d.  1 – 0.31 = 0.69

11.3.14 a.  0.65 = P(I |T) = P(I and T)/0.21 so P(I and T ) = 0.21 × 0.65 = 0.14. a.  0.56 + 0.35 = 0.91 b.  0.35/0.91 = 0.38 11.3.23 a. P(C) = 0.077; P(C|S) = 0.244 b. No because the P(C) ≠ P(C|S) b.  0.14/0.28 = 0.50 c.  0.14 d.  0.65

11.3.15 No. They would be independent if the proportion of Twitter users in the entire population was the same as the proportion of Twitter users among the Instagram users. In other words P(T) = P(T |I).

11.3.16 a.  The probability that a randomly chosen student got the norovirus is 336/1,864 = 0.180. b. The probability that a randomly chosen female got the norovirus is 212/1,261 = 0.168. c.  No, they are not quite independent, as the probability a female got the norovirus is different (lower) than the overall probability that someone got the norovirus. d.  No, the male rate was 124/603 = 0.206 whereas the female rate was 0.168. e.  Yes. If the male rate was the same as the female rate then both of these would have to be the same as the overall rate.

11.3.17 a.  789/3,222 = 0.245 b.  1955/3,222 = 0.607 c.  457/1,955 = 0.234 d.  457/789 = 0.579 e.  No they are not independent because P(So) ≠ P(So|F) [or P(F) ≠ P(F|S)].

11.3.18 a.  (0.61)(0.40) = 0.24 b.  (0.39)(0.30) = 0.12 c.  0.24 + 0.12 = 0.36

11.3.19 a.  0.422 = 0.18 b.  0.582 = 0.34 c.  (0.42)(0.58) + (0.58)(0.42) = 0.49 (or 1 – (0.18 + 0.34) = 0.48. Note that the two answers are a bit different because of rounding error.

11.3.20 a.  0.425 = 0.013 b. 0.585 = 0.066 c.  1 – 0.066 = 0.934

11.3.21 a.  0.110 b.  0.910 = 0.35 c.  1 – 0.35 = 0.65 b.  If there is no association, the two conditional proportions of survival must be the same. The only way they can be the same is if they are the same as the overall proportion of survival. So we would need all three values to equal 0.48 in the penguin study. If surviving is independent of being banded then the probability of surviving given banded needs to be the same as the probability of surviving. This is what we said had to happen if there was no association; we used the idea of proportion instead of probability. Again, these are both 0.48 in the penguin study. b. The event when 2 coins land heads up. c.  The probability exactly 2 of the 3 coins land heads up is 3/8.

The probability 2 or fewer of the 3 coins land heads up is 7/8.

(4,4) c. The event where the sum of the numbers on the dice is more than 3. d. The probability that the sum is more than 3 when the two dice are thrown is 1 – P(X = 2) – P(X = 3) = 1 – (1/16) – (2/16) = 13/16. The only outcomes not included are (1,1), (1,2), and (2,1). Keep in mind that with discrete random variables, this is not the same as P(X ≥ 3). 11.4.8 a. The probability that a randomly chosen family household has exactly 4 members is 0.1936. b. The probability that a randomly chosen family household has more than 4 members is 0.0906 + 0.0341 + 0.0192 = 0.1439. c.  The probability that a randomly chosen family household has 4 or more members is 0.1936 + 0.1439 = 0.3375.

11.4.9 a.  The probability that a randomly chosen nonfamily household has 3 or 4 members is 0.0323. b.  The probability that a randomly chosen nonfamily household has 3 or fewer members is 0.9888. c.  The probability that a randomly chosen nonfamily household has fewer than 4 members (which is the same as 3 or fewer) is 0.9888.

11.4.10 a. 9/24 b.  15/24 c. If 3 people draw their own names then the only name left to draw must be that of the person than hasn’t drawn. So if we know 3 people have drawn their own names, we also know all four of them must.

11.4.11 a.  3.1149 b.  1.2633

11.4.12 a.  1.2393 b.  0.5698

11.4.13 a.  1 b.  1 b. 1 c.  1 d. They are the same.

11.4.14 a.

11.4.15 a.  The expected payout of wheel B is $2.33, the same as that of wheel A. b.  The SD of wheel B is $1.1785, much smaller than that of wheel A. c.  Answers will vary. Both will give the same average outcome in the long run, but you are just spinning it once. Those that like taking chances may tend to pick wheel A, as there is a chance to win a larger amount of money than anything on wheel B. Those that are more cautious may pick wheel B, as there is a greater chance that you will win something.

11.4.16 a. $1.0017 b.  For wheel A the SD is $6.8475 and the average distance is $3.775. For wheel B the SD is $1.1785 and the average distance is $1.0017. So for wheel A they are not similar (because the $25 payout is an outlier), but for wheel B they are similar because there are no outliers.

11.4.17 a. b. The expected value is 1. This means that if you were to pick two puppies at random from this litter many, many times, you should expect to get 1 male puppy, on average, each time.

11.4.18 a. b. The expected value is 90/36 = 2.50, meaning that you can expect to win about $2.50 per game, on average, in the long run. c. You would have to pay $2.50 per game.

11.4.19 a. b.  7/27 11.4.20 a.  0.477 b.  0.523 c.  3.441 d.  2.462 a. 7 b.  2.4152 c.  Mean of 9 and SD of 2.4152 d.  Mean of 14 and SD of 4.8304

1/6 b. Mean of 3.5 and SD of 1.7078 c.  Mean: 3.5 + 3.5 = 7, SD = √1.70782 + 1.70782 = 2.4152. The mean and the standard deviation are the same. d. 2X is describing rolling a fair six-sided die and then doubling the result. It is not the same as the sum of the numbers on two dice. For example, you can get a 3 when you are finding the sum, but that can’t be obtained when doubling the numbers. e. The distribution of 2X has a mean of 2(3.5) = 7 (the same as that of X + X) and a SD of 2(1.7078) = 3.4156 (not the same as that of X + X). So in this case, X + X ≠ 2X! The variability is larger with X + X, because there are now two sources of randomness rather than only one.

11.5.9 a. 16/36 b.  20/36 c.  E(X) = (1)(16/36) + (–1)(20/36) = −4/36. You can expect to lose $0.11, on average, per game.

11.5.10 a.  −2/38 = −$0.0526 (you are expected to lose about 5 cents per game, in the long run) b. $0.9986 c. −$0.1052 d. $1.412

11.5.11 a.  −2/38 = −$0.0526 (you are expected to lose about 5 cents) b. $5.7626 c.  −$0.1578 d.  $9.9811 a. c.  $8.150

11.5.14  Mean of 19.88 inches and SD of 0.75 inches

11.5.15  Mean of 19.76 inches and SD of 0.67 inches

11.5.16 Mean of 8.72 and SD of 2.45

11.5.17 Mean of 3.14 and SD of 0.79

11.5.18 a. If we let S be a score, then the first method is S/2 – 1 and the second method is (S – 1)/2 = S/2 – 1/2. We can see by our equations that the second method will give scores (and thus a mean) that is a half point higher than the first method. b. Because both methods divide by 2 (or multiply by 1/2) they will give identical standard deviations (when the shift occurs does not impact the standard deviation).

11.5.19  Mean of 36.725°F and SD of 0.388°F

11.5.20  Mean of 36.886°F and SD of 0.413°F

Section 11.6

11.6.1 A.

11.6.2 C.

11.6.3

B.  In C, the probability of a yes response presumably differs in the two locations.

11.6.4 C.

11.6.5

A.  would be better modeled with a binomial distribution because the probability of success (picking a female) is much closer to the same value for each student picked because the population size is large compared to the sample size.

11.6.6 a.  0.2201 b. 0.3499 c.  0.9127 d. Not necessarily. The probability that someone has a college degree on a certain street in a certain town could be much different than what it is nationally.

11.6.7 a.  0.0548 b. 0.2516 c.  0.7712 d.  Not necessarily. The probability that a household in a single neighborhood has Internet access could be much different than what it is nationally.

11.6.8 a. 0.00098 b. 0.00098 c.  when π = 0.50, the binomial distribution is symmetric and P(X = 10) = P(X = 0) d. 0.3770

11.6.9 a.  i. 1 − 0.357 = 0.643 ii. 0.6434 = 0.1709 iii. 1 − 0.1709 = 0.8291 b. 0.829156 = 0.0000276 c. The estimate should be larger, as there are many sets of 56 consecutive games in 1941 not just one.

11.6.10 a. X follows a binomial distribution with n = 4 and π = 0.51.

11.6.11  The mean number of boys is 0(0.0576) + 1(0.2400) + 2(0.3747) + 3(0.2600) + 4(0.0677) = 4(0.51) = 2.04 and the SD is √4(0.51)(0.49) ≈ 0.9998. The mean number of girls is 1.96 and the SD is 0.9998.

11.6.12 a.  Let X represent the number of no-shows. X follows a binomial probability distribution with n = 44 and π = 0.05. We want to find b.  P(X ≥ 2) = 0.6529 c.  E(X) = 44(0.95) = 41.8 passengers

P(X < 2) = P(X ≤ 1) = 0.3471.

11.6.13 a.  Let X represent the number of left-handers. X follows a geometric distribution with π = 0.10. We want P(X = 5) = (0.94)(0.1) = 0.06561 b. P(X ≤ 5) = 0.40951 c.  We want to find k so P(X ≤ k) is greater than or equal to 0.5. P(X ≤ 7) = 0.5217, so need 7 picks.

11.6.14 a.  Let X represent the number of winners in 3 cups. X follows a binomial distribution with n = 3 and π = 0.1667. P(X = 1) = 0.3473. b. 1 − P(X = 0) = 0.4214 c.  Let Y represent the number of cups before the first winner. Y follows a geometric distribution with π = 0.1667. P(Y = 3) = 0.1158. d. P(Y ≤ 3) = 0.4214

11.6.15 a. 0.4219 b.  0.6836 c.  0.1055 d. 0.6836 e. No, a 1 in 4 chance does not guarantee that one in every four boxes is a winner, but that if you buy infinitely boxes, the proportion of winners will approach 0.25. Though the chance on a winner in four boxes is 0.6836, there is still about a 32% chance you won’t receive a winner in the first four boxes.

11.6.16 a.  E(X) = 4(0.25) = 1 b.  SD(X) = √4(0.25)(0.75) = 0.8660 c.  1/0.25 = 4

11.6.17 a.  0.4043 b.  0.7293 c.  0.0809 d. 0.7293

11.6.18 a. 1.15 b.  4.348

11.6.19  3 years. The probability of getting in exactly once in 1, 2, or 3 years (if you enter until you get in and then stop) is 0.5435.

11.6.20 a. 18/38 b.  0.2762, 0.9233 c.  0.9233

11.6.21 X will equal 1 (with probability 18/38) and –1 (with probability 20/38). E(X) = 18/38 – 20/38. You can expect to lose about $0.0556 per game in the long run.

11.6.22 a.  i. 0.2373 ii.  0.9990 b. i. 0.0117 ii. P(X < 4) = 0.9844

Section 11.7

11.7.1 B

11.7.2

D.

11.7.3 D.

11.7.4 B.

11.7.5 A.

11.7.6 E.

11.7.7 a. 95% b.  2.5% c.  2.5%

11.7.8 a. 68% b. 95% c.  95%/2 = 47.5% d.  47.5% + 34% = 81.5%

11.7.9 a.  0.0228 or about 2.5% b. 0.0646 c.  119

11.7.10 a.  0.00135 b.  130.8 and above c.  No

11.7.11 a.  About 0.0131 b. About 2,760 g and below

11.7.12 a. 0.01066 b.  About 4,547 g and above

11.7.13 ACT because its standardized statistic (0.56) is larger than that of the SAT (0.41)

11.7.14 a.  82nd (.8233) b.  1,339.1 or above

11.7.15 a.  95th b. 27.92 or above

11.7.16 a.  0.9 b.  66.3″

11.7.17 a.  0 b.  0.1265 c.  P(X = 3) = 10(0.1265)3(0.8735)2 = 0.0154 d.  P(X ≥ 3) = 0.0166

11.7.18 a.  0.1841 b.  P(X = 3) = 4(0.1841)3(0.8159)1 = 0.0204 c.  P(X = 3) + P(X = 4) = 0.0204 + (0.1841)4 = 0.0215

11.7.19 a.  Mean of 133.1″, SD of 4.1″ b. Mean of 266.2″, SD of 5.8″ c.  Mean of 335.5″ SD of 6.5″ 11.7.20 a.  0.3821 b.  38th b.  Assuming a larger percentage, the distribution will shift to the right. Then the peak of the distribution will no longer be in the middle of our interval, so the area under the curve between 0.17 and 0.21 will decrease (assuming not much change in SD).

11.8.11 a.  0.4564 b.  The probability is smaller because the region of interest is no longer centered at the mean of the distribution.

11.8.12 a. 0.5708 b.  0.8253 c.  The probability in part (b) is larger because the interval of interest is still centered at the mean of the distribution but is now wider.

11.8.13 a.  0.2525 b.  0.0175 (with SD = 4.743)

11.8.14 a.  0.1333 b.  0.0065 (with SD = 201.25 g) a.  Answers will vary, but it should be close to 0.058. See figure for Solution 11.8.15a. b.  0.0579 c. a. b. a. b. a. b. a. b. a.  Kayla makes a larger proportion of her shots compared to Jose. b. Yes, for each player we expect more than 10 successes (60, 70) and more than 10 failures (40, 30). c. Mean = 0.10; SD = 0.067 d.  –1.49 e.  0.932 11.8.21 a.  It would increase. b.  It would increase 11.8.22 a.  The mean time for Karen to grade the questions is larger than that for John. b.  Yes, because the times for each individual follows a normal distribution. c.  Mean = –30 seconds; SD = 11.40 seconds d.  2.63 e.  0.004

Solution 11.8.15a