Chapter 6 Answers 1

Chapter 6 Answers

Lesson 6.1 11.. li, lo, ln, ls, il, io, in, is, ol, oi, on, os, nl, ni, no, ns, sl, si, so, sn

22.. 5, 4, 5 × 4 = 20, 6 × 5 = 30 33.. (1,2) (1,3) (1,4) (1,5) (1,6) (1,7) (1,8) (1,9) (2,3) (2,4) (2,5) (2,6) (2,7) (2,8) (2,9) (3,4)

(3,5) (3,6) (3,7) (3,8) (3,9) (4,5) (4,6) (4,7) (4,8) (4,9) (5,6) (5,7) (5,8) (5,9) (6,7) (6,8) (6,9) (7,8) (7,9) (8,9), 36.

44.. 9, 8, 9 × 8/2 = 36 55.. (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), there are 36.

66.. 6 × 6 = 36

77.. 5 × 4 = 20

88.. a. 8 × 7 = 56

b. 8 × 7 × 6 = 336

99.. Modify by dividing by 2: 9! 8

2 = 36.

1100.. Because the order of selection doesn’t matter: selecting 3, then 5 is the same as selecting 5, then 3.

1111.. 25! 24

2 == 300

1122.. Each die can fall in six ways. Two dice together can fall in 6 × 6 = 36 ways.

Chapter 6 Answers 2

1133..

1144..

1155.. a. Win $1: 10 ways, win $2: 1 way, lose $1: 25 ways.

b. You would lose money because you would lose $25 for every $12 you won in the long run.

1166.. 6 × 6 × 6 = 216

1177.. 10 × 10 × 10 = 1,000

Chapter 6 Answers 3

1188.. One of two chromosomes can be inherited from the mother and one of two from the father, thus there are 2 × 2 = 4 different ways of inheriting a chromosome. If X1 and X2 are the mother’s chromosomes, and X3 and Y are the father’s chromosomes, then the possibilities are: (X1, X3), (X2, X3), (X1, Y), and (X2, Y).

Chapter 6 Answers 4

Lesson 6.2 11.. 10!/6!; 5040 22.. P(25, 5) = 6,375,600

33.. a. 3 × 5 = 15 b. A particular front sprocket and a particular rear sprocket

44.. a. P(8, 1) + P(8, 2) + … + P(8, 8) = 109,600 b. 548

55.. a. 233103 = 12,167,000 b. 1/233 = 1/12,617

66.. a. P(9, 9) = 9! = 362,880 b. 362,880/365.25 or about 994 years

77.. a. 6 × 3 × 2 × 2 × 1 × 1 = 72 b. 1/2 c. 0 d. 1

88.. 1 99.. a. 105 – 1 = 99,999

b. 5 × 5 × 5 × 5 × 5 – 1 = 3,125 – 1 = 3,124

c. 5 × 5 × 3 × 1 × 1 – 1 = 74

1100.. a. 52 × 52 = 2,704

b. 52 × 51 = 2,652 1111.. a. 47

b.

c. Sample answer: When two events are mutually exclusive, the number of

things in each can be added. However, if the events are not mutually exclusive, the number of things in both events must be subtracted from this sum because the sum counts those things in both events twice.

Football Basketball

8 732

Chapter 6 Answers 5

d. a + b – c e. 8. Add 37 and 14 then subtract 43.

1122.. 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1 × 1 = 720

1133.. a. 30! or about 2.6525 × 1032

b. About 1.0827 × 1028; the number of seating arrangements is about 24,500 times as large.

1144.. a. 3 × 2 = 6 b. A road from Claremont to Upland and a road from Upland to Pasadena.

c. 3 × 3 = 9

1155.. 2 × 26 × 26 + 2 × 26 × 26 × 26 = 36,504

1166.. P(68, 6), or about 7.8806 × 1010.

1177.. a. 8 × 2 × 9 = 144

b. 8 × 10 × 9 = 720 1188.. a. 105 = 100,000 manufacturers

b. 105 = 100,000 products

1199.. a. 220 = 1,048,576 b. About 10 years; about 262 feet.

2200.. a. 63 = 216

b. Win $1: 1 × 5 × 5 + 5 × 1 × 5 + 5 × 5 × 1 = 75, win $2: 1 × 1 × 5 + 1 × 5 × 1 + 5 × 1 × 1 = 15, win $3: 1 × 1 × 1 = 1, lose $1: 5 × 5 × 5 = 125.

c. Lose money. You would lose $125 for every 75 + (2 × 15) + (3 × 1) = $108 you win in the long run.

2211.. In one type of bet, the order of choosing matters, but in the other it does not. When the order matters, the probability of winning is 1/1000; when order does not matter, the probability of winning is 3/1000.

2222.. f(n) = n × f(n – 1).

Chapter 6 Answers 6

Lesson 6.3 11.. They are equal. 22.. 16, 8, 32. The sum for 6 would be 26, or 64. In general, the sum of all possible

combinations of k things is 2k.

33.. a. C(8, 1) × C(9, 3) = 672

b. C(8, 3) × C(9, 1) = 504 c. 2,380. They are the same.

44.. a. C(10, 4) = 210 b. C(10, 6) = 210 c. 210 = 1,024

55.. a. C(52, 2) = 1,326 b. C(26, 2) = 325

c.

3251,326

or about 0.245.

66.. a. 8 b. C(8,2) = 28 c. 8 + 28 = 36

77.. P(10, 3) = 720 88.. a. C(5, 2) = 10

b. C(5, 3) = 10 99.. a. C(44, 6) = 7,059,052

b. About 245 weeks, or a little less than 5 years. c. About 353 feet. d. C(49, 6) = 13,983,816

e.

80,00013,983,816

or about .00572.

f. The probability of winning today is a little more than half (about 0.6) what it was before the change.

1100.. a. C(6, 5) × C(47, 1) = 282

b. C(6, 4) × C(47, 2) = 16,215

c. C(6, 3) × C(47, 3) = 324,300 1111.. a. P(7, 7) = 7! = 5,040

b. C(7, 3) = 35

1122.. C(7, 3) × C(8, 2) × C(5, 2) × C(12, 3) = 2,156,000

Chapter 6 Answers 7

1133.. a. C(9, 2) = 36 b. Drawing a chord between two points is the same as marking the pair on the

lottery card. c. C(10, 2) = 45

1144.. 511, which can be found by evaluating C(9, 1) + C(9, 2) + … + C(9, 9) or by evaluating 29 – 1.

1155.. 15, because C(15, 2) = 105. 1166.. a. A team requires a center and two forwards and two guards.

b. 180

1177.. a. 10 × 10 × 10 × 10 = 10,000 b. 10 c. 107 = 10,000,000

1188.. a. C(21, 3) = 1,330 b. P(21, 3) = 7,980

1199.. a. 256 ≈ 7.21 × 1016 or about 72,000,000,000,000,000. b. Sample answer: No, because from 1998 to 2004 is 6 years, which is four 18-

month periods. Therefore, a claim of 1/16th seems more reasonable. c. Sample answer: Each doubling adds 1.5 years to the time, and a 128-bit system

doubles the number of keys 72 times over a 56-bit system. Therefore, 72 × 1.5 = 108, and 1998 + 108 = 2,106.

d. Sample answer: The order of the bits in a key is important. Therefore, the word “permutation” seems more appropriate than “combination.”

2200.. a. 7 b. C(7, 2) = 21 c. 7 + C(7, 2) = 28 d. A domino has one number of spots on one half and another number of spots

on the other half, or it has the same number of spots on both halves. e. 7/28 or 1/4 f. 13 + C(13, 2) = 91

2211.. C(4, 1) + C(4, 2) + C(4, 3) + C(4, 4) = 15 2222.. a. C(52, 5) = 2,598,960

b. C(4, 3) = 4

c. 13 × C(4, 3) = 52

d. 13 × C(4, 2) = 78

e. 13 × C(4, 3) × 12 × C(4, 2) = 3,744

Chapter 6 Answers 8

2233.. The number of ways of entering is C(39, 5) = 575,757. Ten tickets per week are 520 per year, so one would expect to win about once every 575757 ÷ 520 ≈ 1,107 years.

2244.. a. 104 = 10,000 b. Sample answer: No. “Permutation lock” would be more appropriate.

2255.. a. 1/6 b. 5/6 c. Sample answer: The odds in favor of Falling Sky are 1-36 or 1/50 or .028. The

probability of Falling Sky’s winning is 1/37 or about .027. The odds in favor of an event and the probability of the event’s occurring are close when the event is unlikely to occur.

Chapter 6 Answers 9

Lesson 6.4 11.. a. 520/1000 = .52

b. 196/360, or about .544. c. No, but they are fairly close. d. 360/1000 = .36 e. 196/1000 = .196. The product is .1872.

f. .544 × .36 = .196. They are the same. g. No.

22.. a. Sample answer: probably not since many teams have better records at home than they do overall.

b. The team would have to be either more or less likely to win in bad weather than to win in general. Alternately, the team would have to have a better or worse record in bad weather than the team has overall.

33.. a. 684/1000 = .684 b. .36 + .52 = .88, which is larger than .684. c. No, because there are sophomores who are males.

44.. a. By adding the three probabilities in the previous paragraph. b. That the three events mentioned in the previous paragraph are mutually

exclusive. 55.. a.

b. 1

66.. a. 4/52 = 1/13 b. 1/13 c. Yes. Drawing an ace from the diamonds has the same probability as drawing

an ace from the entire deck. d. 13/52 = 1/4 e. 1/52. They are the same.

Chapter 6 Answers 10

f. 16/52 = 4/13 g. No.

h. They are not independent because (1/13) × (12/52 ≠ 1/13). They are not mutually exclusive because some cards (the kings) are both kings and face cards.

77.. a. 26/52 = 1/2 b. 25/51

c. (26/52) × (25/51) = 25/102 d. They are the same. e.

f. 26/102 + 26/102 = 52/102 = 26/51

88.. a. 1/2 b. 1/2

c. 1/2 × 1/2 = 1/4 d.

e. 1/4 + 1/4 = 1/2 f. This exercise.

Chapter 6 Answers 11

99.. a. .9 × .7 = .63 b. That the outcomes of the two games are independent. Probably not because,

for example, the outcome of the first game could have a psychological effect on the team in the second game.

1100. a.

b. .12 c. .18 + .14 = .32 d. .56 e. No, because the probability of rain tomorrow depends on today’s weather.

1111.. a.

b. 196 + 1,996 = 2,192 c. 196 d. 196/2192, or about .089. e. Because of the conditional probabilities that are known. For example, since it

is known that the test is positive 98% of the time when a person has the disease, disease incidence must be on the first branch.

1122.. a. Yes. One die has no effect on the other.

b. 1/6 × 1/6 = 1/36

Chapter 6 Answers 12

c.

d. 5/36 + 5/36 = 10/36 e. 25/36. This probability and the other two have a sum of 1. f. Lose money. For example, you lose a dollar 25 times out of 36 and you win a

dollar 10 times out of 36 and win two dollars one time out of 36.

1133.. 1/6 × 1/6 × 1/6 = 1/216 1144.. .95, or about .59. 1155.. (1/37)3 = 1/50653 1166.. a. .9776, or about .870.

b. The first is .870, and the second is 1 – 1/6 = 5/6, or about .83.

c. .023 × .023 = .000529 d. 1 – .000529 = .999471, .9994716 = .99683

1177.. a. .99999 b. About 32,850. c. .9999932850, or about .72.

1188.. a.

b. 10,000 × 0.08 = 800

c. 10,000 × 0.32 = 3,200

Chapter 6 Answers 13

d. 10,000 × 0.54 = 5,400

e. 10,000 × 0.06 = 600 f. 800/6200, or about 12.9%. g. None is independent. Selecting a Hobbit and selecting a human are mutually

exclusive, as are selecting someone who wears shoes and selecting someone who doesn’t.

1199.. a. (1/8) × (1/10) = 1/80 b. That the two events are independent. c. The probability of selecting a man with red hair is .1; the probability of

selecting a man who owns a blue car is .125; and the probability of selecting a man who has red hair and owns a blue car is about .0124. The product of the first two probabilities is .0125, so the events are quite close to being independent.

2200.. a. 6 × 6 = 36 b. 1/36

c. 1/36 × 1/36 = 1/1296 d. 35/36

e. 35/36 × 35/36 = 1225/1296 f. (35/36)21, or about .5534.

2211.. p(A) =

x + yx + y + z

; p(A and B) =

yx + y + z

; p(A from B) =

yx + y

. p(A) × p(A from B) =

x + yx + y + z

! yx + y

= yx + y + z

2222.. Sample answer: Per million, a person has a risk of 19 ÷ 442 + 73 ÷ 365 ≈ 0.243 of dying from terrorist action or auto accident on a given day. Of course, this assumes that the two are mutually exclusive, a reasonable assumption unless a terrorist attack on a vehicle is considered both terrorism and auto accident, which seems unlikely. For a seven-day trip, the risk is then 0.243 × 7 ≈ 1.7 per million.

Chapter 6 Answers 14

Lesson 6.5 11.. a. The probability of each outcome is 1/2, and successive applications are

independent. b. The probability of choosing the third answer is 1/2, while the probability of

choosing each of the others is 1/4; successive applications are independent. c. Each possibility probably has a 1/4 chance of occurring, but successive

applications may not be independent if, for example, one finger is injured. d. Each possibility has the same chance of occurring, and successive applications

are independent. 22.. a. C(5, 3) = 10

b. C(5, 3)(.5)3(.5)2 = .3125 c. .03125, .15625, .3125, .3125, .15625, .03125 d. .00243, .02835, .1323, .3087, .36015, .16807

33.. a. C(10, 6)(.5)6(.5)4 ≈ .2051 b. C(10, 7)(.5)7(.5)3 ≈ .1172 c. C(10, 8)(.5)8(.5)2 ≈ .0439 d. C(10, 9)(.5)9(.5)1 ≈ .0098 e. (.5)10 ≈ .00098 f. .3770

44.. a. C(3, 1)(1/6)1(5/6)2 ≈ .3472 b. .5787, .3472, .0694, .00463 c. –1(.5787) + 1(.3472) + 2(.0694) + 3(.00463) = –.079 d. Lose about $7.90.

55.. a. C(6, 3)(.4)3(.6)3 ≈ .2765 b.

Number of Women

0 1 2 3 4 5 6

Probability .0467 .1866 .3110 .2765 .1382 .0369 .0041

c. The probability of fewer than two women is .2333, which probably isn’t unlikely enough for an accusation of discrimination.

66.. a. Each probability is 1/4. b. 25%, 50%, 25% c. .2373, .3955, .2637, .0879, .0146, .00098 d. 1.25. The average number of children with sickle cell anemia in families with

five children is about 1.25.

Chapter 6 Answers 15

77.. a.

Number of Defective 0 1 2 3 Probability .512 .384 .096 .008

b. Probably not. There is a 51% chance that the engineer won’t detect the problem. This could be improved by testing more widgets. For example, the probability of no defective widgets in a sample of 10 is only about 10%.

88.. a.

17,059,052

; 7,059,0517,059,052

b. $2.82, but this assumes that the jackpot is not shared with another party. c.

Amount Won 27,000,000 –5,000,000

Probability

5,000,0007,059,052

2,059,0527,059,052

The expectation is $17,665,933, but this assumes the jackpot is not shared. 99.. a.

Number Escaped 0 1 2 3 Probability .614 .325 .057 .003

b. .003. 1100.. a.

Amount Won –1 1 20 Probability 21/36 14/36 1/36

b. $0.36 c. No, the council would lose about 36 cents per play. One way of correcting this

would be to give no prize for matching a single number. In fact, the jackpot could then be increased but should be kept under $35.

1111.. a. C(10, 5)(.5)5(.5)5 = .2461 b. C(20, 10)(.5)10(.5)10 = .1762 c.

Number of Heads 4 5 6 Probability .2051 .2461 .2051

d.

Number of Heads 8 9 10 11 12 Probability .1201 .1602 .1762 .1602 .1201

e. In 10, the probability is .6563, and in 20 it is .7368, so it is more likely in 20.

Chapter 6 Answers 16

1122.. a. 1/5 b.

Number Correct 0 1 2 3 4 5 Probability .3277 .4096 .2048 .0512 .0064 .00032

c. .00672 1133.. a. 1/4

b.

Amount Won –$0.50 $1.00 Probability 3/4 1/4

c. –$0.13 d. Yes, about 87 cents per play.

1144.. a.

Shots Made 0 1 2 3 4 5 6 7 Probability .00114 .01307 .06396 .17393 .28378 .27781 .15109 .03522

b. That the individual field goal attempts are independent. At least one study in the NBA supported the belief that individual attempts are independent, although many players disagree.

c. 4.34, the average number of shots Sara could expect to make in 7 attempts. 1155.. a. The expected gain is –5(.3) – 10(.1) + 3(.4) + 7(.2) = $0.10. Since the expected

gain is positive, one could argue in favor of buying the stock. However, since the expected gain is a relatively small percentage of the stock’s value, there are likely to be better investments.

b. Since C(42, 6) = 5,245,786; the jackpot must be more than $5,245,785. c. Answers on this question are a matter of opinion. However, there are

mathematicians who buy lottery tickets only when the expected value is positive.

1166.. a. If the teams are evenly matched, the probability a given team wins a given

game is 12

, and the probability that a given team wins 4 games in a row is

12

!"#

$%&

4

. Multiplication by 2 is necessary because there are two teams.

b. The second student is correct. If the teams are called A and B, then the number of ways A can win a 5-game series are: BAAAA, ABAAA, AABAA, and AAABA. C(5, 4) also counts AAAAB, which is a 4-game series won by A.

The probability is therefore 2 × 4 ×

12

!"#

$%&

5

= .25.

Chapter 6 Answers 17

c. C(6, 4) is too large because it counts the number of series that end in 4 or 5 games, which can be corrected by subtracting the counts in parts a and b. The number of ways a given team can win a 6-game series is C(6, 4) – 4 – 1 = 10.

Thus, the probability a series ends in 6 games is 2 × 10 ×

12

!"#

$%&

6

= .3125.

d. .3125, which can be found by subtracting the previous three probabilities

from 1 or by calculating directly: 2 × (C(7, 4) Ð 15) ×

12

!"#

$%&

7

.

1177.. a. The next three rows are: 1 4 6 4 1, 1 5 10 10 5 1, 1 6 15 20 15 6 1. b. 1, 5, 10, 10, 5, 1. c. They are the values in the fifth row of the triangle. d.

.

Chapter 6 Answers 18

Chapter 6 Review 11.. A reasonable answer should include the following points:

Ways of counting: the fundamental principle, permutations and combinations, the addition principle and mutually exclusive events, the multiplication principle and independent events.

Probability, including how the addition and multiplication principles are applied and, in particular, conditional probability. Binomial probability and probability distributions should also be discussed.

22.. a. 7900/46900, or about .168. b. 2300/13700 or about .168. c. 13700/46900 or about .292. d. 2300/46900 or about .049. e. 19300/46900 or about .412. f. Yes, because, as shown in parts a and b, the cat owners are as common among

the professionals as among the community as a whole. g. No, because there are people who are both cat owners and professionals.

33.. a. 26 × 26 × 10 × 10 × 26 × 26 = 45,697,600 b One might conclude that the old system permitted only about 14 million

different plates. Technically, the use of the word “combination” by the minister is inappropriate—“permutation” is correct.

44.. C(5, 1) = 5, C(5, 2) = 10, C(5, 3) = 10, C(5, 4) = 5, C(5, 5) = 1. 55.. a. (1/2)3 = 1/8

b. (1/2)3 = 1/8 c. 1/8 + 1/8 = 1/4

66.. a. P(6, 6) = 6! = 720

b. 2 × 1 × 4 × 3 × 2 × 1 = 48 c. 48/720 = 1/15 d. The math books can be in positions 1 and 2, or in positions 2 and 3, or in

positions 3 and 4, or in positions 4 and 5, or in positions 5 and 6. 5 × 48 = 240. e. 240/720 = 1/3

77.. a.

Amount Won $2 $1 –$1 Probability 1/4 1/4 1/2

b. 2(1/4) + 1(1/4) – 1(1/2) = 1/4 c. Win about $25.

Chapter 6 Answers 19

88.. a. C(8, 1) +C(8, 2) + C(8, 3) + C(8, 4) +C(8, 5) + C(8, 6) + C(8, 7) + C(8, 8) = 28 – 1 = 255

b. One calculation that is close to 40,312 is 8! = 40,320—perhaps MacDonald’s subtracted 8 from this for some reason such as trying to account for 0 items.

99.. a. 3!5! = 720 b. About 28.

1100.. There are C(39, 5) = 575,757 different winning tickets possible in the first and C(36, 6) = 1,947,792 ways of winning in the second, so the probability of winning in the first is between three and four times as great as in the second. About five in the first and one or two in the second.

1111.. a. 13/52 = 1/4 b. 12/51 = 4/17

c. 1/4 × 4/17 = 1/17 d.

e. No. Since the first card is not put back, the probability of, say, a heart, on the

second draw depends on the outcome of the first draw. 1122.. a. 43 because C(43, 5) = 962,598.

b. 1 in 962,598 is the probability, not the odds. The odds in favor of winning are 1-to-962,597.

1133.. a. C(11, 3) = 165

b. C(5, 1) × C(6, 2) = 75 c. 75/165

d. C(5, 1) × C(6, 2) + C(5, 2) × C(6, 1) = 135 e. 135/165

Chapter 6 Answers 20

1144.. a.

b. .045 × 50000 + .07 × 50000 = 5,750 c. 3,500 d. 3500/5750

1155.. a. (1/10)3 = 1/1000 b. (9/10)3 = 729/1000

c. 9/10 × 8/10 × 7/10 = 504/1000

1166.. a. 10 × 10 = 100 b. P(10, 2) = 90. c. C(10, 2) = 45.

1177.. a. C(5, 2)(1/6)2(5/6)3, or about .161. b. C(5, 0)(1/6)0(5/6)5 + C(5, 1)(1/6)1(5/6)4 + C(5, 2)(1/6)2 (5/6)3, or about .965. c. C(5, 0)(1/6)0(5/6)5, or about .402.

Chapter 6 Answers 21

1188.. A tree diagram helps in this analysis.

a. Percentage of all email labeled spam is 100 × (.54 + .08) = 62.

b. The probability that an email labeled spam actually is spam is .54.62

or about

.871.

c. The probability that an email labeled good is actually spam is .06.38

or about

.158. About 16% of email that passes the filter is spam. 1199.. C(4, 1) + C(4, 2) + C(4, 3) + C(4, 4) = 15 2200.. a. .1 + .1 = .2

b. .1 c. .1 + .1 + .4 = .6

d.

.1.1+ .4

= 15

or .2

e. No. If they were, the probability in the overlap would be 0, not .1.

f. Yes. For example, p(A) × p(B) = .2 × .5 = .1, which is p(A and B). 2211.. a. Sample answer: Yes. The probability a silo is missed by both bombs is

.05 × .05 = .0025 or .25%, which is less than .5%. b. From the answer to part a, the probability a given silo survives is .9975.

The probability that at least one of 10 survives is 1 – .997510, or about .0247. Therefore, the actual probability is about 2.5%.

2222.. a. .004 × .9 + .996 × .05 = .0534, or about 5%.

b. .0498.0534

or about 93%.

c. 20 × .0498, or about 1.

.6

.4

Spam

Non-spam

.9

.1

.2

.8

Labeled spam p(Spam and Labeled)

Not labeled p(Spam and Not Labeled)

Labeled spam p(Non-spam and Labeled)

Not labeled p(Non-spam and Not labeled)

Chapter 6 Answers 22

2233.. Mutually exclusive: rolling a number divisible by 5 and rolling a number divisible by 3, rolling a number divisible by 5 and rolling a number divisible by 2. However rolling a number divisible by 2 and rolling a number divisible by 3 are not mutually exclusive because 6 is divisible by both 2 and 3.

The probability of rolling a number divisible by 2 is 1/2, and the probability of rolling a number divisible by 3 is 1/3. Since the product of these two probabilities, 1/6, is the same as the probability of rolling a number divisible by both 2 and 3, these two events are independent.

Since rolling a 2 and rolling a 5 are mutually exclusive, the probability of doing both is 0. The product of the individual probabilities is not 0, so the events are not independent. A similar argument holds for rolling a 3 and rolling a 5.

2244.. a. No. For the events to be mutually exclusive, the study would have found no places where a gun is kept in the home and a gunshot death occurred. Clearly this is not the case.

b. No. The study found that the existence of a gun in the home increases the probability of gunshot death. If the events were independent, one would have no effect on the probability of the other.

2255.. a. C(59, 5) × C(35, 1) = 175,223,510, so the probability is

1175223510

.

b. Per 1,000 the death rate in a 1-hour period is about

.00839365! 24

, or 0.000000958.

This probability is about 168 times the probability of winning the jackpot. c. No more than about 60 ÷ 168 ≈ 0.52, or about a third of a minute before the

drawing.