chapter 11 counting methods and probability theory€¦ · the number of choices decreases by 1...

Chapter 11 Counting Methods and Probability Theory

Copyright © 2015 Pearson Education, Inc. 464

Check Points 11.1

1. Multiply the number of choices for each of the two courses of the meal:

Appetizers : Main Courses:

15 15010

2. Multiply the number of choices for each of the two courses:

Psychology : Social Science:

4 4010

3. Multiply the number of choices for each of the three decisions:

Topping:Size : Crust :

2 3 305

4. Multiply the number of choices for each of the five options:

Electric/Gas: Onboard Computer: Global Positioning System:Color: A/C:

10 2 1602 22

5. Multiply the number of choices for each of the six questions:

6

Question #1: Question #2: Question #3: Question #4: Question #5: Question #6:

3 3 3 3 3 3 3 729

6. Multiply the number of choices for each of the five digits:

1 9 0 9 0 9 0 9 0 9

Digit 1: Digit 2: Digit 3: Digit 4: Digit 5:

9 10 10 10 10 90,000

Concept and Vocabulary Check 11.1

1. M N

2. multiplying; Fundamental Counting

3. false

4. true

Exercise Set 11.1

1. 8 · 10 = 80

2. 9 · 3 = 27

3. 3 · 4 = 12

4. 5 · 6 = 30

Section 11.1 The Fundamental Counting Principle


5. 3 · 2 = 6

6. 26 · 9 = 234


Drink: Size: Flavor:

2 4 5 40


Size : Crust : Topping :

3 4 6 72

9. Multiply the number of choices for each of the four menu categories:

Main Course: Vegetables: Beverages: Desserts:

4 3 4 3 144

This includes, for example, an order of ham and peas with tea and cake. This also includes an order of beef and peas with milk and pie.

10. Multiply the number of choices for each of the four apartment options:

Option A: Option B: Option C: Option D:

3 2 2 3 36

This includes, for example, a first floor, golf course view apartment with one bedroom and one bathroom. This also includes a first floor, lake view apartment with two bedrooms and one bathroom.

11. Multiply the number of choices for each of the three categories:

Gender: Age: Payment method:

2 2 2 8

12. Multiply the number of choices for each of the three groups of highways.

A to B: B to C: C to D:

3 2 4 24


Color: A/C: Transmission: Windows: CD Player:

6 2 2 2 2 96


Color: A/C: Sun Roof: Transmission: Brakes:

9 2 2 2 2 144

15. Multiply the number of choices for each of the five questions:

Question 1: Question 2: Question 3: Question 4: Question 5:

3 3 3 3 3 243

16. This situation involves making choices with eight groups of items. Each question is considered a group and each group

has 3 choices. Multiply choices: 83 3 3 3 3 3 3 3 3 6561

17. Multiply the number of choices for each of the three digits:

Digit 1: Digit 2: Digit 3:

8 2 9 144


466 Copyright © 2015 Pearson Education, Inc.

18. Multiply the number of choices for each of the four digits:

Digit 4: Digit 5: Digit 6: Digit 7:

10 10 10 10 10,000

19. Multiply the number of choices for each of the letters and digits:

Letter 1: Letter 2: Digit 1: Digit 2: Digit 3:

26 26 10 10 10 676,000

20. Multiply the number of choices for each of the four letters:

Letter 1: Letter 2: Letter 3: Letter 4:

2 26 26 26 35,152

21. This situation involves making choices with seven groups of items. Each stock is a group, and each group has three

choices. Multiply choices: 73 3 3 3 3 3 3 3 2187

22. This situation involves making choices with nine groups of items. Each digit is a group, and each group has ten choices.

Multiply choices: 910 10 10 10 10 10 10 10 10 10 1,000,000,000

26. makes sense

27. makes sense

28. does not make sense; Explanations will vary. Sample explanation: There are 26! or 403,291,461,126,605,635,584,000,000 ways to arrange these letters.

29. makes sense

30. Multiply the number of choices for each of the four digits:

1, 3, 5, 7, 91 9 0 9 0 9


9 10 10 5 4500

31. Multiply the number of choices for each of the four groups of items:

Bun: Sauce: Lettuce: Tomatoes:

12 30 4 3 4320

Total time 10 4320 43,200 minutes, which is 43,200 60 720 hours.

Check Points 11.2

1. There are 5 men to choose from for the first joke. This leaves 5 choices for the second joke. The number of choices then decreases by 1 each time a joke is selected. 1st joke : 2nd joke: 3rd joke: 4th joke: 5th joke: 6th joke:

6005 5 4 3 2 1

2. The number of choices decreases by 1 each time a book is selected.

1st Book: 2nd Book: 3rd Book: 4th Book: 5th Book:

5 4 3 2 1 120

Section 11.2 Permutations


3. a. 9! 9 8 7 6! 9 8 7 6!

6! 6!

6!

9 8 7 504

b. 16! 16 15 14 13 12 11! 16 15 14 13 12 11!

11! 11!

11!

16 15 14 13 12 524,160

c. 100! 100 99! 100 99!

99! 99!

99!

100

4. 7 47! 7! 7 6 5 4 3! 7 6 5 4 3!

(7 4)! 3! 3!P

3!

7 6 5 4 840

5. 9 59! 9! 9 8 7 6 5 4! 9 8 7 6 5 4!

(9 5)! 4! 4!P

4!

9 8 7 6 5 15,120

6. There a 7 letters with 2 O’s and 3 S’s. Thus, ! 7! 7 6 5 4 3!

! ! 2!3!

n

p q

2 1 3!

420


1. factorial; 5; 1; 1

2. !

!

n

n r

3. !

! !

n

p q

4. false

5. false

6. true

7. false

Exercise Set 11.2

1. The number of choices decreases by 1 each time a performer is selected.

1st Performer : 2nd Performer: 3rd Performer: 4th Performer: 5th Performer: 6th Performer:

6 5 4 3 2 1 720

2. The number of choices decreases by 1 each time a singer is selected.

1st Singer: 2nd Singer: 3rd Singer: 4th Singer: 5th Singer:

1205 4 3 2 1



3. The number of choices decreases by 1 each time a sentence is selected.

1st Sentence: 2nd Sentence: 3rd Sentence: 4th Sentence: 5th Sentence:

5 4 3 2 1 120

4. The number of choices decreases by 1 each time a suspect is selected.

1st Person : 2nd: 3rd Person: 4th: 5th Person: 6th: 7th Person: 8th:

8 7 6 5 4 3 2 1 40,320

5. There is only one choice for the 6th performer. The number of choices decreases by 1 each time a performer is selected.

1st Performer : 2nd Performer: 3rd Performer: 4th Performer: 5th Performer: 6th Performer:

5 4 3 2 1 1 120

6. There is only one choice for the 5th singer. The number of choices decreases by 1 each time a singer is selected.

1st Singer: 2nd Singer: 3rd Singer: 4th Singer: 5th Singer:

244 3 2 1 1


1st Book: 2nd: 3rd Book: 4th: 5th Book: 6th: 7th Book: 8th: 9th Book:

9 8 7 6 5 4 3 2 1 362,880


1st Photo: 2nd: 3rd Photo: 4th: 5th Photo: 6th: 7th Photo: 8th: 9th Photo: 10th:

10 9 8 7 6 5 4 3 2 1 3,628,800

9. There is only one choice each for the first and last sentences. For the other values, the number of choices decreases by 1 each time a sentence is selected.


1 3 2 1 1 6

10. There is only one choice each for the first and second sentences. For the other values, the number of choices decreases by 1 each time a sentence is selected.


1 1 3 2 1 6

11. There are two choices for the first movie and one for the second. There is only one choice for the last movie. This leaves two choices for the third movie and one for the fourth.

G rated Other two movies NC-17 Rated

1st Movie: 2nd Movie: 3rd Movie: 4th Movie: 5th Movie:

2 1 2 1 1 4

12. There is only one choice each for the fourth and fifth seats. For the other values, the number of choices decreases by 1 each time a seat is selected.

food fightercounselor

1st Seat: 2nd Seat: 3rd Seat: 4th Seat: 5th Seat: 6th Seat: 7th Seat:

5 4 3 1 1 2 1 120

13. 9! 9 8 7 6!

9 8 7 5046! 6!



14. 12! 12 11 10!

12 11 13210! 10!

15. 29! 29 28 27 26 25!

25! 25!29 28 27 26

570,024

16. 31! 31 30 29 28!

31 30 29 26,97028! 28!

17. 19! 19 18 17 16 15 14 13 12 11!

11! 11!19 18 17 16 15 14 13 12

3,047,466,240

18. 17! 17 16 15 14 13 12 11 10 9!

9! 9!17 16 15 14 13 12 11 10

980,179,200

19. 600! 600 599!

600599! 599!

20. 700! 700 699!

700699! 699!

21. 104! 104 103 102!

104 103 10,712102! 102!

22. 106! 106 105 104!

106 105 11,130104! 104!

23. 7! – 3! = 5040 –6 = 5034

24. 6! – 3! = 720 – 6 = 714

25. (7 – 3)! = 4! = 4 · 3 · 2 · 1 = 24

26. (6 – 3)! = 3! = 3 · 2 · 1 = 6

27. 12

! 3! 3 2 1 64

28. 45

! 5! 5 4 3 2 1 1209

29. 7! 7! 7 6 5!

7 6 42(7 2)! 5! 5!

30. 8! 8!

(8 5)! 3!

8 7 6 5 4 3!

3!8 7 6 5 4

6720

31. 13! 13!

(13 3)! 10!

13 12 11 10!

10!13 12 11

1716

32. 17! 17!

(17 3)! 14!

17 16 15 14!

14!17 16 15

4080

33. 9 49!

(9 4)!

9!

5!9 8 7 6 5!

5!9 8 7 6

3024

P

34. 7 37!

(7 3)!

7!

4!7 6 5 4!

4!7 6 5

210

P

35. 8 58!

(8 5)!

8!

3!8 7 6 5 4 3!

3!8 7 6 5 4

6720

P



36. 10 410!

(10 4)!

10!

6!10 9 8 7 6!

6!10 9 8 7

5040

P

37. 6 66! 6! 6 5 4 3 2 1

720(6 6)! 0! 1

P

38. 9 99!

(9 9)!

9!

0!9 8 7 6 5 4 3 2 1

1362,880

P

39. 8 08! 8!

1(8 0)! 8!

P

40. 6 06! 6!

1(6 0)! 6!

P

41. 10 310!

(10 3)!

10!

7!10 9 8 7!

7!10 9 8

720

P

42. 7 47!

(7 4)!

7!

3!7 6 5 4 3!

3!7 6 5 4

840

P

43. 13 713!

(13 – 7)!

13!

6!13 12 11 10 9 8 7 6!

6!13 12 11 10 9 8 7

8,648,640

P

44. 20 320!

(20 3)!

20!

17!20 19 18 17!

17!20 19 18

6840

P

45. 6 36!

(6 3)!

6!

3!6 5 4 3!

3!6 5 4

120

P

46. 8 38!

(8 3)!P

8!

5!8 7 6 5!

5!8 7 6

336

47. 9 59!

(9 5)!

9!

4!9 8 7 6 5 4!

4!9 8 7 6 5

15,120

P



48. 7 47!

(7 4)!

7!

3!7 6 5 4 3!

3!7 6 5 4

840

P

49. ! 6! 6 5 4 3 2 1

180! ! 2!2! 2 1 2 1

n

p q

50. ! 7! 7 6 5 4 3 2 1

1260! ! 2!2! 2 1 2 1

n

p q

51. ! 11!

! ! ! ! 3!2!2!2!

n

p q r s

11 10 9 8 7 6 5 4 3!

3! 2 1 2 1 2 1

831,600

52. ! 9! 9 8 7 6 5 4!

! ! ! 4!2!2!

n

p q r

4!

37802 1 2 1

53. ! 7! 7 6 5 4!

! ! 4!2!

n

p q

4!

1052 1

54. ! 7! 7 6 5 4 3 2 1

630! ! ! 2!2!2! 2 1 2 1 2 1

n

p q r

55. ! 8! 8 7 6 5 4!

! ! 4!3!

n

p q

4!

2803 2 1

56. ! 9! 9 8 7 6 5!

! ! 5!3!

n

p q

5!

5043 2 1

63. Because the letter B is repeated in the word BABE, the number of permutations is given by

! 4! 4 3 2 112

! 2! 2 1

n

p

64. makes sense

65. makes sense

66. does not make sense; Explanations will vary. Sample explanation: Since the order does not matter, this situation calls for the combination formula.

67. does not make sense; Explanations will vary. Sample explanation: This situation calls for the formula for permutations of duplicate items.

68. 12 1012!

(12 10)!

12!

2!12 11 10 9 8 7 6 5 4 3 2!

2!12 11 10 9 8 7 6 5 4 3

239,500,800

P

69. Multiply the number of ways to select the two first place horses by the number of orders in which the remaining four horses can finish.

6 2 4 4 15 24 360C P

70. First select 3 out of the 8 jazz groups.

There are 8 38! 8!

336(8 3)! 5!

P

ways to arrange

the 1st, 3rd, and 8th performers. This leaves 13 groups (5 remaining jazz groups and 8 rock groups) to be arranged. There are

13 1313! 13!

13! 6,227,020,800(13 13)! 0!

P

ways

to arrange the remaining performers. The total number of arrangements is found by multiplying these values:

12336 6,227,020,800 2.09 10

71. There are 5! ways to arrange the women, and 5! ways to arrange the men. The total number of arrangements is found by multiplying these values: (5!)(5!) = 120 · 120 = 14,400



72. Multiply the number of choices for each of the four digits: 2, 4 2, 4, 6, 7, 8, 9 2, 4, 6, 7, 8, 9 7, 9


2 6 6 2 144

73. 2! ! ! ( 1)( 2) 3 2 1

( 1)( 2) 3( ( 2))! ( 2)! 2! 2n n

n n n n n nP n n n

n n n n

Check Points 11.3

1. a. The order in which you select the DVDs does not matter. This problem involves combinations.

b. Order matters. This problem involves permutations.

2. 7 37! 7! 7 6 5 4! 7 6 5 4!

(7 3)!3! 4!3! 4! 3 2 1C

4!

7 6 535

3 2 13 2 1

35 such combinations are possible.

3. 16 416! 16! 16 15 14 13 12! 16 15 14 13 12!

(16 4)!4! 12!4! 12! 4 3 2 1C

12!

16 15 14 131820

4 3 2 14 3 2 1

1820 such hands can be dealt.

4. Choose the male bears: 6 26! 6! 6 5 4! 6 5 4!

(6 2)!2! 4!2! 4! 2 1C

4!

3015

22 1

Choose the female bears: 7 37! 7! 7 6 5 4! 7 6 5 4!

(7 3)!3! 4!3! 4! 3 2 1C

4!

21035

63 2 1

Multiply the choices: 15 35 525 There are 525 five-bear collections possible.


1. !

! !

n

n r r

2. !r

3. false

4. false

Exercise Set 11.3

1. Order does not matter. This problem involves combinations.

2. Order matters. This problem involves permutations.


Section 11.3 Combinations



5. 6 56! 6! 6 5!

6(6 5)!5! 1!5! 1 5!

C

6. 8 78! 8! 8 7!

8(8 7)!7! 1!7! 1 7!

C

7. 9 59! 9! 9 8 7 6 5!

126(9 5)!5! 4!5! 4 3 2 1 5!

C

8. 10 610! 10! 10 9 8 7 6!

210(10 6)!6! 4!6! 4 3 2 1 6!

C

9. 11 411! 11! 11 10 9 8 7!

330(11 4)!4! 7!4! 7! 4 3 2 1

C

10. 12 512! 12! 12 11 10 9 8 7!

792(12 5)!5! 7!5! 7! 5 4 3 2 1

C

11. 8 18! 8! 8 7!

8(8 1)!1! 7!1! 7!1

C

12. 7 17! 7! 7 6!

7(7 1)!1! 6!1! 6!1

C

13. 7 77! 7!

1(7 7)!7! 0!7!

C

14. 4 44! 4!

1(4 4)!4! 0!4!

C

15. 30 330! 30! 30 29 28 27!

4060(30 3)!3! 27!3! 27! 3 2 1

C

16. 25 425! 25! 25 24 23 22 21!

12,650(25 4)!4! 21!4! 21! 4 3 2 1

C

17. 5 05! 5!

1(5 0)!0! 5!0!

C

18. 6 06! 6!

1(6 0)!0! 6!0!

C

19. 7! 7! 7 6 5 4!

(7 3)!3!7 3 4!3! 4! 3 2 15! 5! 5 4!

5 4 (5 4)!4! 1!4! 1 4!

357

5

C

C



20. 10! 10! 10 9 8 7!

(10 3)!3!10 3 7!3! 7! 3 2 16! 6! 6 5 4!

6 4 (6 4)!4! 2!4! 2 1 4!

1208

15

C

C

21. 7 37 3

7! 7!7! 7! 7! 7!(7 3)! 4! 0

3! 3! (7 3)!3! 3! 4!3! 4!3! 4!3!

PC

22. 20 220 2

20!20! 20! 20!(20 2)!

02! 2! (20 2)!2! (20 2)!2! (20 2)!2!

PC

23. 3 2

4 3

3! 3!3! 3! 1 3(3 2)! 1!1 1 1 1 1

4! 4! 4! 4 3! 4 4(4 3)! 1!

P

P

24. 5 3

10 4

5! 5!5! 6! 5 4 3 2!(5 3)! 2!1 1 1 1

10! 10! 2! 10!(10 4)! 6!

P

P

6!2! 10 9 8 7 6!

60 831

5040 84

25. 7 3

5 4

7!98! 98 97 96!(7 3)!3!

5!96!(5 4)!4!

C

C

96!

7 6 5 4!7!4!3! 950675!

1!4!

4!3 2 15 4!

1! 4!

359506 9506 7 9506 9499

5

26. 10 3

6 4

10! 10! 10 9 8 7! 10 9 846! 46 45 44!(10 3)!3! 7!3! 7!3! 3 2 146 45 2070 2070

6! 6! 6 5 4! 6 544! 44!(6 4)!4! 2!4! 2!4! 2 1

C

C

10 9 8

2070 8 2070 20623 6 5

27. 4 2 6 1

18 3

4! 6! 4 3 2!4! 6!(4 2)!2! (6 1)!1! 2!2! 5!1!

18! 18!(18 3)!3! 15!3!

C C

C

2!

6 5!

2 1

5!1!

18 17 16 15! 15!

36 3

816 68

3 2 1

28. 5 1 7 2

12 3

5! 7! 5 4!5! 7!(5 1)!1! (7 2)!2! 4!1! 5!2!

12! 12!(12 3)!3! 9!3!

C C

C

4!

7 6 5!

1

5! 2 1

12 11 10 9!

9!

1260 21

2640 44

3 2 1

29. 6 36! 6! 6 5 4 3!

20(6 3)!3! 3!3! 3!3 2 1

C

30. 11 411! 11! 11 10 9 8 7!

330(11 4)!4! 7!4! 7!4 3 2 1

C



31. 12 412! 12! 12 11 10 9 8!

495(12 4)!4! 8!4! 8!4 3 2 1

C

32. 14 614! 14! 14 13 12 11 10 9 8!

3003(14 6)!6! 8!6! 8!6 5 4 3 2 1

C

33. 17 817! 17! 17 16 15 14 13 12 11 10 9!

24,310(17 8)!8! 9!8! 9!8 7 6 5 4 3 2 1

C

34. 100 18

19

100! 100!

(100 18)!18! 82!18!

100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82!3.07 10

82!18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

C

35. 53 653! 53! 53 52 51 50 49 48 47!

22,957,480(53 6)!6! 47!6! 47!6 5 4 3 2 1

C

36. 59 659! 59! 59 58 57 56 55 54 53!

45,057,474(59 6)!6! 53!6! 53!6 5 4 3 2 1

C

37. Choose the men: 7 47! 7! 7 6 5 4!

35(7 4)!4! 3!4! 3 2 1 4!

C

Choose the women: 7 57! 7! 7 6 5!

21(7 5)!5! 2!5! 2 1 5!

C

Multiply the choices: 35 · 21 = 735

38. Choose the professors: 5 25! 5! 5 4 3!

10(5 2)!2! 3!2! 3!2 1

C

Choose the students: 15 1015! 15! 15 14 13 12 11 10!

3003(15 10)!10! 5!10! 5 4 3 2 1 10!

C

Multiply the choices: 10 3003 30,030

39. Choose the Republicans: 55 455! 55! 55 54 53 52 51!

341,055(55 4)!4! 51!4! 51! 4 3 2 1

C

Choose the Democrats: 44 344! 44! 44 43 42 41!

13,244(44 3)!3! 41!3! 3 2 1 41!

C

Multiply the choices: 341,055 13,244 4,516,932,420

40. Choose the multiple-choice questions: 10 810! 10! 10 9 8!

45(10 8)!8! 2!8! 2 1 8!

C

Choose the open-ended problems: 5 35! 5! 5 4 3!

10(5 3)!3! 2!3! 2 1 3!

C

Multiply the choices: 45 10 450

41. 6 46!

6 5 4 3 3602!

P ways

42. 40 840!

76,904,68532!8!

C selections



43. 13 613! 13 12 11 10 9 8

7!6! 6 5 4 3 2 1=1716 ways

C

44. 50 350!

50 49 48 177,60047!

P ways

45. 20 320! 20 19 18

114017!3! 3 2 1

C

ways

46. 50 350! 50 49 48

19,60047!3! 3 2 1

C

ways

47. 7 47!

8403!

P passwords

48. 9 59!

9 8 7 6 5 15,1204!

P ways

49. 15 315!

15 14 13 273012!

P cones

50. 31 331! 31 30 29

449528!3! 3 2 1

C

bowls

51. 5 25! 5 4

103!2! 2 1

C

outcomes

52. 6 36! 6 5 4

203!3! 3 2 1

C

groups

53. 3 2 2 12 outcomes

54. 4 24! 4 3

62!2! 2 1

C

outcomes

55. 5 35! 5 4

102!3! 2 1

C

outcomes

56. 4 34! 4

41!3! 1

C ways

57. Choose the Democrats: 4 24! 4 3

62!2! 2 1

C

Choose the Republicans: 5 25! 5 4

103!2! 2 1

C

Multiply the choices: 6 10 60

58. 6 26! 6 5

154!2! 2 1

C

ways

59. 12 512!

12 11 10 9 8 95,040(12 5)!

P

ways



60. 7 27!

7 6 42(7 2)!

P

outcomes

61. 6 66!

6 5 4 3 2 1 720(6 6)!

P

ways

62. 5 55!

5 4 3 2 1 120(5 5)!

P

ways

63. 6 36! 6! 6 5 4 3!

20(6 3)!3! 3!3! 3 2 1 3!

C

ways

64. 6 26! 6! 6 5 4!

15(6 2)!2! 4!2! 2 1 4!

C

ways

65. 4 44!

4 3 2 1 24(4 4)!

P

ways

66. 5 55!

2 2 5 4 3 2 1 240(5 5)!

P

ways

67. 4 24! 4! 4 3 2!

2 2 2 2 12(4 2)!2! 2!2! 2 1 2!

C

ways

68. 4 34! 4! 4 3!

2 2 2 2 8(4 3)!3! 1!3! 1 3!

C

ways

72. does not make sense; Explanations will vary. Sample explanation: Since order matters, the permutation formula is necessary.

73. does not make sense; Explanations will vary. Sample explanation: Since order matters, the permutation formula is necessary.

74. makes sense

75. makes sense

77. Selections for 6/53 lottery: 53 653! 53! 53 52 51 50 49 48 47!

22,957,480(53 6)!6! 47!6! 47!6 5 4 3 2 1

C

Selections for 5/36 lottery: 36 536! 36! 36 35 34 33 32 31!

376,992(36 5)!5! 31!5! 31!5 4 3 2 1

C

The 5/36 lottery is easier to win because there are fewer possible selections.



78. 6

! !6 [apply the cross products principle]

( )! ( )! !

!( )! ! 6 !( )! [divide both sides by !( )!]

!( )!

n r n rP C

n n

n r n r r

n n r r n n r n n r

n n r

!

!( )!

r

n n r6 !( )!n n r

!( )!n n r

! 6

! 3 2 1

3

r

r

r

No, there is not enough information to determine the value of n. The number of permutations will be six times the number of combinations for all values of n when 3r .

79. For a group of 20 people:

20 220! 20! 20 19 18!

190 handshakes(20 2)!2! 18!2! 18!2 1

C

Time 3 190 570 seconds, which gives 570 60 9.5 minutes.

For a group of 40 people:

40 240! 40! 40 39 38!

780 handshakes(40 2)!2! 38!2! 38!2 1

C

Time 3 780 2340 seconds, which gives 2340 60 39 minutes.

80. Since there are 5 defective phones, we must select 4 out of the 15 good phones.

15 415! 15! 15 14 13 12 11!

1365(15 4)!4! 11!4! 11!4 3 2 1

C

Check Points 11.4

1. a. The event of getting a 2 can occur in one way. number of ways a 2 can occur 1

(2)total number of possible outcomes 6

P

b. The event of getting a number less than 4 can occur in three ways: 1, 2, 3. number of ways a number less than 4 can occur 3 1

(less than 4)total number of possible outcomes 6 2

P

c. The event of getting a number greater than 7 cannot occur. number of ways a number greater than 7 can occur 0

(greater than 7) 0total number of possible outcomes 6

P

The probability of an event that cannot occur is 0.

d. The event of getting a number less than 7 can occur in six ways: 1, 2, 3, 4, 5, 6. number of ways a number less than 7 can occur 6

(less than 7) 1total number of possible outcomes 6

P

The probability of any certain event is 1.

Section 11.4 Fundamentals of Probability


2. a. number of ways a ace can occur 4 1

(ace)total number of possibilities 52 13

P

b. number of ways a red card can occur 26 1

(red card)total number of possible outcomes 52 2

P

c. number of ways a red king can occur 2 1

(red king)total number of possible outcomes 52 26

P

3. The table shows the four equally likely outcomes. The Cc and cC children will be carriers who are not actually sick. number of ways or can occur 2 1

(carrier, not sick) ( )total number of possible outcomes 4 2

Cc cCP P Cc

4. a. number of persons never married 74

(never married) 0.31total number of U.S. adults 242

P

b. number of males 118

(male) 0.49total number of U.S. adults 242

P


1. sample space

2. P E ; number of outcomes in E; total number of possible outcomes

3. 52; hearts; diamonds; clubs; spades

4. empirical

5. true

6. false

7. true

8. false

Exercise Set 11.4

1. number of ways a 4 can occur 1


P

2. number of ways a five can occur 1


P

3. number of ways an odd number can occur 3 1

(odd number)total number of possible outcomes 6 2

P



4. number of ways a number of greater than 3 can occur 3 1

(greater than 3)total number of possible outcomes 6 2

P

5. number of ways a number less than 3 can occur 2 1


P

6. number of ways a number greater than 4 can occur 2 1

(greater than 4)total number of possible outcomes 6 3

P

7. number of ways a number less than 20 can occur 6


P



P

9. number of ways a number greater than 20 can occur 0


P



P

11. number of ways a queen can occur 4 1

(queen)total number of possibilities 52 13

P

12. number of ways a jack can occur 4 1

( jack)total number of possibilities 52 13

P

13. number of ways a club can occur 13 1

(club)total number of possibilities 52 4

P

14. number of ways a diamond can occur 13 1

(diamond)total number of possibilities 52 4

P

15. number of ways a picture card can occur 12 3

(picture card)total number of possibilities 52 13

P

16. number of ways a card greater than 3 and less than 7 can occur 12 3

(greater than 3 and less than 7)total number of possibilities 52 13

P

17. number of ways a queen of spades can occur 1

(queen of spades)total number of possibilities 52

P

18. number of ways an ace of clubs can occur 1

(ace of clubs)total number of possibilities 52

P

19. number of ways a diamond and a spade can occur 0

(diamond and spade) 0total number of possibilities 52

P



20. number of ways a card with a green heart can occur 0

(green heart) 0total number of possibilities 52

P

21. number of ways two heads can occur 1

(two heads)total number of possibilities 4

P

22. number of ways two tails can occur 1

(two tails)total number of possibilities 4

P

23. number of ways the same outcome on each toss can occur 2 1

(same on each toss)total number of possibilities 4 2

P

24. number of ways different outcomes on each toss can occur 2 1

(different on each toss)total number of possibilities 4 2

P

25. number of ways a head on the second toss can occur 2 1

(head on second toss)total number of possibilities 4 2

P

26. number of ways at least one head can occur (HH, HT, TH) 3

(at least one head)total number of possibilities 4

P

27. number of ways exactly one female child can occur 3

(exactly one female child)total number of possibilities 8

P

28. number of ways exactly one male child can occur 3

(exactly one male child)total number of possibilities 8

P

29. number of ways exactly two male children can occur 3

(exactly two male children)total number of possibilities 8

P

30. number of ways exactly two female children can occur 3

(exactly two female children)total number of possibilities 8

P

31. number of ways at least one male child can occur 7

(at least one male child)total number of possiblities 8

P

32. number of ways at least two female children can occur 4 1

(at least two female children)total number of possibilities 8 2

P

33. number of ways four male children can occur 0

(four male children) 0total number of possibilities 8

P

34. number of ways fewer than four female children can occur 8

(fewer than four female children) 1total number of possibilities 8

P

35. number of ways two even numbers can occur 9 1

(two even numbers)total number of possibilities 36 4

P



36. number of ways two odd numbers can occur 9 1

(two odd numbers)total number of possibilities 36 4

P

37. number of ways two numbers whose sum is 5 can occur 4 1

(two numbers whose sum is 5)total number of possibilities 36 9

P

38. number of ways two numbers whose sum is 6 can occur 5

(two numbers of whose sum is 6)total number of possibilities 36

P

39. (two numbers whose sum exceeds 12)

number of ways two numbers whose sum exceeds 12 can occur 0= 0

total number of possibilities 36

P

40. (two numbers whose sum is less than 13)

number of ways two numbers whose sum is less than 13 can occur 361

total number of possibilities 36

P

41. number of ways a red region can occur 3

(red region)total number of possibilities 10

P

42. number of ways a yellow region can occur 2 1

(yellow region)total number of possibilities 10 5

P

43. number of ways a blue region can occur 2 1

(blue region)total number of possibilities 10 5

P

44. number of ways a brown region can occur 3

(brown region)total number of possibilities 10

P

45. number of ways a region that is red or blue can occur 5 1

(region that is red or blue)total number of possibilities 10 2

P

46. number of ways a region that is yellow or brown can occur 5 1

(region that is yellow or brown)total number of possibilities 10 2

P

47. number of ways a region that is red and blue can occur 0

(region that is red and blue) 0total number of possibilities 10

P

48. number of ways a region that is yellow and brown can occur 0

(region that is yellow and brown) 0total number of possibilities 10

P

49. number of ways sickle cell anemia can occur 1

(sickle cell anemia)total number of possibilities 4

P

50. number of ways sickle cell trait can occur 2 1

(sickle cell trait)total number of possibilities 4 2

P

51. number of ways a healthy child can occur 1

(healthy)total number of possibilities 4

P



52. number of ways sickle cell anemia can occur 0

(sickle cell anemia) 0total number of possibilities 4

P

53. number of ways sickle cell trait can occur 2 1

(sickle cell trait)total number of possibilities 4 2

P

54. number of ways a healthy child can occur 2 1

(healthy)total number of possibilities 4 2

P

55. number of males 12.5

(male) 0.43total number of Americans living alone 29.3

P

56. number of females 16.8

(female) 0.57total number of Americans living alone 29.3

P

57. number in 25 34 age range 3.8

(25 34 age range) 0.13total number of Americans living alone 29.3

P

58. number in 35 44 age range 4.2

(35 44 age range) 0.14total number of Americans living alone 29.3

P

59. number of women in 15 24 age range 0.8

(woman in 15 24 age range) 0.03total number of Americans living alone 29.3

P

60. number of men in 45 64 age range 4.3

(man in 45 64 age range) 0.15total number of Americans living alone 29.3

P

Table For #61–66

Moved to Same State

Moved to Different

State

Moved to Different Country

Total

Owner 11.7 2.8 0.3 14.8 Renter 18.7 4.5 1.0 24.2 Total 30.4 7.3 1.3 39.0

61. number of owners 14.8

(owner) 0.38total number of Americans who moved in 2004 39.0

P

62. number of renters 24.2

(renter) 0.62total number of Americans who moved in 2004 39.0

P

63. number that moved within state 30.4

(moved within state) 0.78total number of Americans who moved in 2004 39.0

P

64. number that moved to different country 1.3

(moved to different country) 0.03total number of Americans who moved in 2004 39.0

P

65. number of renters who moved to a different state 4.5

(renter who moved to different state) 0.12total number of Americans who moved in 2004 39.0

P



66. number of owners who moved to a different state 2.8

(owner who moved to different state) 0.07total number of Americans who moved in 2004 39.0

P

Table For #67–70

Less Than 4 Years

High School

4 Years High School

Only

Some College

(Less than 4 years)

4 Years College (or

More) Total

Male 3.6 5.0 2.6 3.9 15.1 Female 5.2 8.0 4.1 3.0 20.3 Total 8.8 13.0 6.7 6.9 35.4

67. number with less than 4 years of high school 29 1

(had less than 4 years of high school)total number of Americans aged 25 and older 174 6

P

68. number with 4 years of high school only 56 28

(had 4 years of high school only)total number of Americans aged 25 and older 174 87

P

69. number of women with 4 years of college or more 22 11

(a woman with 4 years of college or more)total number of Americans aged 25 and older 174 87

P

70. number of men with 4 years of college or more 23

(a man with 4 years of college or more)total number of Americans aged 25 and older 174

P

79. does not make sense; Explanations will vary. Sample explanation: Even if there are only two choices, it does not necessarily follow that they are equally likely to be selected.

80. does not make sense; Explanations will vary. Sample explanation: The probability cannot be greater than 1 (100%).

81. makes sense

82. makes sense

83. The area of the target is 2 2(12 in.) 144in.

The area of the yellow region is 2 2 2 2(9 in.) (6 in.) (3 in.) 54 in.

The probability that the dart hits a yellow region is 2

2

54 in.0.375

144 in.

84. First count the number of three-digit numbers that read the same forward and backward: Same as 1st digit1 9 0 9


9 10 1 90

(three-digit number reads the same forward and backward)

number of three-digit numbers that read the same forward and backward 90 1

total number of three-digit numbers 900 10

P

Section 11.5 Probability with the Fundamental Counting Principle, Permutations, and Combinations


Check Points 11.5

1. total number of permutations = 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720 For the given outcome there is 1 choice (first name beginning with G) for the first joke, which would leave 4 choices for the last joke (the 4 remaining men). The remaining jokes have 4, 3, 2, and 1 choice respectively.

4 jokes other than first and lastFirst name begins with G man (other than first joke)

1st: 2nd: 3rd: 4th: 5th: 6th:

1 4 3 2 1 4 96

96 2(first joke is by a man whose name begins with G and the last is by a man)

720 15P

2. Total number of Powerball selections: 59 559!

35 35(59 5)!5!

59!35

54!5!59 58 57 56 55 54!

3554!5 4 3 2 1

5,006,386 35

175,223,510

C

Number of selections that match 4 out of 5 white balls and the Powerball:

match 4 of the 5 any 1 of the 54

selected white balls non-selected white balls the Powerball

5 4 54 1 1 270C C

6

270(matching 4 of the 5 white balls and the powerball)

175,223,510

27

17,522,351

0.000001541

1.541 10

P

3. total number of combinations: 10 310! 10! 10 9 8 7!

120(10 3)!3! 7!3! 7!3 2 1

C

a. total number of combinations of 3 men: 6 36! 6! 6 5 4 3!

20(6 3)!3! 3!3! 3 2 1 3!

C

number of combinations with 3 men 20 1(3 men)

total number of combinations 120 6P

b. Select 2 out of 6 men: 6 26! 6! 6 5 4!

15(6 2)!2! 4!2! 4!2 1

C

Select 1 out of 4 women: 4 14! 4! 4 3! 4 3!

(4 1)!1! 3!1! 3!C

3!

4

total number of combinations of 2 men and 1 woman: 15 4 60 number of combinations with 2 men, 1 woman 60 1

(2 men, 1 woman)total number of combinations 120 2

P




1. permutations; the total number of possible permutations

2. 1; combinations

3. true

4. false

Exercise Set 11.5

1. a. 5! = 5 · 4 · 3 · 2 · 1 = 120

b.

Lee, Nancy, PaulMartha Armando

1st: 2nd: 3rd: 4th: 5th:

1 3 2 1 1 6

c. 6 1

(Martha first and Armando last)120 20

P

2. a. 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

b. 1st woman: 1st man: 2nd woman: 2nd man: 3rd woman: 3rd man:

3 3 2 2 1 1 36

c. 36 1

(first person is a woman and line alternates by gender)720 20

P

3. a. total number of permutations = 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720 number of permutations with E first = 1 · 5 · 4 · 3 · 2 · 1 = 120

number of permutations with E first 120 1(E first)

total number of permutations 720 6P

b. number of permutations with C fifth and B last = 4 · 3 · 2 · 1 · 1 · 1 = 24 number of permutations with C fifth and B last 24 1

(C fifth and B last)total number of permutations 720 30

P

c. number of permutations with order D, E, C, A, B, F 1

(D, E, C, A, B, F)total number of permutations 720

P

d. number of permutations with A or B first = 2 · 5 · 4 · 3 · 2 · 1 = 240 number of permutations with A or B first 240 1

(A or B first)total number of permutations 720 3

P

4. a. total number of permutations = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040 number of permutations with D first = 1 · 6 · 5 · 4 · 3 · 2 · 1 = 720

number of permutations with D first 720 1(D first)

total number of permutations 5040 7P

b. number of permutations with E sixth and B last = 5 · 4 · 3 · 2 · 1 · 1 · 1 = 120 number of permutations with E sixth and B last 120 1

(E sixth and B last)total number of permutations 5040 42

P



c. number of permutations with order C, D, B, A, G, F, E 1

(C, D, B, A, G, F, E)total number of permutations 5040

P

d. number of permutations with F or G first = 2 · 6 · 5 · 4 · 3 · 2 · 1 = 1440 number of permutations with F or G first 1440 2

(F or G first)total number of permutations 5040 7

P

5. a. 9 39! 9! 9 8 7 6!

84(9 3)!3! 6!3! 6!3 2 1

C

b. 5 35! 5! 5 4 3!

10(5 3)!3! 2!3! 2 1 3!

C

c. number of ways to select 3 women 10 5

(all women)total number of possible combinations 84 42

P

6. a. 11 411! 11! 11 10 9 8 7!

330(11 4)!4! 7!4! 7!4 3 2 1

C

b. 6 46! 6! 6 5 4!

15(6 4)!4! 2!4! 2 1 4!

C

c. number of ways to select 4 Republicans 15 1

(all Republicans)total number of possible combinations 330 22

P

7. 56 556! 56! 56 55 54 53 52 51!

46 46 46 46 3,819,816 46 175,711,536(56 5)!5! 51!5! 51!5 4 3 2 1

C

number of ways of winning 1(winning)=

total number of possible combinations 175,711,536P

8. Total number of combinations:

56 556! 56! 56 55 54 53 52 51!

46 46 46 46 3,819,816 46 175,711,536(56 5)!5! 51!5! 51!5 4 3 2 1

C

Number of selections that match 4 out of 5 white balls and the gold Mega Ball:


selected white balls non-selected white balls gold Mega Ball

5 4 51 1 1 5 51 1 255C C

255 85(matching 4 of the 5 white balls and the gold Mega Ball)

175,711,536 58,570,512P


56 556! 56! 56 55 54 53 52 51!

46 46 46 46 3,819,816 46 175,711,536(56 5)!5! 51!5! 51!5 4 3 2 1

C




5 3 51 2 1 10 1275 1 12,750C C

12,750 2125(matching 3 of the 5 white balls and the gold Mega Ball)

175,711,536 29,285,256P




56 556! 56! 56 55 54 53 52 51!

46 46 46 46 3,819,816 46 175,711,536(56 5)!5! 51!5! 51!5 4 3 2 1

C




5 2 51 3 1 10 20,825 1 208,250C C

208,250 14,875(matching 2 of the 5 white balls and the gold Mega Ball)

175,711,536 12,550,824P

11. a. 25 625! 25! 25 24 23 22 21 20 19!

177,100(25 6)!6! 19!6! 19!6 5 4 3 2 1

C

number of ways to choose 6 defective transistors 1(all are defective)= 0.00000565

total number of possible combinations 177,100P

b. 19 619! 19! 19 18 17 16 15 14 13!

27,132(19 6)!6! 13!6! 13!6 5 4 3 2 1

C

number of ways to choose 6 good transistors 27,132 969(none are defective)= 0.153

total number of possible permutations 177,100 6325P

12. a. 13 513! 13! 13 12 11 10 9 8!

1287(13 5)!5! 8!5! 8!5 4 3 2 1

C

6 56! 6! 6 5!

6(6 5)!5! 1!5! 1 5!

C

number of ways to select 5 lawyers 6 2(all lawyers)= 0.00466

total number of possible combinations 1287 429P

b. 7 57! 7! 7 6 5!

21(7 5)!5! 2!5! 2 1 5!

C

number of ways to select 5 teachers 21 7(none are lawyers)= 0.0163


13. total number of possible combinations: 10 310! 10! 10 9 8 7!

120(10 3)!3! 7!3! 7!3 2 1

C

number of ways to select one Democrat: 6 16! 6! 6 5!

6(6 1)!1! 5!1! 5!1

C

number of ways to select two Republicans: 4 24! 4! 4 3 2!

6(4 2)!2! 2!2! 2!2 1

C

number of ways to select one Democrat and two Republicans: 6 1 4 2 6 6 36C C

36 3(one Democrat and two Republicans) 0.3

120 10P



14. total number of possible combinations: 20 420! 20! 20 19 18 17 16!

4845(20 4)!4! 16!4! 16!4 3 2 1

C

number of ways to select two parents: 15 215! 15! 15 14 13!

105(15 2)!2! 13!2! 13!2 1

C

number of ways to select two teachers: 5 25! 5! 5 4 3!

10(5 2)!2! 3!2! 3!2 1

C

number of ways to select two parents and two teachers: 15 2 5 2 105 10 1050C C

1050 70(two parents and two teachers) 0.217

4845 323P

15. a. 52 552! 52! 52 51 50 49 48 47!

2,598,960(52 5)!5! 47!5! 47!5 4 3 2 1

C

b. 13 513! 13! 13 12 11 10 9 8!

1287(13 5)!5! 8!5! 8!5 4 3 2 1

C

c. number of possible 5-card diamond flushes 1287

(diamond flush)= 0.000495total number of possible combinations 2,598,960

P

16. a. 52 552! 52! 52 51 50 49 48 47!

2,598,960(52 5)!5! 47!5! 47!5 4 3 2 1

C

b. 4 44! 4!

1(4 4)!4! 0!4!

C

c. 4 14! 4! 4 3!

4(4 1)!1! 3!1! 3!1

C

d. 4 4 4 1 1 4 4C C

e. number of hands with 4 aces and 1 king 4

(4 aces and 1 king)= 0.00000154total number of possible combinations 2,598,960

P

17. total number of possible combinations: 52 352! 52! 52 51 50 49!

22,100(52 3)!3! 49!3! 49!3 2 1

C

number of ways to select 3 picture cards: 12 312! 12! 12 11 10 9!

220(12 3)!3! 9!3! 9!3 2 1

C

220 11(3 picture cards) 0.00995

22,100 1105P


270,725(52 4)!4! 48!4! 48!4 3 2 1

C

number of ways to select 4 hearts: 13 413! 13! 13 12 11 10 9!

715(13 4)!4! 9!4! 9!4 3 2 1

C

715 11(all 4 are hearts) 0.00264

270,725 4165P




270,725(52 4)!4! 48!4! 48!4 3 2 1

C

number of ways to select 2 queens: 4 24! 4! 4 3 2!

6(4 2)!2! 2!2! 2!2 1

C

number of ways to select 2 kings: 4 2 6C

number of ways to select 2 queens and 2 kings: 4 2 4 2 6 6 36C C

36(2 queens and 2 kings) 0.000133

270,725P


270,725(52 4)!4! 48!4! 48!4 3 2 1

C

number of ways to select 3 jacks: 4 34! 4! 4 3!

4(4 3)!3! 1!3! 1 3!

C

number of ways to select 1 queen: 4 14! 4! 4 3!

4(4 1)!1! 3!1! 3!1

C

number of ways to select 3 jacks and 1 queen: 4 3 4 1 4 4 16C C

16(3 jacks and 1 queen) 0.0000591

270,725P

24. makes sense

25. does not make sense; Explanations will vary. Sample explanation: Each possible combination is equally likely.

26. makes sense

27. makes sense

28. total number of possible combinations: 3 · 2 · 2 · 3 = 36 number of combinations the person wants: 1 · 1 · 1 · 2 = 2

number of combinations the person wants 2 1(what the person wants is available)


29. Other players could also purchase the winning combination and share the prize money, possibly making this person’s share of the prize less than what this person paid for the tickets.

30. total number of possible combinations: Digit 1: Digit 2: Digit 3:

5 4 3 60

number of even numbers greater than 500:

1, 3, and 2 or 45 2 or 4


1 3 2 6

number of even numbers greater than 500 6 1(even and greater than 500)


Section 11.6 Events Involving Not and Or; Odds


31. total number of possible combinations: 52 552! 52! 52 51 50 49 48 47!

2,598,960(52 5)!5! 47!5! 47!5 4 3 2 1

C

number of ways to select one ace: 4 14! 4! 4 3!

4(4 1)!1! 3!1! 3!1

C

Note: one card is an ace, so the other four must not be aces. number of ways to select 4 cards with no face cards and no aces:

36 436! 36! 36 35 34 33 32!

58,905(36 4)!4! 32!4! 32!4 3 2 1

C

number of hands with one ace and no face cards: 4 1 36 4 4 58,905 235,620C C

number of hands with one ace and no face cards 235,620(one ace and no face cards) 0.0907

total number of possible combinations 2,598,960P

Check Points 11.6

1. 13 39 3

(not a diamond) 1 (diamond) 1–52 52 4

P P

2. a. 31 160

(not 50 59) 1 (50 59) 1–191 191

P P

b. 9 182

(at least 20 years old) 1 (less than 20 years) 1–191 191

P P

3. 1 1 2 1

(4 or 5) (4) (5)6 6 6 3

P P P

4. 23 11 7 27

(math or psychology) (math) (psychology) (math and psychology)50 50 50 50

P P P P

5. 4 4 2 6 3

(odd or less than 5) (odd) (less than 5) (odd and less than 5)8 8 8 8 4

P P P P

6. a. These events are not mutually exclusive. (married or female)= (married) (female) (married and female)

130 124 65

242 242 242189

0.78242

P P P P

b. These events are mutually exclusive. (divorced or widowed)= (divorced) (widowed)

24 14

242 24238

24219

0.16121

P P P



7. There are 2 red queens. Number of favorable outcomes = 2, Number of unfavorable outcomes = 50

a. Odds in favor of getting a red queen are 2 to 50 or 2:50 which reduces to 1:25.

b. Odds against getting a red queen are 50 to 2 or 50:2 which reduces to 25:1.

8. number of unfavorable outcomes = 995, number of favorable outcomes = 5 Odds against winning the scholarship are 995 to 5 or 995:5 which reduces to 199:1.

9. number of unfavorable outcomes = 15, number of favorable outcomes = 1 Odds in favor of the horse winning the race are 1 to 15

1 1(the horse wins race) 0.0625 or 6.3%.

1 15 16P


1. 1 P E ; 1 not P E

2. mutually exclusive; P A P B

3. and P A P B P A B

4. E will occur; E will not occur

5. aP E

a b

6. false

7. false

8. true

9. false

Exercise Set 11.6

1. 4 48 12

(not an ace) 1 (ace) 1–52 52 13

P P

2. 4 1 12

(not a 3) 1 (3) 1 152 13 13

P P

3. 13 39 3

(not a heart) 1– (heart) 152 52 4

P P

4. 13 39 3

(not a club) 1 (club) 152 52 4

P P

5. 12 40 10

(not a picture card) 1 (picture card) 152 52 13

P P



6. 6 46 23

(not a red picture card) 1 (red picture card) 152 52 26

P P

7. 36 2,598,924

(not a straight flush) 1 (straight flush) 1 0.9999862,598,960 2,598,960

P P

8. 624 2,598,336

(not four of a kind) 1 (four of a kind) 1 0.9997602,598,960 2,598,960

P P

9. 3744 2,595,216

(not a full house) 1 (full house) 1 0.9985592,598,960 2,598,960

P P

10. 5108 2,593,852

(not a flush) 1 (flush) 1 0.9980352,598,960 2,598,960

P P

11. a. 0.10 (read from graph)

b. 1.00 0.10 0.90

12. a. 0.78 (read from graph)

b. 1.00 0.78 0.22

13. 1080 1920 16

(not age 25 44) 1 (age 25 44) 13000 3000 25

P P

14. 840 2160 18

(not age 45 64) 1 (age 45 64) 13000 3000 25

P P

15. 180 2820 47

(age less than 65) 1 (age 65 74) 13000 3000 50

P P

16. 900 2100 7

(age at least 25) 1 (age 12 24) 13000 3000 10

P P

17. 4 4 8 2

(2 or 3) (2) (3)52 52 52 13

P P P

18. 4 4 8 2

(7 or 8) (7) (8)52 52 52 13

P P P

19. 2 2 4 1

(red 2 or black 3) (red 2) (black 3)52 52 52 13

P P P

20. 2 2 4 1

(red 7 or black 8) (red 7) (black 8)52 52 52 13

P P P

21. 1 1 2 1

(2 of hearts or 3 of spades) (2 of hearts) (3 of spades)52 52 52 26

P P P

22. P(7 or hearts or 8 of spades)1 1 2 1

(7 of hearts)+ (8 of spades)52 52 52 26

P P



23. 8 10 18 9

(professor or instructor) (professor) (instructor)44 44 44 22

P P P

24. 8 4 12

(Independent or Green) (Independent) (Green)67 67 67

P P P

25. 3 4 2 5

(even or less than 5) (even) (less than 5) (even and less than 5)6 6 6 6

P P P P

26. 3 3 2 4 2

(odd or less than 4) (odd) (less than 4) (odd and less than 4) –6 6 6 6 3

P P P P

27. 4 26 2 28 7

(7 or red) (7) (red) (red 7)52 52 52 52 13

P P P P

28. 4 26 2 28 7

(5 or black) (5) (black) (black 5)52 52 52 52 13

P P P P

29. 13 12 3 22 11

(heart or picture card) (heart) (picture card) (heart and picture card)52 52 52 52 26

P P P P

30. (greater than 2 and less than 7, or diamond)

(greater than 2 and less than 7) (diamond) (diamond greater than 2 and less than 7)

16 13 4 25

52 52 52 52

P

P P P

31. 4 5 3 6 3

(odd or less than 6) (odd) (less than 6) (odd and less than 6)8 8 8 8 4

P P P P

32. 4 5 2 7

(odd or greater than 3) (odd)+ (greater than 3) (odd and greater than 3)8 8 8 8

P P P P

33. 4 3 2 5

(even or greater than 5) (even) (greater than 5) (even and greater than 5)8 8 8 8

P P P P

34. 4 3 1 6 3

(even or less than 4) (even) (less than 4) (even and less than 4)8 8 8 8 4

P P P P

35. 19 22 8 33

(professor or male) (professor) (male) (male professor)40 40 40 40

P P P P

36. 19 18 11 26 13

(professor or female) (professor) (female) (female professor)40 40 40 40 20

P P P P

37. 21 18 7 32 4

(teach. assist. or female) (teach. assist.) (female) (female teach. assist.)40 40 40 40 5

P P P P

38. 21 22 14 29

(teaching assistant or male) (teach. assist.) (male) – (male teach. assist.)40 40 40 40

P P P P



39. (Democrat or business major) (Democrat) (business major) (Democrat and business major)

29 11 5 35 7= –

50 50 50 50 10

P P P P

40. 135 85 65 155 31

(math or english) (math) (english) (math and english)200 200 200 200 40

P P P P

41. 45 129 43

(not completed 4 years or more) 1 (completed 4 years or more) 1174 174 58

P P

42. 29 1

(not completed 4 years of high school)174 6

P

43. (completed 4 years of high school only or less than four years of college)

(completed 4 years of high school only) (less than four years of college)

56 44 100 50

174 174 174 87

P

P P

44. (completed less than 4 years of high school or 4 years of high school only)

(completed less than 4 years of high school) (4 years of high school only)

29 56 85

174 174 174

P

P P

45. (completed 4 years of high school only or is a man)

(completed 4 years of high school only) (male) (completed 4 years of high school only and is a man)

56 82 25 113

174 174 174 174

P

P P P

46. (completed 4 years of high school only or is a woman)

(completed 4 years of high school only) (female) (completed 4 years of high school only and is a woman)

56 92 31 117 39

174 174 174 174 58

P

P P P

47. The number that meets the characteristic is 45. The number that does not meet the characteristic is 174 45 129. Odds in favor: 45 to 129 which reduces to 15 to 43 Odds against: 129 to 45 which reduces to 43 to 5

48. The number that meets the characteristic is 29. The number that does not meet the characteristic is 174 29 145. Odds in favor: 29 to 145 which reduces to 1 to 5 Odds against: 145 to 29 which reduces to 5 to 1

49. 490 80 570 85

(not in the Army) 1 (in the Army) 1 11420 1420 142

P P

50. 190 10 200 61

(not in the Marines) 1 (in the Marines) 1 11420 1420 71

P P



51. (in the Navy or a man) (in the Navy) (a man) (in the Navy and a man)

270 50 270 490 190 270 270

1420 1420 1420320 1220 270

1420 1420 14201270

1420127

142

P P P P

52. (in the Army or a woman) (in the Army) (a woman) (in the Army and a woman)

490 80 60 80 10 50 80

1420 1420 1420570 200 80

1420 1420 1420690

142069

142

P P P P

53. 270 60 190 10 530 53

(in the Air Force or the Marines) (in the Air Force) (in the Marines)1420 1420 1420 142

P P P

54. 490 80 270 50 890 89

(in the Army or the Navy) (in the Army) (in the Navy)1420 1420 1420 142

P P P

55. The number that meets the characteristic is 270 + 50 = 320. The number that does not meet the characteristic is 1420 – 320 = 1100. Odds in favor: 320 to 1100 which reduce 16 to 55 Odds against: 1100 to 320 which reduce 55 to 16

56. The number that meets the characteristic is 490 + 80 = 570. The number that does not meet the characteristic is 1420 – 570 = 850. Odds in favor: 570 to 850 which reduce 57 to 85 Odds against: 850 to 570 which reduce 85 to 57

57. The number that meets the characteristic is 10. The number that does not meet the characteristic is 1420 – 10 = 1410. Odds in favor: 10 to 1410 which reduce 1 to 141 Odds against: 1410 to 10 which reduce 141 to 1

58. The number that meets the characteristic is 60. The number that does not meet the characteristic is 1420 – 60 = 1360. Odds in favor: 60 to 1360 which reduce 1 to 68 Odds against: 1360 to 60 which reduce 68 to 1

59. The number that meets the characteristic is 270 + 490 + 190 + 270 = 1220. The number that does not meet the characteristic is 1420 –1220 = 200. Odds in favor: 1220 to 200 which reduce 61 to 10 Odds against: 200 to 1220 which reduce 10 to 61

60. The number that meets the characteristic is 60 + 80 + 10 + 50 = 200. The number that does not meet the characteristic is 1420 – 200= 1220. Odds favor: 200 to 1220 which reduce 10 to 61 Odds in against: 1220 to 200 which reduce 61 to 10



61. number of favorable outcomes = 4, number of unfavorable outcomes = 2 Odds in favor of getting a number greater than 2 are 4:2, or 2:1.

62. number of favorable outcomes = 4, number of unfavorable outcomes = 2 Odds in favor of getting a number less than 5 are 4:2, or 2:1.

63. number of unfavorable outcomes = 2, number of favorable outcomes = 4 Odds against getting a number greater than 2 or 2:4, or 1:2.

64. number of unfavorable outcomes = 2, number of favorable outcomes = 4 Odds against getting a number less than 5 are 2:4, or 1:2.

65. number of favorable outcomes = 9, number of unfavorable outcomes 100 9 91

a. Odds in favor of a child in a one-parent household having a parent who is a college graduate are 9:91.

b. Odds against a child in a one-parent household having a parent who is a college graduate are 91:9.

66. number of favorable outcomes = 29, number of unfavorable outcomes 100 29 71

a. Odds in favor of a child in a two-parent household having parents who are college graduates are 29:71.

b. Odds against a child in a two-parent household having parents who are college graduates are 71:29.

67. number of favorable outcomes = 13, number of unfavorable outcomes = 39 Odds in favor of a heart are 13:39, or 1:3.

68. number of favorable outcomes = 12, number of unfavorable outcomes = 40 Odds in favor of a picture card are 12:40, or 3:10.

69. number of favorable outcomes = 26, number of unfavorable outcomes = 26 Odds in favor of a red card are 26:26, or 1:1.

70. number of favorable outcomes = 26, number of unfavorable outcomes = 26 Odds in favor of a black card are 26:26, or 1:1.

71. number of unfavorable outcomes = 48, number of favorable outcomes = 4 Odds against a 9 are 48:4, or 12:1.

72. number of unfavorable outcomes = 48, number of favorable outcomes = 4 Odds against a 5 are 48:4, or 12:1.

73. number of unfavorable outcomes = 50, number of favorable outcomes = 2 Odds against a black king are 50:2, or 25:1.

74. number of unfavorable outcomes = 50, number of favorable outcomes = 2 Odds against a red jack are 50:2, or 25:1.

75. number of unfavorable outcomes = 47, number of favorable outcomes = 5 Odds against a spade greater than 3 and less than 9 are 47:5.

76. number of unfavorable outcomes = 47, number of favorable outcomes = 5 Odds against a club greater than 4 and less than 9 are 47:5.

77. number of unfavorable outcomes = 980, number of favorable outcomes = 20 Odds against winning are 980:20, or 49:1.

78. number of unfavorable outcomes = 4970, number of favorable outcomes = 30 Odds against winning are 4970:30, or 497:3.



79. The number that meets the characteristic is 18. The number that does not meet the characteristic is 38 – 18 = 20. Odds in favor: 18 to 20 which reduce 9 to 10

80. The number that meets the characteristic is 10. The number that does not meet the characteristic is 38 – 10 = 28. Odds in favor: 10 to 28 which reduce 5 to 14

81. The number that meets the characteristic is 10. The number that does not meet the characteristic is 38 – 10 = 28. Odds against: 28 to 10 which reduce 14 to 5

82. The number that meets the characteristic is 18. The number that does not meet the characteristic is 38 – 18 = 20. Odds against: 20 to 18 which reduce 10 to 9

83. The number that meets the characteristic is 18 + 10 = 28. The number that does not meet the characteristic is 38 – 28 = 10. Odds in favor: 28 to 10 which reduce 14 to 5

84. The number that meets the characteristic is 10 + 10 = 20. The number that does not meet the characteristic is 38 – 20 = 18. Odds in favor: 20 to 18 which reduce 10 to 9

85. The number that meets the characteristic is 10 + 10 = 20. The number that does not meet the characteristic is 38 – 20 = 18. Odds against: 18 to 20 which reduce 9 to 10

86. The number that meets the characteristic is 18 + 10 = 28. The number that does not meet the characteristic is 38 – 28 = 10. Odds against: 10 to 28 which reduce 5 to 14

87. 3 3

(winning)3+4 7

P

88. 3 3

(winning)3+7 10

P

89. 4 4

(miss free throw) 0.16 16%21 4 25

P

In 100 free throws, on average he missed 16, so he made 100 – 16 = 84.

90. 193 193

(still alive at age 70) 41.7%193+270 463

P

91. 1 1

(contracting an airborn illness)1+999 1000

P

92. 1 1

(deep-vein thrombosis)1+28 29

P

100. does not make sense; Explanations will vary. Sample explanation: The two probabilities must add to 1.

101. does not make sense; Explanations will vary. Sample explanation: Since 1 card is a heart and a king, the probability is 4 13 1 16 4

52 52 52 52 13 .

102. does not make sense; Explanations will vary. Sample explanation: The probability of selecting a king or a heart is 4 13 1 16 4

52 52 52 52 13 . The probability of selecting the king of hearts is

1.

52



103. does not make sense; Explanations will vary. Sample explanation: The odds are more likely 1:9.

104. a. (Democrat who is not a business major)P

1 (not a Democrat or Democrat and business major)

=1 [ (not a Democrat) (Democrat and business major)]

21 5=1

50 50

261

5024

5012

25

P

P P

b. P(neither Democrat nor business major) 1 (Democrat or business major)

1 [ (Democrat)+ (business major)– (Democrat and business major)]

29 11 51

50 50 50

351

5015

503

10

P

P P P

105. (driving intoxicated or driving accident)

= (driving intoxicated) (driving accident) (driving accident while intoxicated)

P

P P P

Substitute the three given probabilities and solve for the unknown probability: 0.35 0.32 0.09 (driving accident while intoxicated)

(driving accident while intoxicated) 0.32 0.09 0.35

(driving accident while intoxicated) 0.06

P

P

P

106. ( )

1 ( )

P E a

P E b

1 ( ) 1 ( )( )

1 ( ) 1 1

b P E b P EP E a

P E b

( ) 1 ( )

( ) ( )

( ) ( )

( )( )

( )( )

( )

bP E a P E

bP E a aP E

aP E bP E a

P E a b a

P E a b a

a b a ba

P Ea b



Check Points 11.7

1. 2 2 1 1 1

(green and green) (green) (green) 0.0027738 38 19 19 361

P P P

2. 1 1 1 1 1

(4 boys in a row) (boy and boy and boy and boy)= (boy) (boy) (boy) (boy)2 2 2 2 16

P P P P P P

3. a. 5 5 5 5 625

(hit four years in a row) (hit) (hit) (hit) (hit) 0.00519 19 19 19 130,321

P P P P P

b. Note: 5 14

(not hit in any single year) 1 (hit in any single year) 119 19

P P ,. Therefore,

(not hit in next four years)P

14 14 14 14 38,416(not hit) (not hit) (not hit) (not hit) 0.295

19 19 19 19 130,321P P P P

c. 38,416 91,905130,321 130,321(hit at least once in next four years) 1 (not hit in next four years) 1 0.705P P

4. 4 3 1 1 1

(2 kings) (king) (king given the first card was a king) 0.0045252 51 13 17 221

P P P

5. (3 hearts) (heart) (heart given the first card was a heart) (heart given the first two cards were hearts)P P P P

13 12 11 1 4 11 1 1 11 11

0.012952 51 50 4 17 50 1 17 50 850

6. The sample space is given by { , , , , }.S a e i o u

Of these 5 elements, only a and e precede h.

Thus the probability is 2letter precedes h vowel = .

5P

7. a. The sample space is the set of 13 spades. Of these 13 elements, all 13 cards are black.

Thus the probability is 13black card spade = 1.

13P

b. The sample space is the set of 26 black cards. Of these 26 elements, 13 cards are spades.

Thus the probability is 13 1spade black card = .

26 2P

8. a. 720 9positive mammogram breast cancer = = 0.9.

800 10P

b. 720 45breast cancer positive mammogram = = 0.094.

7664 479P

Section 11.7 Events Involving And; Conditional Probability



1. independent; P A P B

2. the event does not occur

3. dependent; given that occurredP A P B A

4. conditional; P B A

5. false

6. false

7. true

8. true

Exercise Set 11.7

1. 2 3 1 1 1

(green and then red) (green) (red)6 6 3 2 6

P P P

2. 1 2 1 1 1

(yellow and then green) (yellow) (green)6 6 6 3 18

P P P

3. 1 1 1

(yellow and then yellow) (yellow) (yellow)6 6 36

P P P

4. 3 3 1 1 1

(red and then red)= (red) (red)6 6 2 2 4

P P P

5. 3 3 1 1 1

(color other than red each time) (not red) (not red)6 6 2 2 4

P P P

6. 4 4 2 2 4

(color other than green each time) (not green) (not green)6 6 3 3 9

P P P

7. 2 3 1 1 1 1 1

(green and then red and then yellow) (green) (red) (yellow)6 6 6 3 2 6 36

P P P P

8. 3 3 2 1 1 1 1

(red and then red and then green) (red) (red) (green)6 6 6 2 2 3 12

P P P P

9. 3 3 3 1 1 1 1

(red every time) (red) (red) (red)6 6 6 2 2 2 8

P P P P

10. 2 2 2 1 1 1 1

(green every time) (green) (green) (green)6 6 6 3 3 3 27

P P P P



11. 1 1 1

(2 and then 3) (2) (3)6 6 36

P P P

12. 1 1 1

(5 and then 1) (5) (1)6 6 36

P P P

13. 3 4 1 2 1

(even and then greater than 2) (even) (greater than 2)6 6 2 3 3

P P P

14. 3 2 1 1 1

(odd and then less than 3) (odd) (less than 3)6 6 2 3 6

P P P

15. 12 13 3 1 3

(picture card and then heart) (picture card) (heart)52 52 13 4 52

P P P

16. 4 13 1 1 1

( jack and then club) ( jack) (club)52 52 13 4 52

P P P

17. 4 4 1 1 1

(2 kings) (king) (king)52 52 13 13 169

P P P

18. 4 4 1 1 1

(3 each time) (3) (3)52 52 13 13 169

P P P

19. 26 26 1 1 1

(red each time) (red) (red)52 52 2 2 4

P P P

20. 26 26 1 1 1

(black each time) (black) (black)52 52 2 2 4

P P P

21. 1 1 1 1 1 1 1

(all heads) (heads) (heads) (heads) (heads) (heads) (heads)2 2 2 2 2 2 64

P P P P P P P

22. 1 1 1 1 1 1 1 1

(all tails) (tails) (tails) (tails) (tails) (tails) (tails) (tails)2 2 2 2 2 2 2 128

P P P P P P P P

23. 1 2 1

(head and number greater than 4) (head) (number greater than 4)2 6 6

P P P

24. 1 4 1 2 1

(tail and number less than 5) (tail) (number less than 5)2 6 2 3 3

P P P

25. a. 1 1 1

(hit two years in a row) (hit) (hit) 0.0039116 16 256

P P P

b. 1 1 1 1

(Hit three consecutive years) (hit) (hit) (hit) 0.00024416 16 16 4096

P P P P

c. 10 10

10 1 15(not hit in next ten years) [ (not hit)] 1 0.524

16 16P P

d. (hit at least once in next ten years) 1 (not hit in next ten years) 1 0.524 0.476P P



26. a. 1 1 1

(flood two years in a row) (flood) (flood)10 10 100

P P P

b. 1 1 1 1

(flood three consecutive years) (flood) (flood) (flood)10 10 10 1000

P P P P

c. 10 10

10 10 1 9(no flooding in ten years) [ (no flood)] [1 – (flood)] 1 0.349

10 10P P P

d. (flooding at least once in ten years) 1 (no flooding in ten years) 1 0.349 0.651P P

27. (both suffer from depression - from general population) (depression) (depression) 0.12 0.12 0.0144P P P

28. (both suffer from depression - from population of smokers) (depression) (depression) 0.28 0.28 0.0784P P P

29. (all three suffer from frequent hangovers - from population of smokers)

(frequent hangovers) (frequent hangovers) (frequent hangovers) 0.20 0.20 0.20 0.008

P

P P P

30. (all three suffer from frequent hangovers - from general population)

(frequent hangovers) (frequent hangovers) (frequent hangovers) 0.10 0.10 0.10 0.001

P

P P P

31.

(at least one of three suffers from anxiety/panic disorder - from population of smokers)

1 1 (anxiety/panic disorder) 1 (anxiety/panic disorder) 1 (anxiety/panic disorder)

1 [1 0.19] [1 0.19] [1 0

P

P P P .19]

1 [0.81] [0.81] [0.81]

1 0.5314

0.4686

32.

(at least one of three suffers from severe pain - from population of smokers)

1 1 (severe pain) 1 (severe pain) 1 (severe pain)

1 [1 0.14] [1 0.14] [1 0.14]

1 [0.86] [0.86] [0.86]

1 0.6361

0.3639

P

P P P

33. 15 14 1 14 7

(solid and solid) (solid) (solid given first was solid)30 29 2 29 29

P P P

34. 10 9 1 9 3

(two caramel) (caramel) (caramel given first was caramel)30 29 3 29 29

P P P

35. 5 10 1 10 5

(coconut then caramel) (coconut) (caramel given first was coconut)30 29 6 29 87

P P P

36. 5 15 1 15 5

(coconut then solid)= (coconut) (solid given first was coconut)30 29 6 29 58

P P P

37. 5 4 1 2 2

(two Democrats) (Democrat) (Democrat given first was Democrat)15 14 3 7 21

P P P



38. (two Republicans) (Republican) (Republican given first was Republican)

6 5 2 5 1=

15 14 5 14 7

P P P

39. (Independent then Republican) (Independent) (Republican given first was Independent)

4 6 4 3 4

15 14 15 7 35

P P P

40. (Independent then Democrat)= (Independent) (Democrat given first was Independent)

4 5 2=

15 14 21

P P P

41. (no Independents) (not Independent) (not Independent given first was not Independent)

11 10 11 5 11=

15 14 15 7 21

P P P

42. 10 9 2 9 3

(no Democrats) (not Democrat) (not Democrat given first was not Democrat)15 14 3 14 7

P P P

43. P(three cans of apple juice) apple juice given apple juice given first 6 5 4 1

(apple juice) =first was apple juice two were apple juice 20 19 18 57

P P P

44. (three cans of grape juice)

grape juice given grape juice given first 8 7 6 14= (grape juice) =

first was grape juice two were grape juice 20 19 18 285

P

P P P

45. P(grape juice then orange juice then mango juice) orange juice given mango juice given first was grape juice 8 4 2 8

(grape juice) =first was grape juice and second was orange juice 20 19 18 855

P P P

46. P(apple juice then grape juice then orange juice) grape juice given orange juice given first was apple juice 6 8 4 8

(apple juice) =first was apple juice and second was grape juice 20 19 18 285

P P P

47. P(no grape juice) not grape juice given not grape juice given first 12 11 10 11

(not grape juice) =first was not grape juice two were not grape juice 20 19 18 57

P P P

48. (no apple juice)

not apple juice given not apple juice given first 14 13 12 91(not apple juice)

first was not apple juice two were not apple juice 20 19 18 285

P

P P P

49. 13 red =

5P

50. 17 yellow

3P



51. 2even yellow

3P

52. 3odd red

5P

53. 3red odd

4P

54. 1yellow odd

4P

55. 3red at least 5

4P

56. 1yellow at most 3

3P

57. 412,368 68,728surviving wore seat belt 0.999

412,878 68,813P

58. 1601not surviving did not wear seat belt 0.010

164,128P

59. 412,368wore seat belt driver survived 0.717

574,895P

60. 1601did not wear seat belt not surviving 0.758

2111P

61. 24 218 109

(not divorced) 1 (divorced) 1 0.90242 242 121

P P

62. 14 228 114

(not widowed) 1 (widowed) 1 0.94242 242 121

P P

63. 14 24 38 19

(widowed or divorced) (widowed)+ (divorced) 0.16242 242 242 121

P P P

64. 74 24 98 49

(never married or is divorced) (never married)+ (divorced) 0.40242 242 242 121

P P P

65. 118 24 10 132 6

(male or is divorced) (male)+ (divorced) (male and is divorced) 0.55242 242 242 242 11

P P P P

66. (female or is divorced) (female)+ (divorced) (female and is divorced)

124 24 14 134 670.55

242 242 242 242 121

P P P P

67. 10 5male divorced 0.42

24 12P



68. 14 7female divorced 0.58

24 12P

69. 11widowed woman 0.09

124P

70. 10 5divorced man 0.08

118 59P

71. 40 65 105never married or married man 0.89

118 118 118P

72. 34 65 99 99never married or married woman 0.80

124 124 124 124P

73. a. The first person can have any of 365 birthdays. To not match, the second person can then have any of the remaining 364 birthdays.

b. (three different birthdays)P365 364 363

0.992365 365 365

c. (at least two have same birthday) 1 (three different birthdays) 1 0.992 0.008P P

d. (20 different birthdays)P

365 364 363 362 361 360 359 358 357 356 355 354 353 352 351 350 349 348 347 346= 0.589

365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365

(at least two have same birthday) 1 (20 different birthdays) 1 0.589 0.411P P

e. 23 people (determine by trial-and-error using method shown in part d) (23 different birthdays)P

365 364 363 362 361 360 359 358 357 356 355 354 353 352 351 350 349 348 347 346 345 344 343=

365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 365 3650.493

(at least two have same birthday) 1 (23 different birthdays) 1 0.493 0.507P P

82. does not make sense; Explanations will vary. Sample explanation: The previous three children do not affect the odds of the fourth child. The odds are 1:1.

83. does not make sense; Explanations will vary. Sample explanation: The probability of the second selection being a man,

given that the first was a man is 4

.9

84. makes sense

85. does not make sense; Explanations will vary. Sample explanation: ( )P A B does not necessarily equal ( ).P B A

86. 5 5(no one hospitalized) [ (not hospitalized)] (0.9)(0.9)(0.9)(0.9)(0.9) (0.9) 0.59049 59.0%P P

87. 1 5 1 1 5 25

(2 on 1st, 3rd, and 4th rolls only) (2) (not 2) (2) (2) (not 2) 0.003226 6 6 6 6 7776

P P P P P P

Section 11.8 Expected Value


88. There are 5 odd numbered cards and therefore there are 5 2 10C ways to get two odd cards.

There are 4 even numbered cards and therefore there are 4 2 6C ways to get two even cards.

Note that the sum of two odds is an even number and that the sum of two evens is also an even number. Since selecting one even card and one odd card would result in an odd sum, we only need to consider the 16 possible outcomes calculated above.

number of outcomes with both odd and even sum 10 5both odd sum even

number of outcomes where the sum is even 16 8P

89. The sample space has 36 elements. Of these elements, the following 11 fit the given condition: 1&5, 1&6, 3&5, 3&6,

5&1, 5&3, 5&5, 5&6, 6&1, 6&3, 6&5. Thus the probability is 11

.36

Check Points 11.8

1. 1 1 1 1 1 2 3 4 10

1 2 3 4 2.54 4 4 4 4 4

E

2. 1 4 6 4 1 0 4 12 12 4 32

0 1 2 3 4 216 16 16 16 16 16 16

E

3. a. E = $0(0.01)+$2000(0.15) + $4000(0.08) + $6000(0.05) + $8000(0.01) + $10,000(0.70) = $8000 This means that in the long run, the average cost of a claim is expected to be $8000.

b. An average premium charge of $8000 would cause the company to neither lose nor gain money.

4. 1 1 4 1 11 0

5 4 5 5 5E

Since the expected value is 0, there is nothing to gain or lose on average by guessing.

5. Values of gain or loss: Grand Prize: $1000 $2 $998 , Consolation Prize: $50 $2 $48 , Nothing: $0 $2 $2

997 2 1 $1994 $96 $998 $900$2 $48 $998 $0.90

1000 1000 1000 1000 1000E

The expected value for one ticket is $0.90 . This means that in the long run a player can expect to lose $0.90 for each ticket bought. Buying five tickets will make your likelihood of winning five times greater, however there is no advantage to this strategy because the cost of five tickets is also five times greater than one ticket.

6. 20 60 $44 $60 $16$2.20 $1.00 $0.20

80 80 80 80E

This means that in the long run a player can expect to lose an average of $0.20 for each $1 bet.


1. expected; probability; add

2. loss; probability; add

3. false

4. true



Exercise Set 11.8

1. 1 1 1

1 2 3 1.752 4 4

E

2. 1 1 1 1

1 2 3 4 2.8758 8 2 4

E

3. a. E = $0(0.65)+$50,000(0.20) + $100,000(0.10) + $150,000(0.03) + $200,000(0.01) + $250,000(0.01) = $29,000 This means that in the long run the average cost of a claim is $29,000.

b. $29,000

c. $29,050

4. a. E = $0(0.70) + $20,000(0.20) + $40,000(0.06) + $60,000(0.02) + $80,000(0.01) + $100,000(0.01) = $9400 This means that in the long run the average cost of a claim is $9400.

b. $9400

c. $9450

5. E = –$10,000(0.9) + $90,000(0.1) = $0. This means on the average there will be no gain or loss.

6. 4 1

$1500 $38,500 $65005 5

E

. This means an expected gain on the average.

7. 27 9,999,973

$99,999 $1 $0.7310,000,000 10,000,000

E

8. E = –$9,900(0.002) + $100(0.998) = $80

9. Probabilities after eliminating one possible answer: Guess Correctly: 1

4, Guess Incorrectly:

3

4

1 1 3 1 3 11

4 4 4 4 16 16E

expected points on a guess if one answer is eliminated.

Yes, it is advantageous to guess after eliminating one possible answer.

10. Probabilities after eliminating two possible answers: Guess Correctly: 1

3, Guess Incorrectly:

2

3

1 1 2 1 1 11

3 4 3 3 6 6E

expected points on a guess if two answers are eliminated.

Yes, it is advantageous to guess after eliminating two possible answers.

11. First mall: 1 1

$300,000 $100,000 $100,0002 2

E

Second mall: 3 1

$200,000 $60,000 $135,0004 4

E

Choose the second mall.

Section 11.8 Expected Value


12. Site A: E = $80(0.2) – $10(0.8) = $8 million Site B: E = $120(0.1) – $18(0.9) = –$4.2 million Site A has the larger expected profit. $8 million – (–$4.2 million) = $12.2 million Site A’s profit exceeds Site B’s by $12.2 million.

13. a. E = $700,000(0.2) + $0(0.8) = $140,000

b. No

14. 99 1

$80 $270 $76.50100 100

E

15. 1 5 1

$4 $1 $ $0.176 6 6

E

. This means an expected loss of approximately $0.17 per game.

16. 1 1 1 3

–$.25 $.75 $1.75 $1.25 $0.256 6 6 6

E

. This means an expected loss of $0.25 per game.

17. 18 20

$1 $1 $0.05338 38

E

. This means an expected loss of approximately $0.053 per $1.00 bet.

18. 3 1 4 2

$4 $2 $2 – $3 $0.10 10 10 10

E

A player should expect to break even.

19. 1 999

$499 $1 $0.501000 1000

E

. This means an expected loss of $0.50 per $1.00 bet.

26. does not make sense; Explanations will vary. Sample explanation: The expectation does not need to be a whole number.

27. makes sense

28. does not make sense; Explanations will vary. Sample explanation: The likely outcome of playing longer is more losses.

29. does not make sense; Explanations will vary. Sample explanation: The expected value of a lottery game is less than the cost of the ticket.

30. First determine the probabilities.

Total number of possible combinations 35 535!

324,63230!5!

C

Number of ways to select all 5 5 5 1C

Number of ways to select 4 of the 5 winning numbers and 1 of the 30 losing numbers 5 4 30 1 5 30 150C C

Number of ways to select 3 of the 5 winning numbers and 2 of the 30 losing numbers 5 3 30 2 10 435 4350C C

1(all 5) ;

324,632P

150(4 of 5) ;

324,632P

4350(3 of 5)

324,632P ;

324,632 1 150 4350 320,131(losing)

324,632 324,632P

1 150 4350 320,131$49,999 $499 $4 $1 $0.55

324,632 324,632 324,632 324,632E

This means an expected loss of $0.55 per $1.00 ticket.



31. Let x = the charge for the policy. Note, the expected value, E = $60. $60 ( $200,000)(0.0005) ( )(0.9995)

$60 0.0005 $100 0.9995

$160

x x

x x

x

The insurance company should charge $160 for the policy.

Chapter 11 Review Exercises

1. Use the Fundamental Counting Principle with two groups of items. 20 · 40 = 800



4. Use the Fundamental Counting Principle with three groups of items. 5 · 5 · 5 = 125

5. Use the Fundamental Counting Principle with five groups of items. 3 · 3 · 3 · 3 · 3 = 243

6. Use the Fundamental Counting Principle with four groups of items. 5 · 2 · 2 · 3 = 60

7. 16! 16 15 14!

24014! 14!

8. 800! 800 799!

800799! 799!

9. 5! 3! 5 4 3 2 1 3 2 1 120 6 114

10. 11! 11! 11 10 9 8!

990(11 3)! 8! 8!

11. 10 610! 10! 10 9 8 7 6 5 4!

151,200(10 6)! 4! 4!

P

12. 100 2100! 100! 100 99 98!

9900(100 2)! 98! 98!

P

13. 11 711! 11! 11 10 9 8 7!

330(11 7)!7! 4!7! 4 3 2 1 7!

C

14. 14 514! 14! 14 13 12 11 10 9!

2002(14 5)!5! 9!5! 9! 5 4 3 2 1

C




18. Use the Fundamental Counting Principle with six groups of items. 6 · 5 · 4 · 3 · 2 · 1 = 720

19. 15 415! 15! 15 14 13 12 11!

32,760(15 4)! 11! 11!

P



20. 10 410! 10! 10 9 8 7 6!

210(10 4)!4! 6!4! 6!4 3 2 1

C

21. ! 7! 7 6 5 4 3!

! ! 3!2!

n

p q

3!

4202 1

22. 20 320! 20! 20 19 18 17!

1140(20 3)!3! 17!3! 17!3 2 1

C

23. Use the Fundamental Counting Principle with seven groups of items. 1 · 5 · 4 · 3 · 2 · 1 · 1 = 120

24. 20 520! 20! 20 19 18 17 16 15!

1,860,480(20 5)! 15! 15!

P


26. 13 513! 13! 13 12 11 10 9 8!

1287(13 5)!5! 8!5! 8!5 4 3 2 1

C

27. Choose the Republicans: 12 512! 12! 12 11 10 9 8 7!

792(12 5)!5! 7!5! 7!5 4 3 2 1

C

Choose the Democrats: 8 48! 8! 8 7 6 5 4!

70(8 4)!4! 4!4! 4!4 3 2 1

C

Multiply the choices: 792 · 70 = 55,440

28. ! 6! 6 5 4 3!

! ! 3!2!

n

p q

3!

602 1

29. number of ways a 6 can occur 1


P

30. number of ways a number less than 5 can occur 4 2


P



P



P

33. number of ways a 5 can occur 4 1

(5)total number of possible outcomes 52 13

P

34. number of ways a picture card can occur 12 3

(picture card)total number of possible outcomes 52 13

P

35. number of ways a card greater than 4 and less than 8 can occur 12 3

(greater than 4 and less than 8)total number of possible outcomes 52 13

P



36. number of ways a 4 of diamonds can occur 1

(4 of diamonds)total number of possible outcomes 52

P

37. number of ways a red ace can occur 2 1

(red ace)total number of possible outcomes 52 26

P

38. number of ways a chocolate can occur 15 1

(chocolate)total number of possible outcomes 30 2

P

39. number of ways a caramel can occur 10 1

(caramel)total number of possible outcomes 30 3

P

40. number of ways a peppermint can occur 5 1

(peppermint)total number of possible outcomes 30 6

P

41. a. number of ways to be a carrier without the disease 2 1

(carrier without the disease)total number of possible outcomes 4 2

P

b. number of ways to have the disease 0

(disease) 0total number of possible outcomes 4

P

42. 140 7

(employed)240 12

P

43. 124 31

(female)240 60

P

44. 8 1

(unemployed male)240 30

P

45. number of ways to visit in order D, B, A, C = 1 total number of possible permutations = 4 · 3 · 2 · 1 = 24

1(D, B, A, C)

24P

46. number of permutations with C last = 5 · 4 · 3 · 2 · 1 · 1 = 120 total number of possible permutations = 6 · 5 · 4 · 3 · 2 · 1 = 720

120 1(C last)

720 6P

47. number of permutations with B first and A last = 1 · 4 · 3 · 2 · 1 · 1 = 24 total number of possible permutations = 6 · 5 · 4 · 3 · 2 · 1 = 720

24 1(B first and A last)

720 30P

48. number of permutations in order F, E, A, D, C, B = 1 total number of possible permutations = 6 · 5 · 4 · 3 · 2 · 1 = 720

1(F, E, A, D, C, B)

720P



49. number of permutations with A or C first = 2 · 5 · 4 · 3 · 2 · 1 = 240 total number of possible permutations = 6 · 5 · 4 · 3 · 2 · 1 = 720

240 1(A or C first)

720 3P

50. a. number of ways to win = 1 total number of possible combinations:

20 520! 20! 20 19 18 17 16 15!

15,504(20 5)!5! 15!5! 15!5 4 3 2 1

C

1(winning with one ticket) 0.0000645

15,504P

b. number of ways to win = 100 100

(winning with 100 different tickets) 0.0064515,504

P

51. a. number of ways to select 4 Democrats: 6 46! 6! 6 5 4!

15(6 4)!4! 2!4! 2 1 4!

C

total number of possible combinations: 10 410! 10! 10 9 8 7 6!

210(10 4)!4! 6!4! 6!4 3 2 1

C

15 1(all Democrats)

210 14P

b. number of ways to select 2 Democrats: 6 26! 6! 6 5 4!

15(6 2)!2! 4!2! 4!2 1

C

number of ways to select 2 Republicans: 4 24! 4! 4 3 2!

6(4 2)!2! 2!2! 2!2 1

C

number of ways to select 2 Democrats and 2 Republicans = 15 · 6 = 90 90 3

(2 Democrats and 2 Republicans)210 7

P

52. number of ways to get 2 picture cards: 6 26! 6! 6 5 4!

15(6 2)!2! 4!2! 4!2 1

C

number of ways to get one non-picture card = 20 number of ways to get 2 picture cards and one non-picture card = 15 · 20 = 300

total number of possible combinations: 26 326! 26! 26 25 24 23!

2600(26 3)!3! 23!3! 23!3 2 1

C

300 3(2 picture cards)

2600 26P

53. 1 5

(not a 5) 1 (5) 16 6

P P

54. 3 1 1

(not less than 4) 1 (less than 4) 1 1–6 2 2

P P

55. 1 1 2 1

(3 or 5) (3) (5)6 6 6 3

P P P



56. 2 2 1 1 2

(less than 3 or greater than 4) (less than 3) (greater than 4)6 6 3 3 3

P P P

57. (less than 5 or greater than 2) (less than 5) (greater than 2) (less than 5 and greater than 2)P P P P

4 4 2

16 6 6

58. 12 3 10

(not a picture card) 1 (picture card) 1 152 13 13

P P

59. 13 1 3

(not a diamond) 1 (diamond) 1 152 4 4

P P

60. 4 4 1 1 2

(ace or king) (ace) (king)52 52 13 13 13

P P P

61. 2 2 1 1 2 1

(black 6 or red 7) (black 6) (red 7)52 52 26 26 26 13

P P P

62. 4 26 2 28 7

(queen or red card) (queen) (red card) (red queen)52 52 52 52 13

P P P P

63. 13 12 3 22 11

(club or picture card) (club) (picture card) (club and picture card)52 52 52 52 26

P P P P

64. 1 5

(not 4) 1 (4) 16 6

P P

65. 1 5

(not yellow) 1 (yellow) 1–6 6

P P

66. 3 1 1

(not red) 1 (red) 1 16 2 2

P P

67. 3 1 4 2

(red or yellow) (red) (yellow)6 6 6 3

P P P

68. 3 3 0

(red or even) (red) (even) (red and even) 16 6 6

P P P P

69. 3 3 1 5

(red or greater than 3) (red) (greater than 3) (red and greater than 3)6 6 6 6

P P P P

70. (African American or male) (African American) (male) (African American male)P P P P

50 20 50 90 50 160 4

–200 200 200 200 5

71. 20 40 90 40 40 150 3

(female or white) (female) (white) (white female) –200 200 200 200 4

P P P P

72. 252 18

(public college)350 25

P



73. 50 350 50 300 6

(not from high-income family) 1 (from high-income family) 1350 350 350 350 7

P P

74. 160 50 210 3

(from middle-income family or high-income family)350 350 5

P

75. (attended private college or is from a high income family)P

(private college) (high income family) (attended private college and is from a high income family)

98 50 28 120 12

350 350 350 350 35

P P P

76. number of favorable outcomes = 4, number of unfavorable outcomes = 48 Odds in favor of getting a queen are 4:48, or 1:12. Odds against getting a queen are 12:1.

77. number of favorable outcomes = 20, number of unfavorable outcomes = 1980 Odds against winning are 1980: 20, or 99:1.

78. 3 3

(win)3+1 4

P

79. 2 4 1 2 2

(yellow then red) (yellow) (red)6 6 3 3 9

P P P

80. 1 1 1

(1 then 3) (1) (3)6 6 36

P P P

81. 2 2 1 1 1

(yellow both times) (yellow) (yellow)6 6 3 3 9

P P P

82. 2 1 3 1 1 1 1

(yellow then 4 then odd) (yellow) (4) (odd)6 6 6 3 6 2 36

P P P P

83. 4 4 4 2 2 2 8

(red every time) (red) (red) (red)6 6 6 3 3 3 27

P P P P

84. 5

1 1 1 1 1 1 1(five boys in a row) (boy) (boy) (boy) (boy) (boy)

2 2 2 2 2 322P P P P P P

85. a. (flood two years in a row) (flood) (flood) (0.2)(0.2) 0.04P P P

b. (flood for three consecutive years) (flood) (flood) (flood) (0.2)(0.2)(0.2) 0.008P P P P

c. 4 4 4(no flooding for four consecutive years) [1 (flood)] (1 0.2) (0.8) 0.4096P P

d. (flood at least once in next four years) 1 (no flooding for four consecutive years)P P

1 0.4096 0.5904

86. psychology major given 2 4 2 1 1

(music major then psychology major) (music major)first was music major 9 8 9 2 9

P P P

87. 3 2 1 1 1

(two business majors) (bus. major) (bus. major given first was bus. major)9 8 3 4 12

P P P



88. P(solid then two cherry) cherry given cherry given first was solid 30 5 4 3 5 1 1

(solid)first was solid and second was cherry 50 49 48 5 49 12 196

P P P

89. 15 odd

3P

90. 3vowel precedes the letter k

10P

91. a. 2 1odd red

4 2P

b. 2yellow at least 3

7P

92. 11 124 135 27

(does not have TB)9 1 11 124 145 29

P

93. 9 11 20 4

(tests positive)9 1 11 124 145 29

P

94. (does not have TB or tests positive)P

(does not have TB) (tests positive) (does not have TB and tests positive)

11 124 9 11 11

145 145 145144

145

P P P

95. 11 11

(does not have TB positive test)9 11 20

P

96. 11 11

(tests positive does not have TB)11 124 135

P

97. 1 1

(has TB negative Test)1 124 125

P

98. 10 9 1

(two people with TB) (TB) (TB first person selected has TB)145 144 232

P P P

99. 20 19 19

(two people with positive tests) (positive test) (positive test first person has positive test)145 144 1044

P P P

100. 27,336

(male) 0.86531,593

P

101. 11,161

(age 25 44) 0.35331,593

P

Chapter 11 Test


102. 2379 29,214

(less than 75) 1 (greater than or equal to 75) 1 0.92531,593 31,593

P P

103. 4095 11,161 15,256

(age 20 24 or 25 44) (age 20 24) (age 25 44) 0.48331,593 31,593 31,593

P P P

104. (female or younger than 5) (female) (younger than 5) (female and younger than 5)

4257 88 33 43120.136

31,593 31,593 31,593 31,593

P P P P

105. 3684

(age 20 24 male) 0.13527,336

P

106. 2169

(male at least 75) 0.9122379

P

107. 1 1 1 1 1

1 2 3 4 5 3.1254 8 8 4 4

E

108. a. E = $0(0.9999995) + (–$1,000,000)(0.0000005) = –$.50 The insurance company spends an average of $0.50 per person insured.

b. charge $9.50 – (–$0.50) = $10.00

109. 1 3$27,000 $3000 $4500.

4 4E

The expected gain is $4500 per bid.

110. 2 1 1$1 $1 $4 $0.25.

4 4 4E

The expected loss is $0.25 per game.

Chapter 11 Test


2. Use the Fundamental Counting Principle with four groups of items. 4 · 3 · 2 · 1 = 24

3. Use the Fundamental Counting Principle with seven groups of items. 1 · 6 · 5 · 4 · 3 · 2 · 1 = 720

4. 11 311! 11! 11 10 9 8!

990(11 3)! 8! 8!

P

5. 10 410! 10! 10 9 8 7 6!

210(10 4)!4! 6!4! 6!4 3 2 1

C

6. ! 7! 7 6 5 4 3!

! ! 3!2!

n

p q

3!

4202 1

7. 12 6

(freshman)50 25

P



8. 16 8 17

(not a sophomore) 1 (sophomore) 1 150 25 25

P P

9. 20 2 22 11

( junior or senior) ( junior) (senior)50 50 50 25

P P P

10. 20 5

(greater than 4 and less than 10)52 13

P

11. P(C first, A next-to-last, E last)

= P(C) · P(A given C was first) · P(E given C was first and A was next-to-last) 1 1 1 1

7 6 5 210

12. total number of possible combinations: 15 615! 15! 15 14 13 12 11 10 9!

5005(15 6)!6! 9!6! 9!6 5 4 3 2 1

C

50 10(winning with 50 tickets) 0.00999

5005 1001P

13. 2 2 4 1

(red or blue) (red) (blue)8 8 8 2

P P P

14. 2 2 1 1 1

(red then blue) (red) (blue)8 8 4 4 16

P P P

15. 1 1 1 1

(flooding for three consecutive years) (flood) (flood) (flood)20 20 20 8000

P P P P

16. 26 12 6 32 8

(black or picture card) (black) (picture card) (black picture card)52 52 52 52 13

P P P P

17. 10 15 15 5 15 30 3

(freshman or female) (freshman)+ (female) (female freshman)50 50 50 50 5

P P P P

18. 5 4 1 4 1

(both red) (red) (red given first ball was red)20 19 4 19 19

P P P

19. 4

1 1 1 1 1 1(all correct) (correct) (correct) (correct) (correct)

4 4 4 4 4 256P P P P P

20. number of favorable outcomes = 20, number of unfavorable outcomes = 15 Odds against being a man are 15:20, or 3:4.

21. a. Odds in favor are 4:1. b. 4 4

(win)1 4 5

P

22. 18 10 20 12 60 3

(not brown eyes)22 18 10 18 20 12 100 5

P

23. 22 18 18 20 78 39

(brown eyes or blue eyes)22 18 10 18 20 12 100 50

P

Chapter 11 Test


24. (female or green eyes) (female) (green eyes) (female and green eyes)P P P P

18 20 12 10 12 12

100 100 10050 22 12

100 100 10060

1003

5

25. 18 18 9

(male blue eyes)18 20 38 19

P

26. 22 21 7

(two people with green eyes) (green eyes) (green eyes first person has green eyes)100 99 150

P P P

27. E = $65,000(0.2) + (–$15,000)(0.8) = $1000. This means the expected gain is $1000 for this bid.

28. 10 5 3 1 1

( $19) ( $18) ( $15) ( $10) ($80)20 20 20 20 20

E

$190 $90 $45 $10 $80 $255

$12.7520 20

This expected value of $12.75 means that a player will lose an average of $12.75 per play in the long run.

chapter 11 counting methods and probability theory€¦ · the number of choices decreases by 1...

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