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607Algebra 2

Worked-Out Solution Key

Prerequisite Skills (p. 680)

1. The coeffi cient of x2 in the expression 3x3 2 15x2 1 4 is 215.

2. Written as a fraction in lowest terms, the ratio of 18 to 45

is 18

} 45

5 2 } 5 .

3. The expressions x 1 3 and 2x 2 1 are examples of binomials because they have 2 terms.

4. 6 p 5 p 4 p 3

} 2 p 1 5

360 } 2 5 180 5.

13 p 12 p 11 }

10 p 9 p 8 5 1716

} 720 5 143

} 60

6. 8 p 7 p 6 p 5 p 4

}} 5 p 4 p 3 p 2 p 1 5

6720 } 120 5 56

7. (x 1 y)3 5 x3 1 3x2y 1 3xy2 1 y3

8. (5x 1 1)3 5 (5x)3 1 3(5x)2(1) 1 3(5x)(1)2 1 (1)3

5 125x3 1 75x2 1 15x 1 1

9. (3x 2 2y)3 5 (3x)3 2 3(3x)2(2y) 1 3(3x)(2y)2 2 (2y)3

5 27x3 2 54x 2y 1 36xy2 2 8y3

10. Square: A 5 s2 Circle: A 5 πr 2

5 (3)2 5 π(1.5)2

5 9 ø 7.07

9 2 7.07 5 1.93

The area of the shaded region is about 1.93 square feet.

11. Big circle: A 5 πr2 Small circle: A 5 πr2

5 π(8)2 5 π(4)2

5 201.1 5 50.3

201.1 2 50.3 5 150.8

The area of the shaded region is about 151 square meters.

12. Big square: A 5 s2 Small square: A 5 s2

5 (10)2 5 1 Ï}

50 2 2

5 100 5 50

100 2 50 5 50

The area of the shaded region is 50 square inches.

Lesson 10.1

10.1 Guided Practice (pp. 682–686)

1. 3 p 3 5 9

The store offers 9 bicycle choices.

2. a. 26 p 26 p 26 p 10 p 10 p 10 p 10 5 175,760,000

The number of plates would increase to 175,760,000.

b. 26 p 25 p 24 p 10 p 9 p 8 p 7 5 78,624,000

The number of plates would increase to 78,624,000.

3. a. 12! 5 12 p 11 p 10 p 9 p 8 p 7 p 6 p 5 p 4 p 3 p 2 p 1

5 479,001,600

The number of ways to fi nish would increase to 479,001,600.

b. 12 p 11 p 10 5 1320

The number of ways to fi nish would increase to 1320.

4. 5P3 5 5! }

(5 2 3)! 5

5! } 2! 5

120 } 2 5 60

5. 4P1 5 4! }

(4 2 1)! 5

4! } 3! 5

24 } 6 5 4

6. 8P5 5 8! }

(8 2 5)! 5

8! } 3! 5

40,320 } 6 5 6720

7. 12P7 5 12! }

(12 2 7)! 5

12! } 5! 5

479,001,600 } 120 5 3,991,680

8. 4!

} 2!

5 24

} 2 5 12 9. 5! }

2! p 2! 5

120 } 2 p 2 5 30

10. 10! }

2! p 3! p 3! 5

3,628,800 } 2 p 6 p 6 5 50,400

10.1 Exercises (pp. 686–689)

Skill Practice

1. A permutation of n objects is the number of ways n objects can be ordered.

2. Sample answer: When r 5 0, the permutation simplifi es

to n!

} n!

, which is equal to 1. This makes sense because there

is only one way to take 0 objects from a group ofn objects.

3. M long-sleeveM short-sleeveL long-sleeveL short-sleeveXL long-sleeveXL short-sleeve

long-sleeveshort-sleevelong-sleeve

short-sleevelong-sleeve

short-sleeve

M

L

XL

4.

wheat

white

margarine

jam

margarine

jam white, jam

white, margarine

wheat, jam

wheat, margarine

5. corngreen beanpotatocorngreen beanpotatocorngreen beanpotato

chicken

fish

pasta

chicken, cornchicken, green beanchicken, potatofish, cornfish, green beanfish, potatopasta, cornpasta, green beanpasta, potato

6. stainedpaintedunfinishedstainedpaintedunfinishedstainedpaintedunfinishedstainedpaintedunfinished

mahogany

cherry

oak

pine

stained cherrypainted cherryunfinished cherrystained mahoganypainted mahoganyunfinished mahoganystained oakpainted oakunfinished oakstained pinepainted pineunfinished pine

7. 2 p 4 5 8 8 ways 8. 5 p 2 5 10 10 ways

9. 4 p 3 p 5 5 60 10. 3 p 6 p 5 p 2 5 180

60 ways 180 ways

11. a. 26 p 26 p 26 p 26 p 10 p 10 p 10 5 456,976,000

b. 26 p 25 p 24 p 23 p 10 p 9 p 8 5 258,336,000

12. a. 26 p 26 p 10 p 10 p 10 p 10 p 10 5 67,600,000

b. 26 p 25 p 10 p 9 p 8 p 7 p 6 5 19,656,000

13. a. 26 p 26 p 26 p 26 p 10 p 10 5 45,697,600

b. 26 p 25 p 24 p 23 p 10 p 9 5 32,292,000

Chapter 10

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608Algebra 2Worked-Out Solution Key

14. a. 10 p 10 p 10 p 10 p 10 p 26 p 26 p 26 5 1,757,600,000

b. 10 p 9 p 8 p 7 p 6 p 26 p 25 p 24 5 471,744,000

15. a. 10 p 26 p 26 p 26 p 26 p 26 5 118,813,760

b. 10 p 26 p 25 p 24 p 23 p 22 5 78,936,000

16. a. 26 p 26 p 26 p 26 p 26 p 26 5 308,915,776

b. 26 p 25 p 24 p 23 p 22 p 21 5 165,765,600

17. A; 26 p 25 p 10 p 9 p 8 p 7 5 3,276,000

18. 7! 5 7 p 6 p 5 p 4 p 3 p 2 p 1 5 5040

19. 11! 5 11 p 10 p 9 p 8 p 7 p 6 p 5 p 4 p 3 p 2 p 1

5 39,916,800

20. 1! 5 1

21. 8! 5 8 p 7 p 6 p 5 p 4 p 3 p 2 p 1 5 40,320

22. 4! 5 4 p 3 p 2 p 1 5 24

23. 0! 5 1

24. 12! 5 12 p 11 p 10 p 9 p 8 p 7 p 6 p 5 p 4 p 3 p 2 p 1

5 479,001,600

25. 6! 5 6 p 5 p 4 p 3 p 2 p 1 5 720

26. 3! p 4! 5 (3 p 2 p 1) p (4 p 3 p 2 p 1) 5 144

27. 3(4!) 5 3(4 p 3 p 2 p 1) 5 72

28. 8! }

(8 2 5)! 5

8! } 3! 5

40,320 } 6 5 6720

29. 9! }

4! p 4! 5

362,880 } 24 p 24 5 630

30. 4P4 5 4! }

(4 2 4)! 5

4! } 0! 5

24 } 1 5 24

31. 6P2 5 6! }

(6 2 2)! 5

6! } 4! 5

720 } 24 5 30

32. 10P1 5 10! }

(10 2 1)! 5

10! } 9! 5

3,628,800 } 362,880 5 10

33. 8P7 5 8! }

(8 2 7)! 5

8! } 1! 5

40,320 } 1 5 40,320

34. 7P4 5 7! }

(7 2 4)! 5

7! } 3! 5

5040 } 6 5 840

35. 9P2 5 9! }

(9 2 2)! 5

9! } 7! 5

362,880 } 5040 5 72

36. 13P8 5 13! }

(13 2 8)! 5

13! } 5! 5

6,227,020,800 }} 120 5 51,891,840

37. 7P7 5 7! }

(7 2 7)! 5

7! } 0! 5

5040 } 1 5 5040

38. 5P0 5 5! }

(5 2 0)! 5

5! } 5! 5 1

39. 9P4 5 9! }

(9 2 4)! 5

9! } 5! 5

362,880 } 120 5 3024

40. 11P4 5 11! }

(11 2 4)! 5

11! } 7! 5

39,916,800 } 5040 5 7920

41. 15P0 5 15! }

(15 2 0)! 5

15! } 15! 5 1

42. nPn 2 1 5 n! }}

[n 2 (n 2 1)]!

5 n! }

(n 2 n 1 1)!

5 n!

} 1!

5 n!

This is equal to the number of permutations of n objects. This makes sense because you can get an ordering of n objects from an ordering of n 2 1 objects.

43. 3!

} 2!

5 6 } 2 5 3 44.

4! }

2! 5

24 } 2 5 12

45. 5!

} 2!

5 120

} 2 5 60 46. 6!

} 2!

5 720

} 2 5 360

47. 6! 5 720 48. 6!

} 3!

5 720

} 6 5 120

49. 8! }

3! p 2! 5

40,320 } 6 p 2 5 3360 50.

9! }

2! 5

362,880 } 2 5 181,440

51. 8! 5 40,320

52. 8! }

2! p 2! p 2! 5

40,320 } 2 p 2 p 2 5 5040

53. 9! }

2! p 2! 5

362,880 } 2 p 2 5 90,720

54. 11! }

4! p 4! p 2! 5

39,916,800 } 24 p 24 p 2 5 34,650

55. B; 6! }

2! p 2! 5

720 } 2 p 2 5 180

56. The numbers are not replaced, so the number of balls must decrease by 1 after each drawing.

75 p 74 p 73 p 72 5 29,170,800

57. Using the fundamental counting principle, you have n choices for the fi rst object, n 2 1 choices for the second object, and so on. So, the number of ways to choose n objects is n p (n 2 1) p (n 2 2) p . . . p 3 p 2 p 1 5 n!

58. nP4 5 8(nP3

)

n! }

(n 2 4)! 5 8 1 n!

} (n 2 3)!

2

n! }

(n 2 4)! p

(n 2 3)! }

n! 5 8

(n 2 3) p (n 2 4)!

}} (n 2 4)!

5 8

n 2 3 5 8

n 5 11

59. nP6 5 5(nP5

)

n! }

(n 2 6)! 5 5 1 n!

} (n 2 5)!

2

n! }

(n 2 6)! p

(n 2 5)! }

n! 5 5

(n 2 5) p (n 2 6)!

}} (n 2 6)!

5 5

n 2 5 5 5

n 5 10

Chapter 10, continued

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609Algebra 2


60. nP5 5 9(nP4

)

n! }

(n 2 5)! 5 9 1 n!

} (n 2 4)!

2

n! }

(n 2 5)! p

(n 2 4)! }

n! 5 9

(n 2 4) p (n 2 5)!

}} (n 2 5)!

5 9

n 2 4 5 9

n 5 13

61. If an M is not chosen: A, N, A, T, E, E 6! }

2! p 2! 5 180

If an A is not chosen: M, A, N, T, E, E 6!

} 2!

5 360

If an N is not chosen: M, A, A, T, E, E 6! }

2! p 2! 5 180

If a T is not chosen: M, A, N, A, E, E 6! }

2! p 2! 5 180

If an E is not chosen: M, A, N, A, T, E 6!

} 2!

5 360

The total number of distinguishable permutations is 180 1 360 1 180 1 180 1 360 5 1260.

Problem Solving

62. 3 p 6 p 12 5 216

There are 216 possible class rings.

63. 52 p 47 p 45 p 36 p 19 p 3 5 225,678,960

There are 225,678,960 sets of 6 countries that can be represented by the prize winners in a given year.

64. 15! 5 1,307,674,368,000

The photographer can arrange the family members 1,307,674,368,000 ways.

65. 9 p 8 p 7 5 504

The offi ces can be fi lled 504 ways.

66. 8! }

3! p 2! p 3! 5

40,320 } 6 p 2 p 6 5 560

There are 560 ways the instruments can be displayed.

67. a. 5 p 8 p 6 5 240

There are 240 selections of audio components.

b. 7 p 9 p 4 5 252

There are 252 selections of video components.

c. 5 p 8 p 6 p 7 p 9 p 4 5 60,480

There are 60,480 selections of all six components.

68. a. 36 p 36 p 36 p 36 p 36 p 36 5 2,176,782,336

2,176,782,336 passwords are possible.

b. 36 p 35 p 34 p 33 p 32 p 31 5 1,402,410,240

1,402,410,240 passwords are possible.

c. Passwords when characters can be repeated are more secure because there are more possible choices for these passwords, which makes them harder to randomly guess.

69. 8P3 p 7P3 5 8! }

(8 2 3)! p 7!

} (7 2 3)!

5 70,560

There are 70,560 different displays that can be created.

70. The 15 runners can be represented as

AAAAAABBBBBCCCC.

Number of different fi nishing orders:

15! }

6! p 5! p 4! 5

1,307,674,368,000 }} 2,073,600 5 630,630

71. a. 4! 5 24

You can arrange the people around the table 24 ways.

b. (n 2 1)!; because there is no object placed fi rst, second, third, and so on, allow one person to represent a “fi xed” position, so the remaining people (n 2 1), can be arranged (n 2 1)! ways.

Mixed Review

72. (x 2 8)(x 1 8) 5 x2 2 64

73. (4x 2 5)(4x 1 5) 5 16x2 2 25

74. (x 1 7)2 5 x2 1 14x 1 49

75. (5x 2 6y)2 5 25x2 2 60xy 1 36y2

76. (3x 2 2)3 5 (3x)3 2 3(3x)2(2) 1 3(3x)(2)2 2 (2)3

5 27x3 2 54x2 1 36x 2 8

77. (4x 1 3y)3 5 (4x)3 1 3(4x)2(3y) 1 3(4x)(3y)2 1 (3y)3

5 64x3 1 144x2y 1 108xy2 1 27y3

78. y 5 4x 2 9 79. y 5 2x 1 6

x 5 4y 2 9 x 5 2y 1 6

x 1 9 5 4y x 2 6 5 2y

x 1 9

} 4 5 y 2x 1 6 5 y

f 21(x) 5 x 1 9

} 4 f 21(x) 5 6 2 x

80. y 5 4x5 81. y 5 x2

x 5 4y5 x 5 y2

x }

4 5 y5 Ï}

x 5 y

5 Î}

x }

4 5 y f 21(x) 5 Ï

}

x

f 21(x) 5 5 Î}

x }

4

82. y 5 x3 1 5 83. y 5 3x5 2 1

x 5 y3 1 5 x 5 3y5 2 1

x 2 5 5 y3 x 1 1 5 3y5

3 Ï}

x 2 5 5 y x 1 1

} 3 5 y5

f 21(x) 5 3 Ï}

x 2 5 5 Î}

x 1 1

} 3 5 y

f 21(x) 5 5 Î}

x 1 1

} 3


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84. y2 5 224x

x

y

15

5

(26, 12)

(26, 212)

85. x2 1 y2 5 20

x

y

1

1(22 , 0)5 (2 , 0)5

(0, 22 )5

(0, 2 )5

86. x2

} 9 1

y2

} 36 5 1 87. x2

} 81

2 y2

} 121 5 1

x

y

2

2(23, 0) (3, 0)

(0, 6)

(0, 26)

x

y

6

3(29, 0)(9, 0)

(0, 11)

(0, 211)

88. (x 1 3)2 1 y2 5 16

x

y

1

21(27, 0) (23, 0)

(23, 4)

(1, 0)

(23, 24)

89. (y 2 1)2

} 16 2 x2 5 1

x

y

1

1

(24, 1) (4, 1)(0, 1)

(0, 5)

(0, 23)

( y 2 1)2 5 16 2 x2

x2 1 ( y 2 1)2 5 16

Lesson 10.2


1. 8C3 5 8! } 5! p 3! 5

8 p 7 p 6 p 5! } 5! p 3! 5 56

2. 10C6 5 10! } 4! p 6! 5

10 p 9 p 8 p 7 p 6! }} 4! p 6! 5 210

3. 7C2 5 7! } 5! p 2! 5

7 p 6 p 5! } 5! p 2! 5 21

4. 14C5 5 14! } 9! p 5! 5

14 p 13 p 12 p 11 p 10 p 9! }} 9! p 5! 5 2002

5. 10C3 p 10C2 5 10! } 7! p 3! p

10! }

8! p 2!

5 10 p 9 p 8 p 7!

}} 7! p 3! p 10 p 9 p 8! }

8! p 2!

5 120 p 45

5 5400

You can read 5400 sets of plays.

6. n 5 7 (7th row)

1 7 21 35 35 21 7 1

7C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7

There are 7C2 5 21 combinations of representatives for the convention.

7. (x 1 3)5 5 5C0x530 1 5C1x431 1 5C2x332

1 5C3x233 1 5C4x134 1 5C5x035

5 (1)(x5)(1) 1 (5)(x4)(3) 1 (10)(x3)(9)

1 (10)(x2)(27) 1 (5)(x)(81) 1 (1)(1)(243)

5 x5 1 15x4 1 90x3 1 270x2 1 405x 1 243

8. (a 1 2b)4 5 4C0a4(2b)0 1 4C1a3(2b)1 1 4C2a2(2b)2

1 4C3a1(2b)3 1 4C4a0(2b)4

5 (1)(a4)(1) 1 (4)(a3)(2b) 1 (6)(a2)(4b2) 1 (4)(a)(8b3) 1 (1)(1)(16b4) 5 a4 1 8a3b 1 24a2b2 1 32ab3 1 16b4

9. (2p 2 q)4 5 [2p 1 (2q)]4

5 4C0(2p)4(2q)0 1 4C1(2p)3(2q)1

1 4C2(2p)2(2q)2

1 4C3(2p)1(2q)3 1 4C4(2p)0(2q)4

5 (1)(16p4)(1) 1 (4)(8p3)(2q) 1 (6)(4p2)(q2) 1 (4)(2p)(2q3) 1 (1)(1)(q4) 5 16p4 2 32p3q 1 24p2q2 2 8pq3 1 q4

10. (5 2 2y)3 5 [5 1 (22y)]3

5 3C0(5)3(22y)0 1 3C1(5)2(22y)1

1 3C2(5)1(22y)2 1 3C3(5)0(22y)3

5 (1)(125)(1) 1 (3)(25)(22y) 1 (3)(5)(4y2) 1 (1)(1)(28y3) 5 125 2 150y 1 60y2 2 8y3

5 28y3 1 60y2 2 150y 1 125

11. 7C2x5(23)2 5 (21)(x5)(9) 5 189x5

The coeffi cient of x5 is 189.

12. 8C5(2x)3(5)5 5 (56)(8x3)(3125) 5 1,400,000x3

The coeffi cient of x3 is 1,400,000.


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611Algebra 2


10.2 Exercises (pp. 694–697)

Skill Practice

1. The binomial expansion of (a 1 b)n is given by the nth row of Pascal’s triangle.

2. In a permutation, the order of the events is important, but in a combination, the order is not important.

3. 5C2 5 5! } 3! p 2! 5

5 p 4 p 3! } 3! p 2! 5 10

4. 10C3 5 10! } 7! p 3! 5

10 p 9 p 8 p 7! }} 7! p 3! 5 120

5. 9C6 5 9! } 3! p 6! 5

9 p 8 p 7 p 6! } 6! p 3! 5 84

6. 8C2 5 8! } 6! p 2! 5

8 p 7 p 6! } 6! p 2! 5 28

7. 11C11 5 11!

} 0!11! 5 1

8. 12C4 5 12! } 8! p 4! 5

12 p 11 p 10 p 9 p 8! }} 8! p 4! 5 495

9. 7C5 5 7! } 2! p 5! 5

7 p 6 p 5! } 2! p 5! 5 21

10. 14C6 5 14! } 8! p 6! 5

14 p 13 p 12 p 11 p 10 p 9 p 8! }}} 8! p 6! 5 3003

11. The denominator should have been multiplied by 2!;

6! }

(6 2 2)! p 2! 5

720 } 48 5 15

12. The denominator should have been multiplied by

(8 2 3)!; 8! }

(8 2 3)! p 3! 5

40,320 } 720 5 56

13. 12C5 5 12! } 7! p 5! 5

12 p 11 p 10 p 9 p 8 p 7! }} 7! p 5! 5 792 hands

14. 4C4 p 48C1 5 4! } 0! p 4! p

48! }

47! p 1!

5 4! } 0! p 4! p

48 p 47! }

47! p 1!

5 1 p 48

5 48 hands

15. 4C1 p 48C4 5 4! } 3! p 1! p

48! }

44! p 4!

5 4 p 3!

} 3! p 1! p 48 p 47 p 46 p 45 p 44!

}} 44! p 4!

5 4 p 194,580

5 778,320 hands

16. 13C5 1 13C5 5 13! } 8! p 5! 1

13! } 8! p 5!

5 13 p 12 p 11 p 10 p 9 p 8!

}} 8! p 5!

1 13 p 12 p 11 p 10 p 9 p 8!

}} 8! p 5!

5 1287 1 1287

5 2574 hands

17. One queen or zero queens:

(48C4 p 4C1

) 1 (48C5 p 4C0)

5 (194,580 p 4) 1 (1,712,304 p 1)

5 778,320 1 1,712,304

5 2,490,624 hands

18. Total hands 2 hands with no spade:

52C5 2 39C5 5 52 p 51 p 50 p 49 p 48 p 47!

}}} 47! p 5!

2 39 p 38 p 37 p 36 p 35 p 34!

}}} 34! p 5!

5 2,598,960 2 575,757

5 2,023,203 hands

19. 1

2

6

20

70

252

1

3

10

35

126

1

3

10

35

126

1

4

15

56

210

1

4

15

56

210

1

5

21

84

1

5

21

84

1

6

28

120

1

6

28

120

1

7

36

1

7

36

1

8

45

1

8

45

1

9

1

91

101

1

1

101

1

20. (x 1 3)6 5 1x6(3)0 1 6x5(3)1 1 15x4(3)2 1 20x3(3)3

1 15x2(3)4 1 6x(3)5 1 1x0(3)6

5 x6 1 18x5 1 135x4 1 540x3 1 1215x2

1 1458x 1 729

21. ( y 2 3z)10 5 [y 1 (23z)]10

5 1y10(23z)0 1 10y9(23z)1 1 45y8(23z)2

1 120y7(23z)3 1 210y6(23z)4

1 252y5(23z)5 1 210y4(23z)6

1 120y3(23z)7 1 45y2(23z)8

1 10y1(23z)9 1 1y0(23z)10

5 y10 2 30y9z 1 405y8z2 2 3240y7z3

1 17,010y6z4 2 61,236y5z5 1 153,090y4z6

2 262,440y3z7 1 295,245y2z8

2 196,830yz9 1 59,049z10

22. (a 1 b2)8 5 1a8(b2)0 1 8a7(b2)1 1 28a6(b2)2

1 56a5(b2)3 1 70a4(b2)4 1 56a3(b2)5

1 28a2(b2)6 1 8a1(b2)7 1 1a0(b2)8

5 a8 1 8a7b2 1 28a6b4 1 56a5b6

1 70a4b8 1 56a3b10 1 28a2b12

1 8ab14 1 b16

23. (2s 2 t4)7 5 [2s 1 (2t4)]7

5 1(2s)7(2t 4)0 1 7(2s)6(2t 4)1

1 21(2s)5(2t 4)2 1 35(2s)4(2t 4)3

1 35(2s)3(2t 4)4 1 21(2s)2(2t 4)5

1 7(2s)1(2t 4)6 1 1(2s)0(2t 4)7

5 128s7 2 448s6t 4 1 672s5t 8

2 560s4t12 1 280s3t16 2 84s2t 20

1 14st24 2 t28


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24. (x 1 2)3 5 3C0x3(2)0 1 3C1x2(2)1

1 3C2x1(2)2 1 3C3x0(2)3

5 (1)(x3)(1) 1 (3)(x2)(2)

1 (3)(x)(4) 1 (1)(1)(8)

5 x3 1 6x2 1 12x 1 8

25. (c 2 4)5 5 [c 1 (24)]5

5 5C0c5(24)0 1 5C1c4(24)1 1 5C2c3(24)2

1 5C3c2(24)3 1 5C4c1(24)4 1 5C5c0(24)5

5 (1)(c5)(1) 1 (5)(c4)(24) 1 (10)(c3)(16)

1 (10)(c2)(264) 1 (5)(c)(256)

1 (1)(1)(21024)

5 c5 2 20c4 1 160c3 2 640c2

1 1280c 2 1024

26. (a 1 3b)4 5 4C0a4(3b)0 1 4C1a3(3b)1 1 4C2a2(3b)2

1 4C3a1(3b)3 1 4C4a0(3b)4

5 (1)(a4)(1) 1 (4)(a3)(3b) 1 (6)(a2)(9b2) 1 (4)(a)(27b3) 1 (1)(1)(81b4) 5 a4 1 12a3b 1 54a2b2 1 108ab3 1 81b4

27. (4p 2 q)6 5 [4p 1 (2q)]6

5 6C0(4p)6(2q)0 1 6C1(4p)5(2q)1

1 6C2(4p)4(2q)2 1 6C3(4p)3(2q)3

1 6C4(4p)2(2q)4 1 6C5(4p)1(2q)5

1 6C6(4p)0(2q)6

5 (1)(4096p6)(1) 1 (6)(1024p5)(2q)

1 (15)(256p4)(q2) 1 (20)(64p3)(2q3) 1 (15)(16p2)(q4) 1 (6)(4p)(2q5) 1 (1)(1)(q6) 5 4096p6 2 6144p5q 1 3840p4q2

2 1280p3q3 1 240p2q4 2 24pq5 1 q6

28. (w3 2 3)4 5 [w3 1 (23)]4

5 4C0(w3)4(23)0 1 4C1

(w3)3(23)1

1 4C2(w3)2(23)2 1 4C3

(w3)1(23)3

1 4C4(w3)0(23)4

5 (1)(w12)(1) 1 (4) 1 w9 2 (23) 1 (6)(w6)(9)

1 (4)(w3)(227) 1 (1)(1)(81)

5 w12 2 12w9 1 54w6 2 108w3 1 81

29. (2s4 1 5)5 5 5C0(2s4)5(5)0 1 5C1

(2s4)4(5)1

1 5C2(2s4)3(5)2 1 5C3

(2s4)2(5)3

1 5C4(2s4)1(5)4 1 5C5

(2s4)0(5)5

5 (1)(32s20)(1) 1 (5)(16s16)(5)

1 (10)(8s12)(25) 1 (10)(4s8)(125)

1 (5)(2s4)(625) 1 (1)(1)(3125)

5 32s20 1 400s16 1 2000s12

1 5000s8 1 6250s4 1 3125

30. (3u 1 v2)6 5 6C0(3u)6(v2)0 1 6C1(3u)5(v2)1

1 6C2(3u)4(v2)2 1 6C3(3u)3(v2)3

1 6C4(3u)2(v2)4 1 6C5(3u)1(v2)5

1 6C6(3u)0(v2)6

5 (1)(729u6)(1) 1 (6)(243u5)(v2) 1 (15)(81u4)(v4) 1 (20)(27u3)(v6) 1 (15)(9u2)(v8) 1 (6)(3u)(v10) 1 (1)(1)(v12) 5 729u6 1 1458u5v2 1 1215u4v4

1 540u3v6 1 135u2v8 1 18uv10 1 v12

31. (x3 2 y2)4 5 [x3 1 (2y2)]4

5 4C0(x3)4(2y2)0 1 4C1 1 x3 2 3(2y2)1

1 4C2(x3)2(2y2)2 1 4C3 1 x3 2 1(2y2)3

1 4C4(x3)0(2y2)4

5 (1)(x12)(1) 1 (4)(x9)(2y2) 1 (6)(x6)( y4) 1 (4)(x3)(2y6) 1 (1)(1)( y8) 5 x12 2 4x9y2 1 6x6y4 2 4x3y6 1 y8

32. 10C5x5(22)5 5 252x2(232) 5 28064x5


33. 5C2(3x)3 1 22 2 5 10 1 27x3 2 (4) 5 1080x3


34. 8C5 1 x2 2 3(23)5 5 56x6(2243) 5 213,608x6

The coeffi cient of x6 is 213,608.

35. A; 7C3x4(23)3 5 35x4(227) 5 2945x4


36. Row 0: 1, row 1: 2, row 2: 4, row 3: 8, row 4: 16, row n: 2n

37. The sum along each diagonal segment is equal to the sum of the two previous diagonal segment sums.

38. Permutations

12P2 5 12! }

(12 2 2)! 5

12! } 10! 5

479,001,600 } 3,628,800 5 132 ways

39. Combinations

280C5 5 280! } 275! p 5! 5

280 p 279 p 278 p 277 p 276 p 275! }}} 275! p 5!

5 13,836,130,056 ways

40. C; 20P4 5 20! }

(20 2 4)! 5

20! } 16! 5

29 p 19 p 18 p 17 p 16! }} 16!

5 116,280 ways

41. There is only one combination of n objects taken from a group of n objects, so nCn 5 1.

1 5 nCn

5 n! }

(n 2 n)! p n!

5 n! } 0! p n!

5 1 } 0!

So, 0! must equal 1.

42. nC0 5 n! } n! p 0! 5 1 43. nCn 5

n! } 0! p n! 5 1


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613Algebra 2


44. nCr p rCm 0 nCm p n 2 mCr 2 m

n! }

(n 2 r)! p r! p r!

} (r 2 m)! p m!

0 n! }}

(n 2 m)! p m! p

(n 2 m)! }}}

[(n 2 m) 2 (r 2 m)]! p (r 2 m)!

n! }}

(n 2 r)! p (r 2 m)! p m! 5

n! }}

m! p (n 2 r)! p (r 2 m)! ✓

45. nC1 0 nP1

n! }

(n 2 1)! p 1! 0

n! }

(n 2 1)!

n! }

(n 2 1)! 5

n! }

(n 2 1)! ✓

46. nCr 0 nCn 2 r

n! }

(n 2 r)! p r! 0

n! }}

[n 2 (n 2 r)]! p (n 2 r)!

n! }

(n 2 r)! p r! 5

n! }

r! p (n 2 r)! ✓

47. n 1 1Cr 0 nCr 1 nCr 2 1

(n 1 1)! }}

(n 1 1 2 r)! p r! 0

n! }

(n 2 r)! p r! 1

n! }}

[n 2 (r 2 1)]! p (r 2 1)!

0 n! }

(n 2 r)! p r! 1

n! }}

(n 1 1 2 r)! p (r 2 1)!

0 n! p (n 1 1 2 r)

}} (n 2 r)! p r! p (n 1 1 2 r)

1 n! p r

}} (n 1 1 2 r)! p (r 2 1)! p r

0 n! p (n 1 1 2 r) 1 n! p r

}} (n 1 1 2 r)! p r!

0 n! p (n 1 1 2 r 1 r)

}} (n 1 1 2 r)! p r!

0 n! p (n 1 1)

}} (n 1 1 2 r)! p r!

5 (n 1 1)! }}

(n 1 1 2 r)! p r! ✓

Problem Solving

48. 5C2 p 3C1 5 5! } 3! p 2! p

3! }

2! p 1! 5

5 p 4 p 3! } 3! p 2! p 3 p 2!

} 2! p 1!

5 10 p 3 5 30

You can choose 30 sets of music types.

49. 18C3 5 18! } 15! p 3! 5

18 p 17 p 16 p 15! }} 15! p 3! 5 816

You can use 816 combinations of fl ower types in your bouquet.

50. 20C14 1 20C15 1 20C16 1 20C17 1 20C18 1 20C19 1 20C20

5 38,760 1 15,504 1 4845 1 1140 1 190 1 20 1 1

5 60,460

You can play 60,460 combinations of arcade games.

51. a. 20C5 5 20! } 15! p 5!

5 20 p 19 p 18 p 16 p 17 p 16 p 15!

}}} 15! p 5! 5 15,504

During the fi rst episode, 15,504 combinations of singers can be eliminated.

b. 15C5 5 15! } 10! p 5! 5

15 p 14 p 13 p 12 p 11 p 10! }}} 10! p 5! 5 3003

During the second episode, 3003 combinations of singers can be eliminated.

c. 10C5 5 10! } 5! p 5! 5

10 p 9 p 8 p 7 p 6 p 5! }} 5! p 5! 5 252

During the third episode, 252 combinations of singers can be eliminated.

5C2 5 5! } 3! p 2! 5

5 p 4 p 3! } 3! p 2! 5 10

During the fourth episode, 10 combinations of singers can be eliminated.

3C2 5 3! } 2! p 1! 5

3 p 2! } 2! p 1! 5 3

During the fi fth episode, 3 combinations of singers can be eliminated.

d. 20C5 p 15C5 p 10C5 p 5C2 p 3C2 5 351,982,350,700

There are 351,982,350,700 ways in which the 20 original contestants can be eliminated to produce a winner.

52. a. WWWWWRRRRRRRMMM

Number of different ways to assign 5 W’s, 7 R’s, and 3 M’s to the 15 students:

15! }

5! p 7! p 3! 5 360,360 possible job assignments

b. Ways to choose 5 students from 15:

15C5 5 15! } 10! p 5! 5 3003

Ways to choose 7 students from 10:

10C7 5 10! } 3! p 7! 5 120

Ways to choose 3 students from 3:

3C3 5 3! } 0! p 3! 5 1

3003 p 120 p 1 5 360,360 possible job assignments

c. The results from parts (a) and (b) are the same. Part (a) shows the different permutations of 5 W’s, 7 R’s, and 3 M’s to a group of 15 students. Part (b) shows the ways to select 5 students for washing, 7 students for repainting, and 3 students for maintenance. They are two different ways to count the same thing.

53 a. The number of line segments that join pairs of verticesis nC2.

nC2 5 n! }

(n 2 2)! p 2! 5

n p (n 2 1) p (n 2 2)! }}

(n 2 2)! p 2! 5

n(n 2 1) } 2

b. Because the sides of a polygon are not considered diagonals, subtract the number of sides from part (a).

n(n 2 1)

} 2 2 n 5

n2 2 n 2 2n } 2 5

n2 2 3n } 2 5

n(n 2 3) } 2

A formula for the number of diagonals of an n-sided

convex polygon is n(n 2 3)

} 2 .


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Mixed Review

54. A 5 πr2 5 π(16)2 ø 804.25

The area is about 804 square inches.

55. A 5 lw 5 (8.25)(12.1) 5 99.825

The area is 99.825 square feet.

56. A 5 1 } 2 bh 5

1 } 2 (15)(18) 5 135

The area is 135 square centimeters.

57. A 5 1 } 2 h(b1 1 b2

) 5 1 } 2 (9)(12 1 16) 5 126

The area is 126 square meters.

58. 8 Ï}

4x 2 5 5 19 59. (x 2 2)3/2 5 216

8 Ï}

4x 5 24 F (x 2 2)3/2 G 2/3 5 2162/3

Ï}

4x 5 3 x 2 2 5 (2161/3)2

( Ï}

4x )2 5 32 x 2 2 5 (3 Ï}

216 )2

4x 5 9 x 2 2 5 62

x 5 9 } 4 x 2 2 5 36

x 5 38

60. ln(x 1 4) 5 ln 5

x 1 4 5 5

x 5 1

61. 104x 2 5 5 11

104x 5 16

log 104x 5 log 16

4x 5 log 16

x 5 log 16

} 4

x ø 0.301

62. x }

x 2 2 5

x 1 3 } x 1 1

x(x 1 1) 5 (x 2 2)(x 1 3)

x2 1 x 5 x2 1 x 2 6

0 5 26

No solution

63. 1 }

x 2 3 1 3 5

2x } x 1 3

(x 2 3)(x 1 3) 1 1 }

x 2 3 1 3 2 5 1 2x

} x 1 3 2 (x 2 3)(x 1 3)

(x 1 3) 1 3(x 2 3)(x 1 3) 5 2x(x 2 3)

x 1 3 1 3 1 x2 2 9 2 5 2x(x 2 3)

x 1 3 1 3x2 2 27 5 2x2 2 6x

x2 1 7x 2 24 5 0

x 5 27 6 Ï

}}

72 2 4(1)(224) }}

2(1)

x 5 27 6 Ï

}

145 } 2

64. (24, 22), (6, 2)

Midpoint: 1 24 1 6 }

2 ,

22 1 2 }

2 2 5 (1, 0)

m 5 2 2 (22)

} 6 2 (24)

5 4 } 10 5

2 } 5

2 1 } m 5 2

5 } 2

Equation of perpendicular bisector:

y 2 0 5 2 5 } 2 (x 2 1)

y 5 2 5 } 2 x 1

5 } 2

65. (9, 22), (3, 6)

Midpoint: 1 9 1 3 }

2 ,

22 1 6 }

2 2 5 (6, 2)

m 5 6 2 (22)

} 3 2 9 5 8 }

26 5 2 4 } 3

2 1 } m 5

3 } 4


y 2 2 5 3 } 4 (x 2 6)

y 2 2 5 3 } 4 x 2

9 } 2

y 5 3 } 4 x 2

5 } 2

66. (28, 213), (7, 10)

Midpoint: 1 28 1 7 }

2 ,

213 1 10 }

2 2 5 1 2

1 } 2 , 2

3 } 2 2

m 5 10 2 (213)

} 7 2 (28)

5 23

} 15

2 1 } m 5 2

15 } 23


y 2 1 2 3 } 2 2 5 2

15 } 23 1 x 2 1 2

1 } 2 2 2

y 1 3 } 2 5 2

15 } 23 1 x 1

1 } 2 2

y 1 3 } 2 5 2

15 } 23 x 2

15 } 46

y 5 2 15

} 23 x 2 42

} 23

67. (6, 9.3), (0, 8.2)

Midpoint: 1 6 1 0 }

2 ,

9.3 1 8.2 }

2 2 5 1 3,

175 }

20 2

m 5 8.2 2 9.3

} 0 2 6 5 11

} 60

2 1 } m 5 2

60 } 11


y 2 175

} 20 5 2 60

} 11 (x 2 3)

y 2 175

} 20 5 2 60

} 11 x 1 180

} 11

y 5 2 60

} 11 x 1 1105

} 44


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615Algebra 2


Quiz 10.1–10.2 (p. 697)

1. a. 26 p 26 p 10 p 10 p 10 5 676,000 license plates

b. 26 p 25 p 10 p 9 p 8 5 468,000 license plates

2. a. 10 p 10 p 10 p 26 p 26 p 26

5 17,576,000 license plates

b. 10 p 9 p 8 p 26 p 25 p 24 5 11,232,000 license plates

3. 4!

} 2!

5 24

} 2 5 12

4. 5! 5 120

5. 6! }

2! p 2! 5

720 } 4 5 180

6. 9! }

4! p 2! p 2! 5

362,880 } 96 5 3780

7. 8C6 5 8! } 2! p 6! 5

8 p 7 p 6! } 2! p 6! 5 28

8. 7C4 5 7! } 3! p 4! 5

7 p 6 p 5 p 4! } 3! p 4! 5 35

9. 9C0 5 9! } 9! p 0! 5 1

10. 12C11 5 12! } 1! p 11! 5

12 p 11! } 1 p 11! 5 12

11. (x 1 5)5 5 5C0x5(5)0 1 5C1x4(5)1 1 5C2 x3(5)2

1 5C3x2(5)3 1 5C4x1(5)4 1 5C5x0(5)5

5 (1)(x5)(1) 1 (5)(x4)(5) 1 (10)(x3)(25)

1 (10)(x2)(125) 1 (5)(x)(625)

1 (1)(1)(3125)

5 x5 1 25x4 1 250x3 1 1250x2 1 3125x 1 3125

12. (2s 2 3)6 5 [2s 1 (23)]6

5 6C0(2s)6(23)0 1 6C1(2s)5(23)1

1 6C2(2s)4(23)2

1 6C3(2s)3(23)3 1 6C4(2s)2(23)4

1 6C5(2s)1(23)5 1 6C6(2s)0(23)6

5 (1)(64s6)(1) 1 (6)(32s5)(23)

1 (15)(16s4)(9)

1 (20)(8s3)(227) 1 (15)(4s2)(81)

1 (6)(2s)(2243) 1 (1)(1)(729)

5 64s6 2 576s5 1 2160s4 2 4320s3

1 4860s2 2 2916s 1 729

13. (3u 1 v)4 5 4C0(3u)4v0 1 4C1(3u)3v1 1 4C2(3u)2v2

1 4C3(3u)1v3 1 4C4(3u)0v4

5 (1)(81u4)(1) 1 (4)(27u3)(v) 1 (6)(9u2)(v2) 1 4(3u)(v3) 1 (1)(1)(v4) 5 81u4 1 108u3v 1 54u2v2 1 12uv3 1 v4

14. (2x3 2 3y)5 5 [2x3 1 (23y)]5

5 5C0(2x3)5(23y)0 1 5C1

(2x3)4(23y)1

1 5C2(2x3)3(23y)2 1 5C3

(2x3)2(23y)3

1 5C4(2x3)1(23y)4 1 5C5

(2x3)0(23y)5

5 (1)(32x15)(1) 1 (5)(16x12)(23y)

1 (10)(8x9)(9y2) 1 (10)(4x6)(227y3)

1 (5)(2x3)(81y4) 1 (1)(1)(2243y5) 5 32x15 2 240x12y 1 720x9y2 2 1080x6y3

1 810x3y4 2 243y5

15. (x 1 2)9

The seventh term is 9C6x3(2)6 5 84x3(64) 5 5376x3.

The coeffi cent of x3 is 5376.

16. 5C1 p 10C1 p 6C2 5 5 p 10 p 15 5 750

There are 750 different variations of the pizza special.

Lesson 10.3


1. P(perfect square) 5 Perfect squares from 1 to 20

}}} Numbers from 1 to 20

5 4 } 20

5 1 } 5

2. P(factor of 30) 5 Factors of 30 from 1 to 20

}} Numbers from 1 to 20

5 7 } 20

3. a. P(alphabetical order) 5 1 } 9! 5

1 } 362,880

The probability that 9 musicians perform in

alphabetical order decreases to 1 }

362,880 .

b. P(fi rst 2 performers are your friends) 5 4C2

} 9C2

5 6 } 36 5

1 } 6

The probability that the fi rst 2 performers are your

friends would decrease to 1 }

6 .

4. Odds in favor of drawing a heart

5 Number of hearts

}} Number of non-hearts

5 13

} 39

5 1 } 3

5. Odds aganst drawing a queen

5 Number of non-queens

}} Number of queens

5 48

} 4 5 12

} 1

6. a. Of those surveyed, 463 1 1085 1 879 5 2427 would prefer to be at most 39 years old.

P(at most 39 years old) 5 2427

} 3516 ø 0.69

b. Of those surveyed, 879 1 551 1 300 1 238 5 1968 would prefer to be at least 30 years old.

P(at least 30 years old) 5 1968

} 3516 ø 0.56

7. P(5 points) 5 Area of 5-point region

}} Area of entire board

5 (π p 62) 2 (π p 32)

}} 182

5 36 π 2 9 π

} 324 5 27 π

} 324

5 π

} 12 ø 0.262

P(0 points) 5 Area outside largest circle


5 182 2 (π p 92)

}} 182

5 324 2 81 π

} 324 5 4 2 π

} 4 ø 0.215

Because 0.262 > 0.215, you are more likely to get 5 points.


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10.3 Exercises (pp. 701-704)

Skill Practice

1. A probability that is the ratio of two lengths, areas, or volumes is called a geometric probability.

2. A theoretical probability is based on the number of outcomes of the event and the total number of possible outcomes if all outcomes are equally likely to occur. An experimental probability is determined through an experiment, survey, or historical data about an event.

The theoretical probability of rolling a 5 using

a 6-sided die is 1 }

6 . If you actually rolled a 6-sided die

100 times, the experimental probability of rolling a 5

would be the number of fi ves you rolled divided by 100.

3. P(even number) 5 Even numbers from 1 to 50


5 25

} 50 5 1 } 2

4. P(number less than 35) 5 Numbers from 1 to 50 less than 35


5 34

} 50 5 17

} 25

5. P(perfect square) 5 Perfect squares from 1 to 50


5 7 } 50

6. P(prime number) 5 Primes from 1 to 50


5 15

} 50 5 3 } 10



5 10

} 50 5 1 } 5

8. P(multiple of 4) 5 Multiples of 4 from 1 to 50


5 12

} 50 5 6 } 25

9. P(two digit number) 5 Two digit numbers from 1 to 50


5 41

} 50

10. P(perfect cube) 5 Perfect cubes from 1 to 50


5 3 } 50

11. P(king of diamonds) 5 Kings of diamonds

}} Number of cards

5 1 } 52

12. P(king) 5 Number of kings

}} Number of cards

5 4 } 52 5

1 } 13

13. P(spade) 5 Number of spades

}} Number of cards

5 13

} 52 5 1 } 4

14. P(black card) 5 Number of black cards

}} Number of cards

5 26

} 52 5 1 } 2

15. P(card other than a 2) 5 Number of non-twos

}} Number of cards

5 48

} 52 5 12

} 13

16. P(face card) 5 Number of face cards

}} Number of cards

5 12

} 52 5 3 } 13

17. P(6 correct numbers) 5 Number of correct combinations

}}} Ways to choose 6 numbers from 48

5 1 }

48C6 5

1 } 12,271,512

18. P(4 correct numbers) 5 Number of correct permutations

}}} Ways to choose 4 numbers

5 1 }

104 5 1 } 10,000

19. C ; P(both months have 31 days)

5 Ways to choose 2 months with 31 days

}}} Ways to choose 2 months

5 7C2

} 12C2

5 21

} 66

ø 0.318

20. Odds in favor of choosing white

5 Number of whites

}} Number of non-whites

5 4 } 24 5

1 } 6

21. Odds in favor of choosing blue 5 Number of blues

}} Number of non-blues

5 6 } 22 5

3 } 11

22. Odds against choosing red 5 Number of non-reds

}} Number of reds

5 20

} 8 5 5 } 2

23. Odds against choosing black 5 Number of non-blacks

}} Number of blacks

5 18

} 10 5 9 } 5

24. When calculating the odds against an event, the denominator should describe the number of outcomes in the event, 2, not the total number of outcomes, 6;

Odds against 5 or 6 5 4 } 2 =

2 }

1 .

25. The outcomes not in the event, 4, should be in the

numerator and the outcomes in the event, 2, should be in

the denominator. Odds against 5 or 6 5 4 } 2 5

2 } 1 .

26. Answers will vary.

27. If the probability of event A is 0.3, or 3 }

10 , then there are

3 outcomes in event A out of 10 total outcomes. So, 7 of

the outcomes are not in event A, and the odds in favor of

event A are 3 } 7 .

28. Experimental probability: 27

} 150

5 0.18

Theoretical probabiltiy: 1 }

6 ø 0.17

29. Experimental probability: 22 1 26 1 30

}} 150

5 78

} 150 5 0.52

Theoretical probability: 1 }

2 5 0.5

30. Experimental probability: 27 1 22 1 18 1 26

}} 150

5 93

} 150

5 0.62

Theoretical probability: 2 } 3 ø 0.67

31. Experimental probability: 150 2 18

} 150

5 132

} 150 5 0.88

Theoretical probability: 5 }

6 ø 0.83

32. A; 37

} 80

5 0.4625


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617Algebra 2


33. y 5 x2 2 6x 1 c

x–coordinate of vertex: x 5 2 b } 2a 5 2

6 }

2(1) 5 3

y–coordinate of vertex: y 5 32 2 6(3) 1 c 5 29 1 c

When c > 9, the vertex is above the x-axis.

So, P(vertex is above x-axis) 5 11

} 20

34. a. Target A

P(circle) 5 Area of circle


5 π p 62

} 12 5 36 π

} 144 5 π

} 4

Target B

P(circle) 5 Area of 4 circles


5 4( π p 32)

} 122 5

4(9π ) } 144 5

36 π } 144 5

π } 4

Target C

P(circle) 5 Area of 9 circles


5 9( π p 22)

} 122 5

9(4π ) } 144 5

36 π } 144 5

π } 4

b. For a square target with sides 12 inches long

containing n2 identical circles arranged in n rows and

n columns the probability that a dart lands inside a

circle will always be π

} 4 . The radius of one circle is

6 } n , so the area of one circle is π 1 6 }

n 2 2, or

36 π }

n2 . Because

there are n2 identical circles, the total area of all the

circles is n2 p 36 π }

n2 , or 36π. So, the probability that a

dart lands inside one of these circles on a square target

with sides 12 inches long is 36 π

} 122 5

36 π } 144 5

π } 4 .

Problem Solving

35. P(shaded region) 5 Area of triangle

}} Area of square

5 1 } 2 (10)(10)

} 102 5

50 } 100 5

1 } 2

The probability of a dart hitting the shaded region is 1 }

2 .

36. P(shaded region) 5 Area of triangle

}} Area of circle

5 1 } 2 (16)(8)

} π 82 5

64 } 64 π 5

1 } π

The probability of a dart hitting the shaded region is 1 } π� .

37. P(shaded region) 5 Area of square 2 Area of circle

}}} Area of square

5 142 2 ( π p 72)

}} 142 5

196 2 49 π } 196

5 1 2 π

} 4

The probability of a dart hitting the shaded region is

1 2 π

} 4 .

38. There are 30C12 ways to choose 12 people for the jury.

Of these, 18C12 are 12 women. So, the probability is:

P(12 women) 5 18C12

} 30C12

5 18,564

} 86,493,225 ø 0.000215

39. P(center circle) 5 Area of center circle

}} Area of entire target

5 π p 82

} π p 402 5

64 π } 1600 π 5

1 } 25

The probability of an arrow shot at the target hitting the

center circle is 1 }

25 .

40. a. P(bus) 5 376,900

} 1,181,100 5 3769

} 11,811

b. 3769

} 11,811

ø 0.319

c. 0.319 5 31.9% or about 32%

d. Odds in favor of the trip having been on a bus

5 3769 }} 11,811 – 3769

5 3769

} 8042

41. The length of the shoreline along the Gulf of Mexico is

367 1 397 1 44 1 53 1 770 5 1631 miles.

a. P(Texas) 5 367

} 1631

The probability that the ship lands in Texas is 367

} 1631

.

b. P(Florida) 5 770

} 1631 5 110

} 233

The probability that the ship lands in Florida is 110

} 233

.

c. P(Alabama) 5 53 } 1631

The probability that the ship lands in Alabama is 53 }

1631 .

42. a. P(3 and 4) 5 Number of correct combinations

}}} Ways to choose 2 numbers from 5

5 1 }

5C2 5

1 } 10

The probability of selecting the cards 3 and 4 is 1 }

10 .

b. P(2 correct cards) 5 Number of correct guesses

}}} Ways to guess 2 numbers from 5

5 1 }

5C2 5

1 } 10

The probability that the magician can correctly identify

two numbers by guessing is 1 }

10 .

c. No. Sample answer: If the magician were really a mind reader, the magician would be able to identify the numbers 100% of the time.


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43. a. P(all 4 pegs correct) 5 Number of correct permutations

}}} Number of possible permutations

5 1 }

4! 5

1 } 24

The probability that the second player has all 4 pegs

correct is 1 }

24 .

b. There are 4 ways to place a correct peg. After the correct peg is place, there are only 2 ways to arrange the remaining 3 pegs so that none are correct. So, there are 4 p 2 5 8 ways to have exactly one peg correct.

P(exactly 1 peg correct) 5 8 } 4! 5

8 } 24 5

1 } 3

c. There is only one way to choose the 2 pegs that are placed incorrectly. The number of ways to choose 2 pegs from 4 is 4C2 5 6.

P(all pegs now correct) 5 Ways to pick the 2 incorrect pegs

}}} Ways to pick 2 pegs from 4

5 1 }

4C2 5

1 } 6

Mixed Review

44. y 5 4(0.75)x 45. f (x) 5 3e22x

1

y

21 x

1

y

21 x

46. y 5 ln x 1 2 47. y 5 1 3 } 2 2 x

1

y

21 x

2

y

21 x

48. g(x) 5 21

} x 1 1 2 2 49. y 5 3x 1 1

} x2 2 4

1

y

23 x

1

y

21 x

50. log464

43 5 64, so log4 64 5 3

51. ln e

1n e 5 loge e 5 1

52. log6 36

62 5 36, so log6 36 5 2

53. log2 512

29 5 512 so log 2512 5 9

54. ln e2.9

ln e2.9 5 loge e2.9 5 2.9

55. log1/3 9

1 1 } 3 2

22 5 9, so log1/3 9 5 22

56. log9 27

93/2 5 27, so log9 27 5 3 } 2

57. log4 1 }

32

425/2 5 1 } 32 , so log4

1 }

32 5 2

5 } 2

58. 8C3 5 8! } 5! p 3! 5

8 p 7 p 6 p 5! } 5! p 3! 5 56

59. 10C9 5 10!

} 1!9! 5 10 p 9!

} 1! p 9! 5 10

60. 7C4 5 7! } 3!4! 5

7 p 6 p 5 p 4! } 3! p 4! 5 35

61. 12C5 5 12!

} 7!5! 5 12 p 11 p 10 p 9 p 8 p 7!

}} 7! p 5! 5 792

Mixed Review of Problem Solving (p. 705)

1. a. 5P5 5 5!

} 0! 5 120

} 1 5 120

There are 120 ways to seat 5 people in 5 empty seats.

b. 8P5 5 8!

} 3! 5 40,320

} 6 5 6720

There are 6720 ways to seat 5 people in 8 empty seats.

c. The number of ways to seat 5 people in n empty seats

is nP5 5 n! }

(n 2 5)!

d. nP5 ≥ 1,000,000, so n 5 18 because 17P5 5 742,560 and 18P5 5 1,028,160.

2. a. The arrival times can be represented by 0 ≤ x ≤ 30 and 0 ≤ y ≤ 30.

b. If your arrival times must not exceed 10 minutes of each other, then x 2 y ≤ 10 and y 2 x ≤ 10.

c.

Your arrival time(minutes after 9:00)

Frie

nd

’s a

rriv

al t

ime

(min

ute

s af

ter

9:00

)

0 10 20 305 15 25 35 x

y

0

10

20

30

5

15

25

35


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619Algebra 2


d. P(you and friend meet) 5 Area of shaded region

}} Area of square

5 302 2 2 1 1 }

2 2 (20)(20)

}} 302

5 900 2 400

} 900 5 500

} 900 5 5 } 9

The probability that you and your friend will meet at

the gym is 5 }

9 .

3. 8C3 5 8! } 5! p 3! 5

8 p 7 p 6 p 5! } 5! p 3! 5 56

You can make 56 different fruit smoothies using 3 of the 8 fruits.

4. F So So J J Sr Sr Sr

Number of different ways for 1 freshman, 2 sophomores, 2 juniors, and 3 seniors to line up:

8! }

2! p 2! p 3! 5 1680 possible ways

5. a. P(Track and Field) 5 233

} 1013 ø 0.23

The probability of a U.S. adult wanting to participate in Track and Field is about 0.23.

b. This is an example of experimental probability because the answer is based on the results of a survey.

c. Odds in favor of Gymnastics 5 101 } 1013 2 101 5

101 } 912

The odds in favor of a US adult preferring to

participate in gymnastics is 101

} 912

.

6. Finding the number of possible course selections will involve combinations because the order in which you must select the courses is not specifi ed.

30C18 5 30! } 12! p 18! 5 86,493,225

There are 86,493,225 ways to select 18 of the 30 courses.

7. Sample answer: You are buying your mother a bouquet of fl owers for her birthday. You want to include 3 of the 7 types of roses and 4 of the 8 types of calla lilies. How many different bouquets of fl owers do you have to choose from?

7C3 p 8C4 5 7! }

4! p 3! p 8!

} 4! p 4!

5 2450

8. 31C3 5 31! } 28! p 3! 5

31 p 30 p 29 p 28! }} 28! p 3! 5 4495

There are 4495 different ice cream cones that can be made with 3 of the 31 fl avors.

Lesson 10.4

Investigating Algebra Activity 10.4 (p. 706)

Answers will vary.


1. P(ace or 8) 5 P(ace) 1 P(8)

5 4 } 52 1

4 } 52 5

8 } 52 5

2 } 13

2. P(10 or diamond) 5 P(10) 1 P(diamond)

2 P(10 and diamond)

5 4 } 52 1

13 } 52 2

1 } 52 5

16 } 52 5

4 } 13

3. P(band or honor roll) 5 P(band) 1 P(honor roll)

2 P(both)

64

} 200

5 32

} 200 1 51

} 200 2 P(both)

P(both) 5 32

} 200 1 51

} 200 2 64

} 200

P(both) 5 19

} 200 5 0.095

The probability that a randomly selected senior is both in the band and on the honor roll is 0.095.

4. P (A) 5 1 2 P(A) 5 1 2 0.45 5 0.55

5. P(A) 5 1 2 P(A) 5 1 2 1 } 4 5

3 } 4

6. P(A) 5 1 2 P(A) 5 1 2 1 5 0

7. P(A) 5 1 2 P(A) 5 1 2 0.03 5 0.97

8. P(at least 2 are the same) 5 1 2 P(none are the same)

5 1 2 100 p 99 p 98 p 97 p 96

}} 1005

ø 0.097

If there are only 100 different hidden messages inside the fortune cookies, the probability that at least 2 of the 5 people have the same message increases to about 0.097.

10.4 Exercises (pp. 710–713)

Skill Practice

1. The union or intersection of two events is called a compound event.

2. Yes; the events in A are those events that are not in A.

Sample answer:

Event: You go on a rafting trip.

Complement: You do not go on a rafting trip.

3. P(A or B) 5 P(A) 1 P(B) 5 0.3 1 0.1 5 0.4

4. P(A or B) 5 P(A) 1 P(B) 5 0.55 1 0.2 5 0.75

5. P(A or B) 5 P(A) 1 P(B) 5 0.41 1 0.24 5 0.65

6. P(A or B) 5 P(A) 1 P(B) 5 2 } 5 1

3 } 5 5

5 } 5 5 1

7. P(A or B) 5 P(A) 1 P(B) 5 1 } 3 1

1 } 4 5

4 } 12 1

3 } 12 5

7 } 12

8. P(A or B) 5 P(A) 1 P(B) 5 2 } 3 1

1 } 5 5

10 } 15 1

3 } 15 5

13 } 15

9. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)

5 0.5 1 0.35 2 0.2 5 0.65


0.7 5 0.6 1 0.2 2 P(A and B)

P(A and B) 5 0.6 1 0.2 2 0.7 5 0.1


0.71 5 0.28 1 0.64 2 P(A and B)

P(A and B) 5 0.28 1 0.64 2 0.71 5 0.21


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5 0.46 1 0.37 2 0.31 5 0.52


5 2 } 7 1

4 } 7 2

1 } 7 5

5 } 7


7 } 11 5

6 } 11 1

3 } 11 2 P(A and B)

P(A and B) 5 6 } 11 1

3 } 11 2

7 } 11 5

2 } 11

15. B; P(A or B) 5 P(A) 1 P(B) 2 P(A and B)

5 0.41 1 0.53 2 0.27 5 0.67

16. P(A) 5 1 2 P(A) 5 1 2 0.5 5 0.5

17. P(A) 5 1 2 P(A) 5 1 2 0 5 1

18. P(A) 5 1 2 P(A) 5 1 2 1 } 3 5

2 } 3

19. P(A) 5 1 2 P(A) 5 1 2 5 } 8 5

3 } 8

20. P(king and diamond) 5 1 } 52

21. P(king or diamond) 5 P(king) 1 P(diamond)

2 P(king and diamond)

5 4 } 52 1

13 } 52 2

1 } 52 5

16 } 52 5

4 } 13

22. P(spade or club) 5 P(spade) 1 P(club)

5 13

} 52 1 13

} 52 5 26

} 52 5 1 } 2

23. P(4 or a 5) 5 P(4) 1 P(5)

5 4 } 52 1

4 } 52 5

8 } 52 5

2 } 13

24. P(6 and a face card) 5 0 } 52 5 0

25. P(not a heart) 5 1 2 P(heart)

5 1 2 13

} 52 5 1 2 1 } 4 5

3 } 4

26. P(heart or face card) are overlapping events, so the intersection must be subtracted from the sum of the separate probabilites.

P(heart or face card) 5 P(heart) 1 P(face card)

2 P(heart and face card)

5 13

} 52 1 12

} 52 2 3 } 52 5

22 } 52 5

11 } 26

27. The intersection must be subtracted, not added.

P(club or 9) 5 P(club) 1 P(9) 2 P(club and 9)

5 13

} 52 1 4 } 52 2

1 } 52 5

16 } 52 5

4 } 13


0.50 5 0.25 1 0.4 2 P(A and B)

P(A and B) 5 0.25 1 0.4 2 0.50 5 0.15

A and B are not disjoint.


5 0.6 1 0.32 2 0.25 5 0.67



0.65 5 P(A) 1 0.38 2 0

0.27 5 P(A)

A and B are disjoint.


3 } 5 5

8 } 15 1 P(B) 2

2 } 15

3 }

15 5 P(B)

1 } 5 5 P(B)



2 }

3 5

1 } 2 1

1 } 6 2 P(A and B)

P(A and B) 5 1 } 2 1

1 } 6 2

2 } 3 5

0 } 6 5 0

A and B are disjoint.


32% 5 16% 1 P(B) 2 8%

32% 5 P(B) 1 8%

24% 5 P(B)


34. Sample answer: Two disjoint events: Jordan will either go biking or swimming at 4 P.M.; Two overlapping events:Mr. Peterson has 23 students enrolled in his martial arts courses. 18 of these students are enrolled in Karate and 12 are enrolled in Muay Thai. Some students are enrolled in both.

35. P(sum is 3 or 4) 5 P(3) 1 (4)

5 2 } 36 1

3 } 36 5

5 } 36

36. P(sum is not 7) 5 1 2 P(sum is 7)

5 1 2 6 } 36 5

30 } 36 5

5 } 6

37. P(sum ≥ 5) 5 1 2 P(sum < 5)

5 1 2 6 } 36 5

30 } 36 5

5 } 6

38. P(sum < 8 or > 11) 5 P( < 8) 1 P( > 11)

5 [1 2 P( ≥ 8)] 1 P( > 11)

5 1 1 2 15

} 36 2 1 1 } 36

5 21

} 36 1 1 } 36 5

22 } 36 5

11 } 18

39. C; P (sum is prime)

5 P(2) 1 P(3) 1 P(5) 1 P(7) 1 P(11)

5 1 } 36 1

2 } 36 1

4 } 36 1

6 } 36 1

2 } 36

5 15

} 36 5 5 } 12


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621Algebra 2


40. When P(A) and P(B) are added, the outcomes in the intersection of A and B are counted twice. So, P(A and B) must be added to P(A or B) to make the equation true.

A

P(A)

B A B A B A B1 5 1

1 5 1P(B) P(A or B) P(A and B)

41. To fi nd P(A or B or C), fi rst you add the probabilities of events A, B, and C. Then, you subtract the probabilitites of the common outcomes for A and B, B and C, and A and C. Finally, you must add the probability of the common outcomes of A, B, and C becuase this was subtracted three times in the previous step.

A B

C

A B

C

A B

C

A B

C

A B

C

A B

C

A B

C

A B

C

5 1 1

5 1 1

2 2 2

1

2 2 2

1

P(A or B or C) P(A) P(B) P(C)

P(A and B) P(B and C) P(A and C)

P(A and B and C)

Problem Solving

42. P(you or your best friend) 5 P(you) 1 P(best friend)

5 45% 1 25%

5 70%

The probability that either you or your best friend will win the election is 70%.

43. P(visible or ultraviolet) 5 P(visible) 1 P(ultraviolet)

2 P(visible and ultraviolet)

5 12

} 30 1 15

} 30 2 6 } 30 5

21 } 30 5

7 } 10

The probability that a plant in the experiment receives

either visible light or ultraviolet light is 7 }

10 .

44. D; P(rains on Sunday) 5 80%

P(does not rain on Saturday) 5 100% 2 30% 5 70%

P(rains on Monday) 5 90%

P(does not rain on Friday) 5 100% 2 5% 5 95%

P(does not rain on Friday) is the greatest probability.

45. P(at least 2 bring the same)

5 1 2 P(none are the same)

5 1 2 10 p 9 p 8 p 7 p 6 p 5

}} 106

5 1 2 0.1512 5 0.8488

The proability that at least 2 of the 6 cast members bring the same item is 0.8488.

46. P(at least 2 have the same code)


5 1 2 4096 p 4095 p 4094 p 4093 p 4092 p 4091

}}}} 40966

ø 1 2 0.99634 5 0.00366

The probability that at least 2 of the 6 houses have the same code is about 0.00366.

47. a. R 5 tomato is partially rotten

I 5 tomato has been fed on by insects

P(R or I) 5 P(R) 1 P(I) 2 P(R and I)

5 40% 1 30% 2 12% 5 58%

The probability that a randomly selected tomato is partially rotten or has been fed on by insects is 58%.

b. B 5 tomato has bite marks

R 5 tomato is partially rotten

P(B or R) 5 P(B) 1 P(R) 2 P(B and R)

5 20% 1 40% 2 7% 5 53%

The probability that a randomly selected tomato has bite marks or is partially rotten is 53%.

c. P(I or R or B) 5 P(I) 1 P(R) 1 P(B) 2 P(I and R)

2 P(R and B) 2 P(I and B)

1 P(I and R and B)

5 30% 1 40% 1 20% 2 12%

2 7% 2 6% 1 ?

There is not enough information. You need to know what percent of the tomatoes have been fed on by insects and are partially rotten and have bite marks from a chipmunk.

48. a. P (at least 2 people share the same birthday)


5 1 2 365 p 364 p 363 p 362 p 361 p 360

}}} 3656

ø 1 2 0.96 5 0.04

The probability that at least 2 of the 6 people share the same birthday is about 0.04.

b. P (at least 2 people share the same birthday)


5 1 2

365 p 364 p 363 p 362 p 361 p 360 p 359 p 358 p 367 p 356

}}}}} 36510

ø 1 2 0.88 5 0.12

The probability that at least 2 of the 10 people share the same birthday is about 0.12.

c. P(x) 5 1 2 365Px

} 365x


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d. X Y1 21 0.44369 22 0.4757 23 0.5073 24 0.53834 25 0.5687 26 0.59824 27 0.62686 X=23

The probability that at least 2 people share the same birthday fi rst exceeds 50% in a group of 23 people.

49. P(female or yellow)

5 P(female) 1 P(yellow) 2 P(female and yellow)

5 9 } 20 1

12 } 20 2

4 } 20 5

17 } 20

The probability that you choose one of the labrador

puppies and it is a female or a yellow labrador retriever

is 17

} 20

.

50. C 5 comedy; D 5 drama; T 5 thriller

To pick at least one DVD of each type, the combinations of movies must be CCDT, CDDT, CDTT (in any order).

P(CCDT or CDDT or CDTT) 5 P(CCDT) 1 P(CDDT)

1 P(CDTT)

5 25C2 p 15C1 p 10C1

}} 50C4

1 25C1 p 15C2 p 10C1

}} 50C4

1 25C1 p 15C1 p 10C2

}} 50C4

5 45,000 1 26,250 1 16,875

}} 230,300

5 88,125

} 230,300 ø 0.38

Mixed Review

51. y 5 kx

20 5 k(25)

24 5 k

An equation is y 5 24x. When x 5 8, y 5 24(8) 5 232.

52. y 5 kx

29 5 k(54)

2 1 } 6 5 k

An equation is y 5 2 x } 6 .

When x 5 8, y 5 2 8 } 6 5 2

4 } 3 5 21

1 }

3 .

53. y 5 k } x

24 5 k } 12

248 5 k

An equation is y 5 2 48

} x . When x 5 8, y 5 2 48

} 8 5 26.

54. y 5 k } x

23 5 k }

22

6 5 k

An equation is y 5 6 } x . When x 5 8, y 5

6 } 8 5

3 } 4 .

55. f (x) 5 3x 2 7

y 5 3x 2 7

x 5 3y 2 7

x 1 7 5 3y

x 1 7

} 3 5 y

f 21(x) 5 x 1 7

} 3

56. f (x) 5 25x 1 3

y 5 25x 1 3

x 5 25y 1 3

x 2 3 5 25y

3 2 x

} 5 5 y

f 21(x) 5 3 2 x

} 5

57. f (x) 5 26x2, x ≤ 0 y 5 26x2

x 5 26y2

x }

26 5 y2

6 Ï}

x }

26 5 y

f 21(x) 5 2 Ï}

x }

26

58. f (x) 5 22.5x5

y 5 22.5x5

x 5 22.5y5

x }

22.5 5 y5

5

Î}

x }

22.5 5 y

f 21(x) 5 5

Î}

x }

22.5

59. f (x) 5 4x2 2 12, x ≥ 0 y 5 4x2 2 12

x 5 4y2 2 12

x 1 12 5 4y2

x 1 12

} 4 5 y2

6 Ï}

x 1 12

} 4 5 y

Ï}

x 1 12

} 2 5 y

f 21(x) 5 Ï}

x 1 12

} 2


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623Algebra 2


60. f (x) 5 0.2x3 1 0.5

y 5 0.2x3 1 0.5

x 5 0.2y3 1 0.5

x 2 0.5 5 0.2y3

x 2 0.5

} 0.2

5 y3

3

Î} x 2 0.5

} 0.2

5 y

f 21(x) 5 3 Î} x 2 0.5

} 0.2

61. 2 p 4 5 8 ways

62. 13 p 7 5 91 ways

63. 3 p 5 p 6 5 90 ways

64. 12 p 11 p 8 p 10 5 10,560 ways

Quiz 10.3–10.4 (p. 713)

1. P(queen of hearts) 5 1 } 52

2. P(ace) 5 4 } 52 5

1 } 13

3. P(diamond) 5 13

} 52 5 1 } 4

4. P(red card) 5 26

} 52 5 1 } 2

5. P(not 10) 5 48

} 52 5 12

} 13

6. P(6 of clubs) 5 1 } 52

7. Odds in favor of choosing blue 5 12

} 28 5 3 } 7

8. Odds in favor of choosing black or white 5 15

} 25 5 3 } 5

9. Odds aganst choosing red 5 27

} 13

10. Odds aganst choosing red or white 5 20

} 20

5 1 } 1


5 0.6 1 0.35 2 0.2 5 0.75


0.56 5 P(A) 1 0.44 2 0.12

0.56 5 P(A) 1 0.32

0.24 5 P(A)


0.83 5 0.75 1 P(B) 2 0.25

0.83 5 P(B) 1 0.5

0.33 5 P(B)


41% 5 8% 1 33% 2 P(A and B)

P(A and B) 5 8% 1 33% 2 41% 5 0%

15. P(at least 2 defective)

5 1 2 P(none defective or one defective)

The probability that none of the 8 chips are defective is (0.99)(0.99)(0.99)(0.99)(0.99)(0.99)(0.99)(0.99) 5 (0.99)8.

To fi nd the probability that one of the 8 chips is defective, consider 8 cases. In the fi rst case, the fi rst chip selected is defective, in the second case, the second chip selected is defective, and so on. Each of these 8 probabilities is equal to (0.01)(0.99)7. So, the total probability is equal to their sum, or 8(0.01)(0.99)7.

P(none defective or one defective)

5 P(none defective) 1 P(one defective)

5 (0.99)8 1 8(0.01)(0.99)7

ø 0.9973

So, the probability that at least 2 out of 8 chips are defective is about 1 2 0.9973 5 0.0027.

Problem Solving Workshop 10.4 (p. 714)

Answers will vary.

10.4 Extension (pp. 715–716)

1. A < B 5 {2,3,5,7,11,13,17} < {1,4,9,16}

5 {1,2,3,4,5,7,9,11,13,16,17}

2. A ù B 5 {2,3,5,7,11,13,17} ù {1,4,9,16}

5 [

3. A 5 {2,3,5,7,11,13,17}

5 {1,4,6,8,9,10,12,14,15,16,18,19,20}

4. B 5 {1,4,9,16}

5 {2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,20}

5. A < B < C 5 {2,3,5,7,11,13,17}<{1,4,9,16}

<{2,5,8,11,14,17,20}

5 {1,2,3,4,5,7,8,9,11,13,14,16,17,20}

6. A ù C 5 {2,3,5,7,11,13,17}ù{2,5,8,11,14,17,20}

5 {1,4,6,8,9,10,12,14,15,16,18,19,20}

ù{2,5,8,11,14,17,20}

5 {8,14,20}

7. C < B 5 {2,5,8,11,14,17,20}<{1,4,9,16}

5 {1,2,4,5,8,9,11,14,16,17,20}

5 {3,6,7,10,12,13,15,18,19}

8. B < (A ù C) 5 {1,4,9,16}<({2,3,5,7,11,13,17}ù

{2,5,8,11,14,17,20}) 5 {1,4,9,16} < {2,5,11,17}

5 {1,2,4,5,9,11,16,17}


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9. No; Not every element of A is an element of B, because π is not an element of B. So, A is not a subset of B.

10. Yes; Every element of A is an element of C. So, A is a subset of C.

11. Yes;

Note that A ù B 5 {25, π,10} ù {25,1, Ï}

5 ,10}

5 {25, 10}

Every element of (A ù B) is an element of C. So, (A ù B) is a subset of C.

12. The subsets of the set A 5 {22, 4, 9}are [, {22},{4},{9},{22, 4},{22, 9},{4, 9}, and {22, 4, 9}.

13. X < Z 5 {April, June, September, November}

< { September, October, November, December}

5 { April, June, September, October, November,December}

14. X ù Y 5 {April, June, September, November}

ù { January, March, May, July, August, October, December} 5 [

15. Z 5 {September, October, November, December}

5 { January, February, March, April, May, June, July, August}

16. X < Y 5 {April, June, September, November}

<{ January, March, May, July, August,October, December}

5 {February}

17. Yes; no; an irrational number is a real number but it is not an integer.

18. A ù B 5 {Weston, Centerville, Fairview, Jackson,

Lakeville, Midland, Newton} ù

{Centerville, Jackson, Lakeville, Midland,

Newton, Norton}

5 {Centerville, Jackson, Lakeville, Midland,

Newton}

Centerville, Jackson, Lakeville, Midland, and Newton can receive a signal from both of the towers.

Lesson 10.5


1. P(A and not B) 5 P(A) p P(not B) 5 P(A) p (1 2 P(B))

5 5 } 150 p 195

} 200

5 1 } 30 p 39

} 40

5 39 } 1200 5

13 } 400

2. P(three perfect squares) 5 F P(perfect square) G 3

5 1 3 } 10 2 3 5 27 } 1000 5 0.027

3. P(not hearing song) 5 11C4

} 12C4

P(hearing song) 5 1 2 [P(not hearing song)]5

5 1 2 1 11C4 }

12C4 2

5

ø 0.87

If there are only 12 songs on the CD, the probability of hearing your favorite song at least once during the week increases to about 0.87.

4. a. P (tropical storm) 5 Number of tropical storms

}} Total number of cyclones

5 598

} 1575 ø 0.38

The probability that a future tropical cyclone is a tropical storm is about 0.38.

b. P (tropical stormSouthern Hemisphere)

5 Number of tropical storms in Southern Hemisphere

}}}} Total number of cyclones in Southern Hemisphere

5 200

} 433 ø 0.46

The probability that a future tropical cyclone in the Southern Hemisphere is a tropical storm is about 0.46.

5. a. P(spade and club) 5 P(spade) p P(club)

5 13

} 52 p 13 }

52 5

169 } 2704 5

1 } 16

b. P(spade and club) 5 P(spade) p P(clubspade)

5 13

} 52 p 13 }

51 5

169 } 2652 5

13 } 204

6. a. P(jack and jack) 5 P(jack) p P(jack)

5 4 } 52 p 4 }

52 5

16 } 2704 5

1 } 169

b. P(jack and jack) 5 P(jack) p P(jackjack)

5 4 } 52 p 3 }

51 5

12 } 2652 5

1 } 221

7. P(A and B and C) 5 P(A) p P(BA) p P(CA and B)

5 20

} 20 p 19 }

20 p 18

} 20

5 6840

} 8000 5 0.855

If the store sellls 20 different costumes, the probability that you and your friends choose different costumes is 0.855.

8. Event A

Team leadsat halftime

Event BTeam does notlead at halftime

0.80

0.200.60

0.40 0.10

0.90

Event CTeam winsEvent DTeam loses

Event CTeam wins

Event DTeam loses

P(C) 5 P(A and C) 1 P(B and C)

5 P(A) p P(CA) 1 P(B) p P(CB) 5 (0.60)(0.80) 1 (0.40)(0.10) 5 0.52 5 52%

The probability that the team wins a particular game during the season is 52%.


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625Algebra 2


10.5 Exercises (pp. 721–723)

Skill Practice

1. The probability that B will occur given that A has occurred is called the conditional probability of B given A.

2. Independent events are two or more events where the occurrence of one has no effect on the occurrence of the other(s).

Example: Flipping a coin and spinning a wheel.

Dependent events are two or more events where the occurrence of one affects the occurrence of the other(s).

Example: Choosing a marble from a bag of different colored marbles and then choosing another marble without replacing the fi rst marble.

3. P(A and B) 5 P(A) p P(B) 5 (0.4)(0.6) 5 0.24

4. P(A and B) 5 P(A) p P(B) 5 (0.3)(0.4) 5 0.12

5. P(A and B) 5 P(A) p P(B)

0.2 5 0.25 p P(B)

0.8 5 P(B)


0.1 5 0.5 p P(B)

0.2 5 P(B)


0.6 5 P(A) p 0.8

0.75 5 P(A)


0.45 5 P(A) p 0.9

0.5 5 P(A)

9. P(green, blue) 5 P(green) p P(blue)

5 1 4 } 16 2 1 3 } 16

2 ø 0.047

10. P(red, yellow) 5 P(red) p P (yellow)

5 1 5 } 16 2 1 4 } 16

2 ø 0.078

11. P (blue, red) 5 P (blue) p P (red)

5 1 3 } 16 2 1 5 } 16

2 ø 0.059

12. P (yellow, green) 5 P (yellow) p P (green)

5 1 4 } 16 2 1 4 } 16

2 ø 0.063

13. P (blue, green, red) 5 P(blue) p P(green) p P(red)

5 1 3 } 16 2 1 4 } 16

2 1 5 } 16

2 ø 0.015

14. P(green, red, yellow) 5 P(green) p P(red) p P(yellow)

5 1 4 } 16 2 1 5 } 16

2 1 4 } 16

2 ø 0.020

15. A; P (A and B) 5 P(A) p P(B) 5 (0.3)(0.2) 5 0.06

16. P (A and B) 5 P(A) p P(BA) 5 (0.3)(0.6) 5 0.18

17. P (A and B) 5 P(A) p P (BA) 5 (0.7)(0.5) 5 0.35

18. P (A and B) 5 P(A) p P (BA) 0.32 5 0.8 p P(BA) 0.4 5 P(BA) 19. P(A and B) 5 P(A) p P(BA) P(A and B) 5 P(A) p P(BA) 0.45 5 0.6 p P(BA) 0.75 5 P(BA)

20. P(A and B) 5 P(A) p P(BA) 0.2 5 P(A) p 0.4

0.5 5 P(A)

21. P(A and B) 5 P(A) p P(BA) 0.63 5 0.7 p P(BA) 0.9 5 P(BA)

22. P(2even) 5 Number of 2’s

}} Number of even numbers

5 1 } 10

23. P(5< 8) 5 Number of 5’s

}} Numbers less than 8

5 1 } 7

24. P(prime2 digits) 5 Number of primes

}}} Number of 2-digit numbers

5 4 } 11

25. P(oddprime) 5 Number of odd numbers

}} Number of primes

5 7 } 8

26. a. P(club and spade) 5 P(club) p P(spade)

5 13

} 52 p 13 }

52 5

169 } 2704 5

1 } 16

b. P(club and spade) 5 P(club) p (spadeclub)

5 13

} 52 p 13 }

51 5

169 } 2652 5

13 } 204

27. a. P(queen and ace) 5 P(queen) p P(ace)

5 4 } 52 p 4 }

52 5

16 } 2704 5

1 } 169

b. P(queen and ace) 5 P(queen) p P(acequeen)

5 4 } 52 p 4 }

51 5

16 } 2652 5

4 } 663

28. a. P (face card and 6) 5 P (face card) p P(6)

5 12

} 52 p 4 } 52

5 48 } 2704 5

3 } 169

b. P(face card and 6) 5 P(face card) p P(6face card)

5 12

} 52 p 4 } 51

5 48 } 2652 5

4 } 221

29. a. P(10 and 2) 5 P(10) p P(2)

5 4 } 52 p 4 }

52 5

16 } 2704 5

1 } 169


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b. P(10 and 2) 5 P (10) p P(210)

5 4 } 52 p 4 }

51 5

16 } 2652 5

4 } 663

30. a. P(king and queen and jack) 5 P(king) p P(queen)

p P(jack)

5 4 } 52 p 4 }

52 p 4 }

52 5

64 } 140,608 5

1 } 2197

b. P(king and queen and jack)

5 P(king) p P(queenking) p P(jackking and queen)

5 4 } 52 p 4 }

51 p 4 }

50 5

64 } 132,600 5

8 } 16,575

31. a. P(spade and club and spade)

5 P(spade) p P(club) p P(spade)

5 13

} 52 p 13 }

52 p 13

} 52

5 2197

} 140,608 5 1 } 64

b. P(spade and club and spade)

P(spade) p P(clubspade) p P(spadespade and club)

5 13

} 52 p 13 }

51 p 12

} 50

5 2028

} 132,600 5 13

} 850

32. B; P(heart and heart and heart)

5 P(heart) p P(heartheart) p P(heartheart and heart)

5 13

} 52 p 12 }

51 p 11

} 50

5 1716

} 132,600 ø 0.0129

33. The probabilites P(A) and P(B) must be multiplied to fi nd P(A and B), not added; P(A and B) 5 (0.4)(0.5) 5 0.2


35. If A and B are independent events, then P(A) has no effect on P(BA). So, P(BA) 5 P(B).

36. Rolling two six-sided dice are independent events, so the theoretical probability of rolling two 6’s on each roll is:

P(6 and 6) 5 P(6) p P(6) 5 1 } 6 p 1 }

6 5

1 } 36

P(not 6 and 6) 5 1 2 1 }

36 5

35 }

36

After x rolls, the probability of not rolling two 6’s is

P(not 6 and 6 after x rolls) 5 1 35 }

36 2

x 5 1 2 50% 5

1 }

2

x ø 25

You must roll two six-sided dice 25 times for there to be at least a 50% chance of rolling two 6’s at least once.

Problem Solving

37. P(not early) 5 72%

P(early at least once) 5 1 2 [P(not early)]5

5 1 2 (0.72)5 ø 0.81

The probability that the bus will come early at least once during a 5 day school week is about 81%.

38. a. P(bird) 5 Number of birds

}} Number of species listed

5 91

} 519 ø 0.18

The probability that a listed animal is a bird is about 0.18.

b. P(birdendangered)

5 Number of endangered birds

}}} total number of endangered species

5 77

} 390 ø 0.20

The probability that a listed animal is a bird is about 0.20.

39. Event A

Player servesfirst

Event BOpponent serves

first

0.55

0.450.50

0.50 0.47

0.53

Event CPlayer wins

Event DPlayer losesEvent CPlayer winsEvent DPlayer loses


5 P(A) p P(CA) 1 P(B) p P(CB) 5 (0.50)(0.55) 1 (0.50)(0.47)

5 0.51

The probability that the player wins a given match is 51%.

40. P(blue and reported blue)

5 P(blue) p P(reported blueblue) 5 (0.15)(0.80)

5 0.12

The probability that the car involved in the accident was blue is 0.12.

41. Let Event A 5 scoring 1 point and Event B 5 scoring 2 points.

a. P(winning) 5 0%; If the team makes both the kicks, the score will be tied. If they miss at least one kick, then they will lose. Either way, the probability that the coach’s team will win by going for one point after each touchdown is 0%.

P(losing) 5 P(A and A) 1 P(A and A) 1 P(A and A)

5 P(A) p P(A) 1 P(A) p P(A) 1 P(A) p P(A)

5 (0.99)(0.01) 1 (0.01)(0.99) 1 (0.01)(0.01)

5 0.0099 1 0.0099 1 0.0001

5 0.0199 ø 2%

The probability that the coach’s team loses by going for one point after each touchdown is about 2%.

P(tie) 5 P(A and A)

5 P(A) p P(A)

5 (0.99)(0.99)

5 0.9801 ø 98%

The probability that the coach’s team ties by going for one point and each touchown is about 98%.


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627Algebra 2


b. P(winning) 5 P(B and B)

5 (0.45)(0.45)

5 0.2025 ø 20%

The probability that the coach’s team wins by going for two points after each touchdown is about 20%.

P(losing) 5 P(B and B)

5 P(B) p P(B)

5 (0.55)(0.55)

5 0.3025 ø 30%

The probability that the coach’s team loses by going for two points after each touchdown is about 30%.

P(tie) 5 P(B and B) 1 P(B and B)

5 P(B) p P(B) 1 P(B) p P(B)

5 (0.45)(0.55) 1 (0.55)(0.45)

5 0.2475 1 0.2475

5 0.495 ø 50%

The probability that the coach’s team ties by going for two points after each touchdown is about 50%.

c. Yes; go for 2 points after the fi rst touchdown. If the two points are scored, go for 1 point after the second touchdown. If the two points are not scored, go for 2 points after the second touchdown for the tie. If the two points are not scored after the fi rst or second touchdown, then the team will lose. The probability that the coach’s team wins the game using this strategy is (0.45)(0.99) or about 45%. The probability that they will lose the game is only (0.55)(0.55) or about 30%.

42.Event A

Has diabetes

Event BDoes not have

diabetes

0.98

0.020.059

0.941 0.05

0.95

Event CHas diabetes from test

Event DDoes not have diabetesfrom test

Event CHas diabetes from test

Event DDoes not have diabetesfrom test

P(AC) 5 P(A and C)

} P(C)

5 P(A and C)

}} P(A and C) 1 P(B and C)

5 P(A) p P(CA)

}}} P(A) p P(CA) 1 P(B) p P(CB)

5 0.059 3 0.98

}}} 0.059 3 0.98 1 0.941 3 0.05 5 0.055

Mixed Review

43. (x 1 1)6 5 6C0x6(1)0 1 6C1x5(1) 1 6C2x4(1)2

1 6C3x(1)3 1 6C4x2(1)4

1 6C5x1(1)5 1 6C6x0(1)6

5 (1)(x6)(1) 1 (6)(x5)(1) 1 (15)(x4)(1)

1 (20)(x3)(1) 1 (15)(x2)(1)

1 (6)(x)(1) 1 (1)(1)(1)

5 x6 1 6x5 1 15x4 1 20x3 1 15x2 1 6x 1 1

44. (x 2 3)5 5 [x 1 (23)]5

5 5C0x5(23)0 1 5C1x4(23) 1 5C2x3(23)2

1 5C3x2(23)3 1 5C4x1(23)4 1 5C5x0(23)5

5 (1)(x5)(1) 1 (5)(x4)(23) 1 (10)(x3)(9)

1 (10)(x2)(227)

1 (5)(x)(81) 1 (1)(1)(2243)

5 x5 2 15x4 1 90x3 2 270x2 1 405x 2 243

45. (3x 1 2)7 5 7C0(3x)7(2)0 1 7C1(3x)6(2)1 1 7C2(3x)5(2)2

1 7C3(3x)4(2)3 1 7C4(3x)3(2)4

1 7C5(3x)2(2)5 1 7C6(3x)1(2)6

1 7C7(3x)0(2)7

5 (1)(2187x7)(1) 1 (7)(729x6)(2)

1 (21)(243x5)(4) 1 (35)(81x4)(8)

1 (35)(27x3)(16) 1 (21)(9x2)(32)

1 (7)(3x)(64) 1 (1)(1)(128)

5 2187x7 1 10,206x6 1 20,412x5

1 22,680x4 1 15,120x3

1 6048x2 1 1344x 1 128

46. (5x 2 1)5 5 [5x 1 (21)]5

5 5C0(5x)5(21)0 1 5C1(5x)4(21)1

1 5C2(5x)3(21)2 1 5C3(5x)2(21)3

1 5C4(5x)1(21)4 1 5C5(5x)0(21)5

5 (1)(3125x5)(1) 1 (5)(625x4)(21)

1 (10)(125x3)(1) 1 (10)(25x2)(21)

1 (5)(5x)(1) 1 (1)(1)(21)

5 3125x5 2 3125x4 1 1250x3 2 250x2

1 25x 2 1

47. (4x 1 y)6 5 6C0(4x)6( y)0 1 6C1(4x)5( y)1 1 6C2(4x)4( y)2

1 6C3(4x)3( y)3 1 6C4(4x)2( y)4

1 6C5(4x)1( y)5 1 6C6(4x)0( y)6

5 (1)(4096x6)(1) 1 (6)(1024x5)( y)

1 (15)(256x4)( y2) 1 (20)(64x3)( y3) 1 (15)(16x2)( y4) 1 (6)(4x)( y5) 1 (1)(1)( y6) 5 4096x6 1 6144x5y 1 3840x4y2

1 1280x3y3 1 240x2y4 1 24xy5 1 y6

48. (2x 2 3y)4 5 [2x 1 (23y)]4

5 4C0(2x)4(23y)0 1 4C1(2x)3(23y)1

1 4C2(2x)2(23y)2 1 4C3(2x)1(23y)3

1 4C4(2x)0(23y)4

5 (1)(16x4)(1) 1 (4)(8x3)(23y)

1 (6)(4x2)(9y2) 1 (4)(2x)(227y3) 1 (1)(1)(81y4) 5 16x4 2 96x3y 1 216x2y2

2 216xy3 1 81y4

49. f (x) 1 g(x) 5 x2 1 2 1 x 2 4 5 x2 1 x 2 2

Domain: all real numbers


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50. f (x) 2 g(x) 5 x2 1 2 2 (x 2 4) 5 x2 1 2 2 x 1 4

5 x2 2 x 1 6


51. f (x) p g(x) 5 (x2 1 2)(x 2 4) 5 x3 2 4x2 1 2x 2 8


52. f (x)

} g(x)

5 x2 1 2

} x 2 4

Domain: all real numbers except x 5 4

53. f(g(x)) 5 f (x 2 4) 5 (x 2 4)2 1 2 5 x2 2 8x 1 16 1 2

5 x2 2 8x 1 18


54. g(f(x)) 5 g(x2 1 2) 5 (x2 1 2) 2 4 5 x2 2 2


55. f ( f (x)) 5 f (x2 1 2) 5 (x2 1 2)2

1 2

5 x4 1 4x2 1 4 1 2

5 x4 1 4x2 1 6


56. g(g(x)) 5 g(x 2 4) 5 (x 2 4) 2 4 5 x 2 8


57. 4x 1 1 5 83x

(22)x 1 1 5 (23)3x

22x 1 2 5 29x

2x 1 2 5 9x

2 5 7x

2 } 7 5 x

58. 4 ln x 5 10

ln x 5 10

} 4

eln x 5 e10/4

x ø 12.18

59. 2 }

x 2 3 2

1 } x 1 2 5

x 2 5 } x 1 2

(x 1 2)(x 2 3) 1 2 }

x 2 3 2 2 (x 1 2)(x 2 3) 1 1

} x 1 2 2 5 1 x 2 5

} x 1 2 2 (x 1 2)(x 2 3)

2(x 1 2) 2 (x 2 3) 5 (x 2 5)(x 2 3)

2x 1 4 2 x 1 3 5 x2 2 8x 1 15

0 5 x2 2 9x 1 8

0 5 (x 2 8)(x 2 1)

x 5 8 or x 5 1

60. x }

x 2 2 1

1 } x 1 1 5

2x 1 1 } x 1 1

(x 2 2)(x 1 1) 1 x }

x 2 2 2 1 (x 2 2)(x 1 1) 1 1

} x 1 1

2

5 1 2x 1 1 } x 1 1 2 (x 2 2)(x 1 1)

x(x 1 1) 1 (x 2 2) 5 (2x 1 1)(x 2 2)

x2 1 x 1 x 2 2 5 2x2 2 3x 2 2

0 5 x2 2 5x

0 5 x(x 2 5)

x 5 0 or x 5 5

Lesson 10.6


1. X(sum) 2 3 4 5 6 7 8

outcomes 1 2 3 4 3 2 1

P(X) 1 }

16

1 }

8

3 }

16

1 }

4

3 }

16

1 }

8

1 }

16

Sum of dice

Pro

bab

ility

2 3 4 6 7 85

116

18

14

316

2. The probability is greatest for X 5 5. So, the most likely sum when rolling the two tetrahedral dice is 5; The probability that the sum of the two tetrahedral dice is at most 3 is :

P(X ≤ 3) 5 P(X 5 3) 1 P(X 5 2) 5 1 } 8 1

1 } 16 5

3 } 16

3. P(k 5 0) 5 6C0(0.61)0(0.39)6 ø 0.004

P(k 5 1) 5 6C1(0.61)1(0.39)5 ø 0.033

P(k 5 2) 5 6C2(0.61)2(0.39)4 ø 0.129

P(k 5 3) 5 6C3(0.61)3(0.39)3 ø 0.269

P(k 5 4) 5 6C4(0.61)4(0.39)2 ø 0.316

P(k 5 5) 5 6C5(0.61)5(0.39)1 ø 0.198

P(k 5 6) 5 6C6(0.61)6(0.39)0 ø 0.052

Number of households

Pro

bab

ility

1 2 3 4 5 60

0.3

0.2

0.1

0


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629Algebra 2


4. The probability is greatest when k 5 4. So, the most likely outcome is that 4 of the 6 Swedish households have a soccer ball; the probability that at most 2 households have a soccer ball is:

P(k ≤ 2) 5 P(k 5 2) 1 P(k 5 1) 1 (k 5 0)

ø 0.129 1 0.033 1 0.004

5 0.166

5. a. P(k 5 0) 5 5C0(0.4)0(0.6)5 5 0.07776

P(k 5 1) 5 5C1(0.4)1(0.6)4 5 0.2592

P(k 5 2) 5 5C2(0.4)2(0.6)3 5 0.3456

P(k 5 3) 5 5C3(0.4)3(0.6)2 5 0.2304

P(k 5 4) 5 5C4(0.4)4(0.6)1 5 0.0768

P(k 5 5) 5 5C5(0.4)5(0.6)0 5 0.01024

The distribution is skewed because the values for 0, 1, and 2 are not the same as the values for 5, 4, and 3, so the left and right halves of the histogram will not be symmetric about a vertical line.

b. P(k 5 0) 5 5C0(0.5)0(0.5)5 5 0.03125

P(k 5 1) 5 5C1(0.5)1(0.5)4 5 0.15625

P(k 5 2) 5 5C2(0.5)2(0.5)3 5 0.3125

P(k 5 3) 5 5C3(0.5)3(0.5)2 5 0.3125

P(k 5 4) 5 5C4(0.5)4(0.5)1 5 0.15625

P(k 5 5) 5 5C5(0.5)5(0.5)0 5 0.03125

The distribution is symmetric because the values for 0, 1, and 2 are the same as the values for 5, 4, and 3, so the left and right halves of the histogram will be symmetric about a vertical line.

10.6 Exercises (pp. 727–730)

Skill Practice

1. A probability distribution represented by a histogram is symmetric if you can draw a vertical line dividing the histogram into two parts that are mirror images.

2. Sample answer: A binomial experiment is an experiment that has only two outcomes for each trial: Success and failure. There are n independent trials. The probability of success is the same for each trial. A binomial distribution shows the probabilities of the outcomes of a binomial experiment.

3.X 1 2 3

P(X) 1 }

2

3 }

10

1 } 5

4.W 1 2

P(W) 5 }

26

21 }

26

Label on ball

01 2 3

0.2

0.4

0.6

0.8

Pro

bab

ility

Value of letter

Pro

bab

ility

1 2

0.8

0.6

0.4

0.2

0

5. N 1 2 3

P(N) 0.01 0.09 0.9

Number of digits

01 2 3

0.2

0.4

0.6

1.0

0.8

Pro

bab

ility

6. The probability that X is equal to 1 is 0.1.

7. The most likely value for X is 3.

8. P(X is odd) 5 P(X 5 1) 1 P(X 5 3) 5 0.1 1 0.4 5 0.5

The probability that X is odd is 0.5.

9. C; P(X ≥ 3) 5 P(X 5 3) 1 P(X 5 4) 5 0.4 1 0.2 5 0.6

10. P(1 head) 5 20C1(0.5)1(0.5)19 ø 0.000019

11. P(2 heads) 5 20C2(0.5)2(0.5)18 ø 0.00018

12. P(4 heads) 5 20C4(0.5)4(0.5)16 ø 0.0046

13. P(6 heads) 5 20C6(0.5)6(0.5)14 ø 0.037

14. P(9 heads) 5 20C9(0.5)9(0.5)11 ø 0.1602

15. P(12 heads) 5 20C12(0.5)12(0.5)8 ø 0.12

16. P(15 heads) 5 20C15(0.5)15(0.5)5 ø 0.015

17. P(18 heads) 5 20C18(0.5)18(0.5)2 ø 0.00018

18. P(0 correct) 5 30C0(0.25)0(0.75)30 ø 0.00018

19. P(2 correct) 5 30C2(0.25)2(0.75)28 ø 0.0086

20. P(6 correct) 5 30C6(0.25)6(0.75)24 ø 0.145

21. P(11 correct) 5 30C11(0.25)11(0.75)19 ø 0.055

22. P(15 correct) 5 30C15(0.25)15(0.75)15 ø 0.0019

23. P(21 correct) 5 30C21(0.25)21(0.75)9 ø 0.00000024

24. P(26 correct) 5 30C26(0.25)26(0.75)4 ø 1.93 3 10212

25. P(30 correct) 5 30C30(0.25)30(0.75)0 ø 8.67 3 10219

26. The exponent of p and 1 2 p were reversed;

P(k 5 3) 5 5C3 1 1 } 6 2

3 1 5 } 6 2

5 2 3 ø 0.032

27. The probability was not multiplied by nCk 5 5C3;

P(k 5 3) 5 5C3 1 1 } 6 2

3 1 5 } 6 2

5 2 3 ø 0.032

28. P(k ≤ 3) 5 7C3(0.3)3(0.7)4 1 7C2(0.3)2(0.7)5

1 7C1(0.3)1(0.7)6 1 7C0(0.3)0(0.7)7

ø 0.227 1 0.318 1 0.247 1 0.082

5 0.874

29. P(k ≥ 5)

5 8C5(0.6)5(0.4)3 1 8C6(0.6)6(0.4)2

1 8C7(0.6)7(0.4)1 1 8C8(0.6)8(0.4)0

ø 0.279 1 0.209 1 0.09 1 0.017

5 0.595


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30. P(k ≤ 2)

5 5C2(0.12)2(0.88)3 1 5C1(0.12)1(0.88)4

1 5C0(0.12)0(0.88)5

ø 0.098 1 0.36 1 0.528

5 0.986

31. P(k ≥ 10)

5 15C10(0.75)10(0.25)5 1 15C11(0.75)11(0.25)4

1 15C12(0.75)12(0.25)3 1 15C13(0.75)13(0.25)2

1 15C14(0.75)14(0.25)1 1 15C15(0.75)15(0.25)0

ø 0.165 1 0.225 1 0.225 1 0.156 1 0.067 1 0.013

5 0.851

32. A; P(k 5 0) 5 10C0(0.36)0(0.64)10 ø 0.012

P(k 5 1) 5 10C1(0.36)1(0.64)9 ø 0.065

P(k 5 2) 5 10C2(0.36)2(0.64)8 ø 0.164

P(k 5 3) 5 10C3(0.36)3(0.64)7 ø 0.246

P(k 5 4) 5 10C4(0.36)4(0.64)6 ø 0.242

P(k 5 5) 5 10C5(0.36)5(0.64)5 ø 0.164

P(k 5 6) 5 10C6(0.36)6(0.64)4 ø 0.077

P(k 5 7) 5 10C7(0.36)7(0.64)3 ø 0.025

P(k 5 8) 5 10C8(0.36)8(0.64)2 ø 0.005

P(k 5 9) 5 10C9(0.36)9(0.64)1 ø 0.00065

P(k 5 10) 5 10C10(0.36)10(0.64)0 ø 0.000037

The most likely number of successes is 3.

33. P(k 5 0) 5 3C0(0.3)0(0.7)3 5 0.343

P(k 5 1) 5 3C1(0.3)1(0.7)2 5 0.441

P(k 5 2) 5 3C2(0.3)2(0.7)1 5 0.189

P(k 5 3) 5 3C3(0.3)3(0.7)0 5 0.027

Number of successes

00 1 2 3

0.1

0.2

0.3

0.5

0.4

Pro

bab

ility

The distribution is skewed because it is not symmetric about any vertical line. The most likely number of successes is 1.

34. P(k 5 0) 5 6C0(0.5)0(0.5)6 5 0.015625

P(k 5 1) 5 6C1(0.5)1(0.5)5 5 0.09375

P(k 5 2) 5 6C2(0.5)2(0.5)4 5 0.234375

P(k 5 3) 5 6C3(0.5)3(0.5)3 5 0.3125

P(k 5 4) 5 6C4(0.5)4(0.5)2 5 0.234375

P(k 5 5) 5 6C5(0.5)5(0.5)1 5 0.09375

P(k 5 6) 5 6C6(0.5)6(0.5)0 5 0.015625

Number of successes

Pro

bab

ility

0 1 2 3 4 5 6

0.4

0.3

0.2

0.1

0

The distribution is symmetric because the left half is a mirror image of the right half. The most likely number of successes is 3.

35. P(k 5 0) 5 4C0(0.16)0(0.84)4 ø 0.498

P(k 5 1) 5 4C1(0.16)1(0.84)3 ø 0.379

P(k 5 2) 5 4C2(0.16)2(0.84)2 ø 0.108

P(k 5 3) 5 4C3(0.16)3(0.84)1 ø 0.014

P(k 5 4) 5 4C4(0.16)4(0.84)0 ø 0.00066

Number of successes

00 1 2 3 4

0.1

0.2

0.3

0.5

0.4

Pro

bab

ility

The distribution is skewed because it is not symmetric about any vertical line. the most likely number of successes is 0.


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631Algebra 2


36. P(k 5 0) 5 7C0(0.85)0(0.15)7 ø 0.0000017

P(k 5 1) 5 7C1(0.85)1(0.15)6 ø 0.000068

P(k 5 2) 5 7C2(0.85)2(0.15)5 ø 0.00115

P(k 5 3) 5 7C3(0.85)3(0.15)4 ø 0.0109

P(k 5 4) 5 7C4(0.85)4(0.15)3 ø 0.0617

P(k 5 5) 5 7C5(0.85)5(0.15)2 ø 0.210

P(k 5 6) 5 7C6(0.85)6(0.15)1 ø 0.396

P(k 5 7) 5 7C7(0.85)7(0.15)0 ø 0.321

Number of successes

Pro

bab

ility

0 71 2 3 4 5 6

0.4

0.3

0.2

0.1

0


37. P(k 5 0) 5 8C0(0.025)0(0.975)8 ø 0.817

P(k 5 1) 5 8C1(0.025)1(0.975)7 ø 0.168

P(k 5 2) 5 8C2(0.025)2(0.975)6 ø 0.015

P(k 5 3) 5 8C3(0.025)3(0.975)5 ø 0.00078

P(k 5 4) 5 8C4(0.025)4(0.975)4 ø 0.000025

P(k 5 5) 5 8C5(0.025)5(0.975)3 ø 5.1 3 1027

P(k 5 6) 5 8C6(0.025)6(0.975)2 ø 6.5 3 1029

P(k 5 7) 5 8C7(0.025)7(0.975)1 ø 4.8 3 10211

P(k 5 8) 5 8C8(0.025)8(0.975)0 ø 1.5 3 10213

Number of successes

Pro

bab

ility

0 7 81 2 3 4 5 6

0.8

0.6

0.4

0.2

0


38. P(k 5 0) 5 12C0(0.5)0(0.5)12 ø 0.00024

P(k 5 1) 5 12C1(0.5)1(0.5)11 ø 0.0029

P(k 5 2) 5 12C2(0.5)2(0.5)10 ø 0.016

P(k 5 3) 5 12C3(0.5)3(0.5)9 ø 0.054

P(k 5 4) 5 12C4(0.5)4(0.5)8 ø 0.121

P(k 5 5) 5 12C5(0.5)5(0.5)7 ø 0.193

P(k 5 6) 5 12C6(0.5)6(0.5)6 ø 0.226

P(k 5 7) 5 12C7(0.5)7(0.5)5 ø 0.193

P(k 5 8) 5 12C8(0.5)8(0.5)4 ø 0.121

P(k 5 9) 5 12C9(0.5)9(0.5)3 ø 0.054

P(k 5 10) 5 12C10(0.5)10(0.5)2 ø 0.016

P(k 5 11) 5 12C11(0.5)11(0.5)1 ø 0.0029

P(k 5 12) 5 12C12(0.5)12(0.5)0 ø 0.00024

Number of successesP

rob

abili

ty0 7 81 2 3 4 5 6 11 129 10

0.25

0.20

0.15

0.10

0.05

0

The distribution is symmetric because the left half is a mirror image of the right half. The most likely number of successes is 6.


40. p(k successes) 5 pk

p(n 2 k failures) 5 (1 2 p)n 2 k

p(k successes, n 2 k failures) 5 pk(1 2 p)n 2 k

41. Because order does not matter, the combination of n things taken k at a time nCk is the number of sequences of k successes and n 2 k failures.

42. The probability of one sequence is pk(1 2 p)n 2 k and there are nCk sequences, so the probability of having a sequence of n items with k successes that have probability p is nCk pk(1 2 p)n 2 k, which is the binomial probability formula.

Problem Solving

43. P(k 5 1) 5 25C1(0.01)1(0.99)24 ø 0.196

The probablility that exactly 1 person in a class of 25 is allergic to bee stings is about 0.196.

44. P(k 5 10) 5 15C10(0.927)10(0.073)5 ø 0.0029

The probability that Predrag Stojakovic of the Sacramento Kings will make exactly 10 of his next 15 free throw attempts is about 0.0029.

45. a. P(k 5 5) 5 10C5(0.34)5(0.66)5 ø 0.143

The probability that exactly 5 of the 10 people who donate blood are type A1 is about 0.143.


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b. P(k 5 2) 5 10C2(0.15)2(0.85)8 ø 0.276

The probability that exactly 2 of the 10 people who donate blood are Rh2 (O2 , A2 , B2 , or AB2) is about 0.276.

c. P(k ≤ 2) 5 10C2(0.43)2(0.57)8 1 10C1(0.43)1(0.57)9

1 10C0(0.43)0(0.57)10

ø 0.093 1 0.027 1 0.004

5 0.124

The probability that at most 2 of the 10 people who donate blood are type O (O2 or O1) is about 0.124.

d. P(k ≥ 5) 5 10C5(0.85)5(0.15)5 1 10C6(0.85)6(0.15)4

1 10C7(0.85)7(0.15)3 1 10C8(0.85)8(0.15)2

1 10C9(0.85)9(0.15)1

1 10C10(0.85)10(0.15)0

ø 0.0085 1 0.04 1 0.13 1 0.276

1 0.347 1 0.197

ø 0.999

The probability that exactly 5 of the 10 people who donate blood are Rh1(O1, A1, B1, or AB1) is about 0.999.

46. a. P(k 5 0) 5 10C0(0.35)0(0.65)10 ø 0.013

P(k 5 1) 5 10C1(0.35)1(0.65)9 ø 0.072

P(k 5 2) 5 10C2(0.35)2(0.65)8 ø 0.176

P(k 5 3) 5 10C3(0.35)3(0.65)7 ø 0.252

P(k 5 4) 5 10C4(0.35)4(0.65)6 ø 0.238

P(k 5 5) 5 10C5(0.35)5(0.65)5 ø 0.154

P(k 5 6) 5 10C6(0.35)6(0.65)4 ø 0.069

P(k 5 7) 5 10C7(0.35)7(0.65)3 ø 0.021

P(k 5 8) 5 10C8(0.35)8(0.65)2 ø 0.004

P(k 5 9) 5 10C9(0.35)9(0.65)1 ø 0.0005

P(k 5 10) 5 10C10(0.35)10(0.65)0 ø 0.00003

Number of people

Pro

bab

ility

0 7 81 2 3 4 5 6 9 10

0.25

0.20

0.15

0.10

0.05

0

b. P(k ≤ 4) 5 P(k 5 4) 1 P(k 5 3) 1 P(k 5 2)

1 P(k 5 1) 1 P(k 5 0)

ø 0.238 1 0.252 1 0.176 1 0.072 1 0.013

5 0.751

The probability that at most 4 of the 10 people visited an art museum is about 0.751.

47. a. P(hole in carrot patch) 5 Area of carrot patch

}} Area of farm

5 1 } 2 h(b1 1 b2)

} lw

5 1 } 2 (0.3)(0.6 1 0.3)

}} (0.8)(0.6)

ø 0.28

P(k 5 0) 5 7C0(0.28)0(0.72)7 ø 0.100

P(k 5 1) 5 7C1(0.28)1(0.72)6 ø 0.273

P(k 5 2) 5 7C2(0.28)2(0.72)5 ø 0.319

P(k 5 3) 5 7C3(0.28)3(0.72)4 ø 0.206

P(k 5 4) 5 7C4(0.28)4(0.72)3 ø 0.080

P(k 5 5) 5 7C5(0.28)5(0.72)2 ø 0.019

P(k 5 6) 5 7C6(0.28)6(0.72)1 ø 0.002

P(k 5 7) 5 7C7(0.28)7(0.72)0 ø 0.0001

b. x p(x)

0 0.100

1 0.273

2 0.319

3 0.206

4 0.080

5 0.019

6 0.002

7 0.0001

c.

00 1 2 3 4 5 6 7

0.1

0.2

0.3

0.4

Pro

bab

ility

Number of gopher holesin carrot patch

48. a. The statement, “The fi rst 4 kids were all boys, so the next one will probability be a girl”, is not valid because the events are independent.

b. P(four M’s and an F)

5 P(M) p P(M) p P(M) p P(M) p P(F)

5 F P(M) G 4 p P(F)

5 (0.5)4(0.5) 5 0.03125

c. X 5 number of male childern before having a female child

n 5 total number of children 5 X 1 1

n 5 1: P(X 5 0) 5 (0.5)0(0.5)1 5 0.5

n 5 2: P(X 5 1) 5 (0.5)1(0.5)1 5 0.25

n 5 3: P(X 5 2) 5 (0.5)2(0.5)1 5 0.125

n 5 4: P(X 5 3) 5 (0.5)3(0.5)1 5 0.0625

n 5 5: P(X 5 4) 5 (0.5)4(0.5)1 5 0.0313

n 5 6: P(X 5 5) 5 (0.5)5(0.5)1 5 0.0156

n 5 7: P(X 5 6) 5 (0.5)6(0.5)1 5 0.0078

n 5 8: P(X 5 7) 5 (0.5)7(0.5)1 5 0.0039

n 5 9: P(X 5 8) 5 (0.5)8(0.5)1 5 0.0020

n 5 10: P(X 5 9) 5 (0.5)9(0.5)1 5 0.0010

n 5 11: P(X 5 10) 5 (0.5)10(0.5)1 5 0.0005


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633Algebra 2


Pro

bab

ility

0 7 81 2 3 4 5 6 9 10

0.5

0.4

0.3

0.2

0.1

0

Number of childrena couple already has

49. When n 5 5: P(k ≥ 3) 5 P(k 5 3) 1 P(k 5 4)

1 P(k 5 5)

5 5C3 p3(1 2 p)2 1 5C4 p4(1 2 p)1

1 5C5 p5(1 2 p)0

5 10p3(1 2 2p 1 p2) 1 5p4(1 2 p) 1 p5

5 10p3 2 20p4 1 10p5 1 5p4 2 5p5 1 p5

5 6p5 2 15p4 1 10p3

When n 5 3: P(k ≥ 2) 5 P(k 5 2) 1 P(k 5 3)

5 3C2 p2(1 2 p)1 1 3C3 p3(1 2 p)0

5 3p2(1 2 p) 1 p3

5 3p2 2 3p3 1 p3

5 22p3 1 3p2

6p5 2 15p4 1 10p3 > 22p3 1 3p2

6p5 2 15p4 1 12p3 2 3p2 > 0

3p2(2p3 2 5p2 1 4p 2 1) > 0

3p2 1 p 2 1 } 2 2 (2p2 2 4p 1 2) > 0

6p2 1 p 2 1 } 2 2 (p 2 1)2 > 0 → p >

1 }

2

For values of p > 0.5, a 5-speaker system is more likely to operate than a 3-speaker system.

Mixed Review

50. 8 1 24 4 4 5 8 1 6 5 14

51. 4 p 3 1 28 4 7 5 12 1 28 4 7 5 12 1 4 5 16

52. 35 2 3 p 2 4 8 5 35 2 6 4 8 5 35 2 0.75 5 34.25

53. 6 2 (15 p 2)2 4 9 5 6 2 (30)2 4 9

5 6 2 900 4 9

5 6 2 100

5 294

54. 2 1 48 4 6 p 4 2 5 5 2 1 8 p 4 2 5

5 2 1 32 2 5

5 34 2 5

5 29

55. 14 2 9 4 3 1 40 4 8 5 14 2 3 1 40 4 8

5 14 2 3 1 5

5 11 1 5

5 16

56. 5 2 2x ≤ 12

22x ≤ 7

x ≥ 2 7 } 2

02324 2122

57. 1 < 4x 2 3 < 7

4 < 4x < 10

1 < x < 5 }

2

0 2 3121

58. 22 < 3x 2 5 ≤ 4 3 < 3x ≤ 9 1 < x ≤ 3

0 64222

59. 6x2 ≥ 36

x2 ≥ 6 x ≥ Ï

}

6 or x ≤ 2 Ï}

6

0 4224 22

2 6 6

60. 3x2 1 11x 2 4 < 0

(3x 2 1)(x 1 4) < 0

24 < x < 1 }

3

0 22426 22

13

61. 3x2 1 9x < x2 1 4

2x2 1 9x 2 4 < 0

x 5 29 6 Ï

}}

92 2 4(2)(24) }}

2(2) 5

29 6 Ï}

113 } 4

29 2 Ï

}

113 }

4 < x <

29 1 Ï}

113 }

4

0 22426 22

0.4124.9

62. f (x) 5 x3 2 4x2 2 7x 1 10

1 2 4 2 7 1 10 5 0, so 1 is a zero.

1 1 24 27 10

1 23 210

1 23 210 0

x2 2 3x 2 10 5 0

(x 1 2)(x 2 5) 5 0

x 5 1, x 5 22, or x 5 5


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63. g(x) 5 3x3 2 3x2 1 75x 2 75

3 2 3 1 75 2 75 5 0 so 1 is a zero.

1 3 23 75 275

3 0 75

3 0 75 0

3x2 1 75 5 0

3(x2 1 25) 5 0

3(x 1 5i)(x 2 5i) 5 0

x 5 1, x 5 5i, or x 5 25i

64. h(x) 5 x4 2 x3 2 5x2 2 x 2 6

p }

q 5 61, 62, 63, 66

22 1 21 25 21 26

22 6 22 6

1 23 1 23 0

3 1 23 1 23 0

3 0 3 0

1 0 1 0 0

x2 1 1 5 0

(x 1 i)(x 2 i) 5 0

x 5 22, x 5 3, x 5 i, or x 5 2i

65. f (x) 5 2x4 1 5x3 1 29x2 1 80x 2 48

p } q 5 61, 62, 63, 64, 66, 68, 612, 624, 648, 6

1 } 2 , 6

3 } 2

23 2 5 29 80 248

26 3 264 148

2 21 32 216 0

2x3 2 x2 1 32x 2 16 5 0

x2(2x 2 1) 1 16(2x 2 1) 5 0

(x2 1 16)(2x 2 1) 5 0

(x 1 4i)(x 2 4i)(2x 2 1) 5 0

x 5 23, x 5 4i, x 5 24i, or x 5 1 } 2

Quiz 10.5–10.6 (p. 730)

1. P(red and green) 5 P(red) p P(greenred)

5 6 } 20 p 9 }

19 5

54 } 380 5

27 } 190

2. P(blue and red) 5 P(blue) p P(redblue)

5 5 }

20 p 6 }

19 5

30 } 380 5

3 } 38

3. P(green and green) 5 P(green) p P(greengreen)

5 9 } 20 p 8 }

19 5

72 } 380 5

18 } 95

4. P(0 successes) 5 10C0 1 1 } 6 2

0 1 5 } 6 2 10

ø 0.162

5. P(1 success) 5 10C1 1 1 } 6 2 1 1 5 }

6 2 9 ø 0.323

6. P(4 successes) 5 10C4 1 1 } 6 2 4 1 5 }

6 2 6 ø 0.054

7. P(8 successes) 5 10C8 1 1 } 6 2 8 1 5 }

6 2 2 ø 0.000019

8. P(k 5 0) 5 5C0(0.2)0(0.8)5 ø 0.3277

P(k 5 1) 5 5C1(0.2)1(0.8)4 ø 0.4096

P(k 5 2) 5 5C2(0.2)2(0.8)3 ø 0.2048

P(k 5 3) 5 5C3(0.2)3(0.8)2 ø 0.0512

P(k 5 4) 5 5C4(0.2)4(0.8)1 ø 0.0064

P(k 5 5) 5 5C5(0.2)5(0.8)0 ø 0.0003

Number of successes

Pro

bab

ility

0 1 2 3 4 5

0.4

0.3

0.2

0.1

0

9. P(k 5 0) 5 8C0(0.5)0(0.5)8 ø 0.004

P(k 5 1) 5 8C1(0.5)1(0.5)7 ø 0.031

P(k 5 2) 5 8C2(0.5)2(0.5)6 ø 0.109

P(k 5 3) 5 8C3(0.5)3(0.5)5 ø 0.219

P(k 5 4) 5 8C4(0.5)4(0.5)4 ø 0.273

P(k 5 5) 5 8C5(0.5)5(0.5)3 ø 0.219

P(k 5 6) 5 8C6(0.5)6(0.5)2 ø 0.109

P(k 5 7) 5 8C7(0.5)7(0.5)1 ø 0.031

P(k 5 8) 5 8C8(0.5)8(0.5)0 ø 0.004

Pro

bab

ility

0.3

0.2

0.1

0

Number of successes0 7 81 2 3 4 5 6


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635Algebra 2


10. P(k 5 0) 5 6C0(0.72)0(0.28)6 ø 0.0005

P(k 5 1) 5 6C1(0.72)1(0.28)5 ø 0.007

P(k 5 2) 5 6C2(0.72)2(0.28)4 ø 0.048

P(k 5 3) 5 6C3(0.72)3(0.28)3 ø 0.164

P(k 5 4) 5 6C4(0.72)4(0.28)2 ø 0.316

P(k 5 5) 5 6C5(0.72)5(0.28)1 ø 0.325

P(k 5 6) 5 6C6(0.72)6(0.28)0 ø 0.139

Pro

bab

ility

0.3

0.2

0.1

0

Number of successes0 1 2 3 4 5 6

11. Let events A, B, C, D, and E represent selecting the fi rst, second, third, fourth, and fi fth beverage, respectively. The events are independent. So the probability is:

P(A and B and C and D and E)

5 P(A) p P(B) p P(C) p P(D) p P(E)

5 1 } 5 p 1 } 5 p 1 } 5 p 1 } 5 p 1 } 5 5

1 } 3125 5 0.00032

Graphing Calculator Activity 10.6 (p. 731)

1. n 5 12, p 5 0.29


2. n 5 14, p 5 0.58


3. n 5 15, p 5 0.805


4. The most likely number of adults is 5 and the histogram is spread wider with the probabilities not being as great as the original.

Mixed Review of Problem Solving (p. 732)

1. a. P(vowel) 5 42

} 100 5 0.42

The probability of drawing a vowel is 0.42.

b. P(vowel and vowel) 5 P(vowel) p P(vowelvowel)

5 42

} 100 p 41 }

99 5

1722 } 9900 ø 0.174

The probability of drawing 2 vowels without replacing the fi rst tile is about 0.174.

c. P(7 vowels) 5 42C7

} 100C7

5 26,978,328

}} 16,007,560,800 ø 0.0017

The probability of drawing 7 vowels without replacing the fi rst tile is about 0.0017.

2. P(k 5 0) 5 14C0(0.62)0(0.38)14 ø 1.3 3 1026

P(k 5 1) 5 14C1(0.62)1(0.38)13 ø 3 3 1025

P(k 5 2) 5 14C2(0.62)2(0.38)12 ø 3.2 3 1024

P(k 5 3) 5 14C3(0.62)3(0.38)11 ø 0.002

P(k 5 4) 5 14C4(0.62)4(0.38)10 ø 0.009

P(k 5 5) 5 14C5(0.62)5(0.38)9 ø 0.030

P(k 5 6) 5 14C6(0.62)6(0.38)8 ø 0.0074

P(k 5 7) 5 14C7(0.62)7(0.38)7 ø 0.138

P(k 5 8) 5 14C8(0.62)8(0.38)6 ø 0.197

P(k 5 9) 5 14C9(0.62)9(0.38)5 ø 0.215

P(k 5 10) 5 14C10(0.62)10(0.38)4 ø 0.175

P(k 5 11) 5 14C11(0.62)11(0.38)3 ø 0.104

P(k 5 12) 5 14C12(0.62)12(0.38)2 ø 0.042

P(k 5 13) 5 14C13(0.62)13(0.38)1 ø 0.012

P(k 5 14) 5 14C14(0.62)14(0.38)0 ø 0.001

Number of sports fans

Pro

bab

ility

0 7 81 2 3 4 5 6 9 10 11 121314

0.20

0.15

0.10

0.05

0

b. The most likely number of adults who consider themselves sports fans is 9.

c. P(x ≥ 7) ø 0.138 1 0.197 1 0.215 1 0.175

1 0.104 1 0.042 1 0.012 1 0.001

5 0.884

The probability that at least 7 of the 14 adults consider themselves sports fans is about 0.884.


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3. P(at least 2 have the same code)


5 1 2 1000 p 999 p 998 p 997

}} 10004

ø 0.00599

The probability that at least two of the four briefcases have the same code is about 0.00599. If two more friends buy the same briefcase, the probability will increase.

4. Sample answer: At a certain high school, a teacher is only able to cover 85% of the skills tested on the state exam. Student who take the exam choose the correct answer 69% of the time on the concepts that the teacher did cover and 12% of the time on the concepts not covered. What is the probability that a student who takes the test chooses a correct answer?

Event ACovered by

teacher

Event BNot coveredby teacher

0.69

0.310.85

0.15 0.12

0.88

Event CCorrect answerEvent DIncorrect answer

Event CCorrect answerEvent DIncorrect answer

Conceptson test


5 P(A) p P(CA) 1 P(B) 2 P(CB) 5 (0.85)(0.69) 1 (0.15)(0.12)

5 0.5865 1 0.018

5 0.6045

The probability that a student will choose a correct answer is 0.6045.

5. P(k 5 0) 5 15C0(0.31)0(0.69)15 ø 0.004

P(k 5 1) 5 15C1(0.31)1(0.69)14 ø 0.026

P(k 5 2) 5 15C2(0.31)2(0.69)13 ø 0.081

P(k 5 3) 5 15C3(0.31)3(0.69)12 ø 0.158

P(k 5 4) 5 15C4(0.31)4(0.69)11 ø 0.213

P(k 5 5) 5 15C5(0.31)5(0.69)10 ø 0.210

P(k 5 6) 5 15C6(0.31)6(0.69)9 ø 0.157

P(k 5 7) 5 15C7(0.31)7(0.69)8 ø 0.091

P(k 5 8) 5 15C8(0.31)8(0.69)7 ø 0.041

P(k 5 9) 5 15C9(0.31)9(0.69)6 ø 0.014

P(k 5 10) 5 15C10(0.31)10(0.69)5 ø 0.004

P(k 5 11) 5 15C11(0.31)11(0.69)4 ø 7.9 3 1024

P(k 5 12) 5 15C12(0.31)12(0.69)3 ø 1.2 3 1024

P(k 5 13) 5 15C13(0.31)13(0.69)2 ø 1.2 3 1025

P(k 5 14) 5 15C14(0.31)14(0.69)1 ø 7.8 3 1027

P(k 5 15) 5 15C15(0.31)15(0.69)0 ø 2.3 3 1028

The most likely number of hits the player will have in 15 at bats is 4 hits.

6. a. P(A and B and C) 5 P(A) p P(B) p P(C)

5 (0.1)(0.08)(0.18)

5 0.00144

The probability that all three lawn mowers are unusable on a given day is 0.00144.

b. P(at lease one is usable) 5 1 2 (none are usable)

5 1 2 0.00144

5 0.99856

The probability that at least one of the mowers is usable on a given day is 0.99856.

c. If one of the mowers stops working completely, it will slighty decrease the probability that the lawn mowing business can be productive on a given day.

7. a.

40Design

A

45 90Design

B75

b. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)

5 85

} 250 1 135

} 250 2 45

} 250

5 175

} 250 5 0.7

The probability that a person likes design A or B is 0.7.

c. 1 2 P(A or B) 5 1 2 0.7 5 0.3

The probability that a person does not like either design is 0.3.

d. You can calculate the probability from part (c) by subtracting the probablility in part (b) from 1.

Chapter 10 Review (pp. 734–736)

1. A combination is a selection of r objects from a group of n objects where the order of the objects selected is not important.

2. The probability of an event is a number from 0 to 1 that measures the likelihood the event will occur, and is written as a fraction, decimal, or percent. The odds in favor of an event is a ratio that measures the chances in favor of the event occurring as opposed to it not occurring.

3. No; This is not a binomial experiment because there are more than just two outcomes for each card selection.

4. No; Events A and B are not disjoint events because they have one event in common. A marble that is not red could be green or blue, so selecting a green marble is included in both events.

5. 12! 5 12 p 11 p 10 p 9 p 8 p 7 p 6 p 5 p 4 p 3 p 2 p 1

5 479,001,600

There are 479,001,600 ways to place 12 pictures in the

album; 12P4 5 12! }

(12 2 4)! 5

12! } 8! 5 11,880

There are 11,880 ways to place 4 of the 12 pictures on the fi rst 4 pages.

6. 9P1 5 9! }

(9 2 1)! 5

9! } 8! 5 9

7. 5P5 5 5! }

(5 2 5)! 5

5! } 0! 5 120


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637Algebra 2


8. 6P3 5 6! }

(6 2 3)! 5

6! } 3! 5 120

9. 10P2 5 10! }

(10 2 2)! 5

10! } 8! 5 90

10. (t 1 3)6 5 6C0t6(3)0 1 6C1t5(3)1 1 6C2t4(3)2

1 6C3t3(3)3 1 6C4t2(3)4

1 6C5t1(3)5 1 6C6t0(3)6

5 (1)(t6)(1) 1 (6)(t5)(3) 1 (15)(t4)(9)

1 (20)(t3)(27) 1 (15)(t2)(81)

1 (6)(t)(243) 1 (1)(1)(729)

5 t6 1 18t5 1 135t4 1 540t3 1 1215t2

1 1458t 1 729

11. (2a 1 b2)4 5 4C0(2a)4(b2)0 1 4C1(2a)3(b2)1

1 4C2(2a)2(b2)2 1 4C3(2a)1(b2)3

1 4C4(2a)0(b2)4

5 (1)(16a4)(1) 1 (4)(8a3)(b2) 1 (6)(4a2)(b4) 1 (4)(2a)(b6) 1 (1)(1)(b8) 5 16a4 1 32a3b2 1 24a2b4 1 8ab6 1 b8

12. (w 2 8v)4 5 [w 1 ( 2 8v)]4

5 4C0w4(28v)0 1 4C1w3(28v)1

1 4C2w2(28v)2 1 4C3w1(28v)3

1 4C4w0(28v)4

5 (1)(w4)(1) 1 (4)(w3)(28v)

1 (6)(w2)(64v2) 1 (4)(w)(2512v3) 1 (1)(1)(4096v4) 5 w4 2 32w3v 1 384w2v2

2 2048wv3 1 4096v4

13. (r3 2 4s)5 5 [r3 1 (24s)]5

5 5C0(r3)5(24s)0 1 5C1

(r3)4(24s)1

1 5C2(r3)3(24s)2 1 5C3

(r3)2(24s)3

1 5C4(r3)1(24s)4 1 5C5

(r3)0(24s)5

5 (1)(r15)(1) 1 (5)(r12)(24s)

1 (10)(r9)(16s2) 1 (10)(r6)(264s3) 1 (5)(r3)(256s4) 1 (1)(1)(21024s5) 5 r15 2 20r12s 1 160r9s2 2 640r 6s3

1 1280r 3s4 2 1024s5

14. For each of the 15 fl avors, you can choose to sample or not sample the fl avor, so there are 215 total combinations. If you sample at least 4 of the fl avors, you do not sample only a total of 0, 1, 2, or 3 fl avors. So, the number of combinations of ice cream fl avors you can sample is:

215 2 (15C0 1 15C1 1 15C2 1 15C3)

5 32,768 2 (1 1 15 1 105 1 455)

5 32,192

15. P(even) 5 Even numbers from 1 to 30


5 15

} 30 5 1 } 2

16. P(multiple of 5) 5 Multiples of 5 from 1 to 30


5 6 } 30 5

1 } 5



5 11

} 30

18. P(prime) 5 Prime numbers from 1 to 30


5 10

} 30 5 1 } 3

19. P(arrival at work on time) 5 Days arrived on time

}} Total number of days

5 217

} 250 5 0.868

The probability that the commuter arrived at work on time is 0.868.


5 0.32 1 0.48 2 0.12

5 0.68

21. P(A) 5 1 2 P(A) 5 1 2 0.32 5 0.68

22. P(B) 5 1 2 P(B) 5 1 2 0.48 5 0.52

23. a. P(red and green) 5 P(red) p P(green)

5 5 } 16 p 8 }

16 5

5 } 32

b. P(red and green) 5 P(red) p P(greenred)

5 5 } 16 p 8 }

15 5

1 } 6

24. a. P(blue and red) 5 P(blue) p P(red)

5 3 }

16 p 5 }

16 5

15 } 256

b. P(blue and red) 5 P(blue) p P(redblue)

5 3 } 16 p 5 }

15 5

1 } 16

25. a. P(green and green) 5 P(green) p P(green)

5 8 } 16 p 8 }

16 5

1 } 4

b. P(green and green) 5 P(green) p P(greengreen)

5 8 } 16 p 7 }

15 5

7 } 30

26. P(k 5 6) 5 8C6(0.5)6(0.5)2 ø 0.109

27. P(k 5 4) 5 8C4(0.5)4(0.5)4 ø 0.273

28. P(k 5 7) 5 8C7(0.5)7(0.5)1 ø 0.031

29. P(k 5 0) 5 8C0(0.5)0(0.5)8 ø 0.0039

Chapter 10 Test (p. 737)

1. 5 P2 5 5! }

(5 2 2)! 5

5! } 3! 5 20

2. 8 P3 5 8! }

(8 2 3)! 5

8! } 5! 5 336

3. 12 P7 5 12! }

(12 2 7)! 5

12! } 5! 5 3,991,680

4. 17P10 5 17! }

(17 2 10)! 5

17! } 7! 5 70,572,902,400

5. 4C3 5 4! } 1 p 3! 5

4 p 3! } 1! p 3! 5 4


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6. 7C7 5 7! } 0! p 7! 5 1

7. 18C4 5 18! } 14! p 4! 5

18 p 17 p 16 p 15 p 14! }} 14! p 4! 5 3060

8. 9C5 5 9! } 4! p 5! 5

9 p 8 p 7 p 6 p 5! }} 4! p 5! 5 126

9. (x 1 5)3 5 3C0x3(5)0 1 3C1x2(5)1 1 3C2x1(5)2

1 3C3x0(5)3

5 (1)(x3)(1) 1 (3)(x2)(5) 1 (3)(x)(25)

1 (1)(1)(125)

5 x3 1 15x2 1 75x 1 125

10. (3a 2 3)5 5 F 3a 1 (23) G 5 5 5C0(3a)5(23)0 1 5C1(3a)4(23)1

1 5C2(3a)3(23)2 1 5C3(3a)2(23)3

1 5C4(3a)1(23)4 1 5C5(3a)0(23)5

5 (1)(243a5)(1) 1 (5)(81a4)(23)

1 (10)(27a3)(9)

1 (10)(9a2)(227)(5)(3a)(81)

1 (1)(1)(2243)

5 243a5 2 1215a4 1 2430a3 2 2430a2

1 1215a 2 243

11. (s 1 t2)4 5 4C0s4(t2)0 1 4C1s3(t 2)1 1 4C2s2(t 2)2

1 4C3s1(t2)3 1 4C4S0(t2)4

5 (1)(s4)(1) 1 (4)(s3)(t 2) 1 (6)(s2)(t4) 1 (4)(5)(t 6) 1 (1)(1)(t8)

5 s4 1 4s3t2 1 6s2t4 1 4st 6 1 t8

12. (c3 2 2d2)6 5 [c3 1 (22b2)]6

5 6C0(c3)6(22d2)0 1 6C1

(c3)5(22d 2)1

1 6C2(c3)4(22d2)2 1 6C3

(c3)3(22d2)3

1 6C4(c3)2(22d2)4 1 6C5

(c3)1(22d 2)5

1 6C6(c3)0(22d2)6

5 (1)(c18)(1) 1 (6)(c15)(22d2) 1 (15)(c12)(4d 4) 1 (20)(c9)(28d6) 1 (15)(c6)(16d8) 1 (6)(c3)(232d10) 1 (1)(1)(64d12) 5 c18 2 12c15d2 1 60c12d4 2 160c9d 6

240c6d 8 2 192c3d10 1 64d12

13. P(queen) 5 4 } 52 5

1 } 13

14. P(red king) 5 2 } 52 5

1 } 26

15. P(diamond) 5 13

} 52 5 1 } 4

16. P(not a club) 5 39

} 52 5 3 } 4


5 0.3 1 0.6 2 0.1

5 0.8


80% 5 35 1 P(B) 2 20%

65% 5 P(B)

19. P( }

A ) 5 1 2 P(A)

2 } 5 5 1 2 P(A)

P(A) 5 1 2 2 } 5 5

3 } 5

20. P(A and B) 5 P(A) p P(B) 5 (0.15)(0.6) 5 0.09

21. P(A and B) 5 P(A) p P(BA) 25% 5 60% p P(BA) 41.7% ø P(BA) 22. P(A and B) 5 P(A) p P(BA) 0.36 5 P(A) p 0.4

0.9 5 P(A)

23. P(k 5 4) 5 11C4(0.4)4(0.6)7 ø 0.236

24. P(k 5 2) 5 5C2(0.7)2(0.3)3 5 0.1323

P(k 5 1) 5 5C1(0.7)1(0.3)4 5 0.02835

P(k 5 0) 5 5C0(0.7)0(0.3)5 5 0.00243

P(k ≤ 2) 5P(k 5 2) 1 P(k 5 1) 1 P(k 5 0)

5 0.1323 1 0.02835 1 0.00243

5 0.16308

25. P(k 5 8) 5 9C8(0.9)8(0.1)1 ø 0.3874

P(k 5 9) 5 9C9(0.9)9(0.1)0 ø 0.3874

P(k ≥ 8) 5 P(k 5 8) 1 P(k 5 9)

ø 0.3874 1 0.3874

ø 0.775

26. P(k 5 8) 5 10C8(0.5)8(0.5)2 ø 0.0439

P(k 5 9) 5 10C9(0.5)9(0.5)1 ø 0.0098

P(k 5 10) 5 10C10(0.5)10(0.5)0 ø 0.00098

P(k ≥ 8) ø 0.0439 1 0.0098 1 0.00098 ø 0.055 The probability that at least 8 of the 10 true or false

questions are correct is about 0.055.

27. 15C8 5 15! } 7! p 8! 5

15 p 14 p 13 p 12 p 11 p 10 p 9 p 8! }}} 7! p 8!

5 6435

There are 6435 combinations of council members that could have voted in favor of the budget increase.

28. P(square) 5 Area of square

}}} Area of circular landing area

5 252

} π p 202 5 625

} 400π� ø 0.497

The probability that the parachuter fi rst touches the ground within the square is about 0.497.

29. P(female not in activity) 5 Females not in activity

}} Total number of students

5 325

} 1800 ø 0.18

The probability that a randomly-selected student is a female who is not involved in an activity is about 0.18.


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639Algebra 2


30. P(k 5 0) 5 8C0(0.09)0(0.91)8 ø 0.470

P(k 5 1) 5 8C1(0.09)1(0.91)7 ø 0.372

P(k ≥ 2) 5 1 2 F P(k 5 0) 1 P(k 5 1) G ø 0.158

The probability that at least 2 of the people cite fi shing as their favorite is about 0.158.

Standardized Test Preparation (p. 738)

1. This solution should be scored as partial credit. The explanations are correct, but the probabilities are incorrect. In order to earn a score of full credit, the student would need to re-calculate the areas of the different sections.

Standardized Test Practice (pp. 740–741)

1. a. 109 5 1,000,000,000

There are 1,000,000,000 possible Social Security Numbers.

b. Let x 5 number of years

415,000,000 1 6,000,000x 5 1,000,000,000

6,000,000x 5 585,000,000

x 5 97.5

The Social Security Administration will run out of new numbers in about 97.5 years.

c. Sample answer: The Social Security Administration could assign more digits to the number or include letters.

2. The team with the worst record is A, the second worse is B, and so on.

a. There are 66 total entries.

Teams A B C D E F

Entries (x) 11 10 9 8 7 6

P(x) ø 0.17 0.15 0.14 0.12 0.11 0.09

Teams G H I J K

Entries (x) 5 4 3 2 1

P (x) ø 0.08 0.06 0.05 0.03 0.02

b. There are 1000 total entries.

Teams A B C D E F

Entries (x) 250 200 157 120 89 64

P (x ) ø 0.25 0.2 0.16 0.12 0.09 0.06

Teams G H I J K

Entries (x) 44 29 18 11 7

P(x) ø 0.04 0.03 0.02 0.01 0.01

Teams L M

Entries (x) 6 5

P(x) ø 0.01 0.01

c. The NBA changed the lottery so that the worst teams have a better chance of improving their teams, so that the level of play across the NBA might be more balanced.

3. a. like

school

dislikeschool

0.95

0.050.61

0.390.7

0.3

collegeno college

collegeno college

b. P(A and C) 1 P(B and C)

5 P(A) p P(CA) 1 P(B) p P(CB) 5 (0.61)(0.95) 1 (0.39)(0.70)

5 0.5795 1 0.273

5 0.8525

The probability that a high school student in the U.S. plans to attend college is 0.8525.

c. The probability in part (b) is experimental because the results are based on a survey.

4. a. P(South) 5 Presidents born in South

}} All Presidents

5 7 } 30 .

The probability that a President was born in the south

is 7 }

30 .

b. P(SouthRepublican)

5 Republican Presidents born in South

}}} All Republican Presidents

5 3 } 18 5

1 } 6

The probability that a Republican President was born

in the South is 1 }

6 .

c. No. Sample answer: Even though there have been no Democratic Presidents that were born in the west, this does not mean there never can be in the future.

5. B; 6 p 10 5 60 possible outcomes

6. A; P(spade) 5 13

} 52 5 25%

7. C; 5C3(2x)2(1)3 5 10(4x2)(1) 5 40x2

8. 7!

} 3!

5 840

There are 840 distinguishable premutations of the letters in the word WEEKEND.

9. 9C6(4x)3(21)6 5 (84)(64x3)(1) 5 5376x3

The coeffi cent of x3 is 5376.

10. 5C3 2 5C2 5 5! } 2! p 3! 2

5! } 3! p 2!

5 5 p 4 p 3!

} 2! p 3! 2 5 p 4 p 3!

} 3! p 2!

5 10 2 10

5 0


0.61 5 0.52 1 0.24 2 P(A and B)

P(A and B) 5 0.52 1 0.24 2 0.61 5 0.15


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ompa

ny.


12. P(k 5 15) 5 50C15(0.5)15(0.5)35 ø 0.002

The probability of tossing a coin 50 times and getting heads exactly 15 times is about 0.002.

13. P(k 5 0) 5 15C0(0.30)0(0.70)15 ø 0.0047

P(k 5 1) 5 15C1(0.30)1(0.70)14 ø 0.0305

P(k 5 2) 5 15C2(0.30)2(0.70)13 ø 0.0916

P(k 5 3) 5 15C3(0.30)3(0.70)12 ø 0.17

P(k < 4) 5 P(k 5 0) 1 P(k 5 1) 1 P(k 5 2) 1 P(k 5 3)

ø 0.297

The probability that fewer than 4 of the 15 adults will say that football is their favorite sport is about 0.297.

14. There are 10 possible digits (0 2 9) and 26 possible letters, so there are 36 possible choices for one character. Using the counting principle the number of 6, 7, and 8 character PIN is:

6 characters: 366 5 2,176,782,336

7 characters: 367 5 78,364,164,096

8 characters: 368 5 2,821,109,907,456

So, the total number of different PINs that are between 6 and 8 charecters is:

366 1 367 1 368 5 2,901,650,853,888 PINs.

15. P(often or very often) 5 P(often) 1 P(very often)

5 215

} 614 1 243

} 614

5 458

} 614 ø 0.746

The experimental probability that an adult Internet user uses the Internet to send or receive e-mail often or very often is about 0.746. If the survey had polled computer programmers instead of all adults, the results would be greater because computer programmers would be more likely to use the Internet more often.

16. P(VA or MD or DE or PA or NJ)

5 P(VA) 1 P(MD) 1 P(DE) 1 P(PA) 1 P(NJ)

5 178

} 1907 1 110

} 1907 1 26 } 1907 1

58 } 1907 1

44 } 1907 5

416 } 1907 ø 0.22

The probability that a selected segment of Interstate 95 is in the fi ve states stretching from Virgina to New Jersey is about 0.22. This is calculated by adding the probabililites for each of the fi ve states.


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