Chapter 10meposchools.org/wp-content/uploads/2013/08/chapter-10-worked-out... · Copyright © by...
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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
Worked-Out Solution Key
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
Chapter 10, continued
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610Algebra 2Worked-Out Solution Key
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
10.2 Guided Practice (pp. 690–694)
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.
Chapter 10, continued
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611Algebra 2
Worked-Out Solution Key
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
Chapter 10, continued
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612Algebra 2Worked-Out Solution Key
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
The coeffi cient of x5 is 28064.
33. 5C2(3x)3 1 22 2 5 10 1 27x3 2 (4) 5 1080x3
The coeffi cient of x5 is 1080.
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
The coeffi cient of x4 is 2945.
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
Chapter 10, continued
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613Algebra 2
Worked-Out Solution Key
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 .
Chapter 10, continued
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614Algebra 2Worked-Out Solution Key
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
Equation of perpendicular bisector:
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
Equation of perpendicular bisector:
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
Equation of perpendicular bisector:
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
Chapter 10, continued
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615Algebra 2
Worked-Out Solution Key
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
10.3 Guided Practice (pp. 699–701)
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
}} Area of entire board
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.
Chapter 10, continued
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616Algebra 2Worked-Out Solution Key
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
}} 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
}}} Numbers from 1 to 50
5 34
} 50 5 17
} 25
5. P(perfect square) 5 Perfect squares from 1 to 50
}}} Numbers from 1 to 50
5 7 } 50
6. P(prime number) 5 Primes from 1 to 50
}} Numbers from 1 to 50
5 15
} 50 5 3 } 10
7. P(factor of 150) 5 Factors of 150 from 1 to 50
}} Numbers from 1 to 50
5 10
} 50 5 1 } 5
8. P(multiple of 4) 5 Multiples of 4 from 1 to 50
}} Numbers from 1 to 50
5 12
} 50 5 6 } 25
9. P(two digit number) 5 Two digit numbers from 1 to 50
}}} Numbers from 1 to 50
5 41
} 50
10. P(perfect cube) 5 Perfect cubes from 1 to 50
}} Numbers 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
Chapter 10, continued
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617Algebra 2
Worked-Out Solution Key
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
}} Area of entire board
5 π p 62
} 12 5 36 π
} 144 5 π
} 4
Target B
P(circle) 5 Area of 4 circles
}} Area of entire board
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
}} Area of entire board
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.
Chapter 10, continued
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618Algebra 2Worked-Out Solution Key
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
Chapter 10, continued
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619Algebra 2
Worked-Out Solution Key
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.
10.4 Guided Practice (pp. 708–709)
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
10. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
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
11. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
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
Chapter 10, continued
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620Algebra 2Worked-Out Solution Key
12. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
5 0.46 1 0.37 2 0.31 5 0.52
13. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
5 2 } 7 1
4 } 7 2
1 } 7 5
5 } 7
14. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
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
28. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
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.
29. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
5 0.6 1 0.32 2 0.25 5 0.67
A and B are not disjoint.
30. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
0.65 5 P(A) 1 0.38 2 0
0.27 5 P(A)
A and B are disjoint.
31. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
3 } 5 5
8 } 15 1 P(B) 2
2 } 15
3 }
15 5 P(B)
1 } 5 5 P(B)
A and B are not disjoint.
32. P(A or B) 5 P(A) 1 P(B) 2 P(A and 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.
33. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
32% 5 16% 1 P(B) 2 8%
32% 5 P(B) 1 8%
24% 5 P(B)
A and B are not disjoint.
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
Chapter 10, continued
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621Algebra 2
Worked-Out Solution Key
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 P(none are the same)
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 P(none are the same)
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 P(none are the same)
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
Chapter 10, continued
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622Algebra 2Worked-Out Solution Key
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
Chapter 10, continued
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623Algebra 2
Worked-Out Solution Key
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
11. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
5 0.6 1 0.35 2 0.2 5 0.75
12. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
0.56 5 P(A) 1 0.44 2 0.12
0.56 5 P(A) 1 0.32
0.24 5 P(A)
13. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
0.83 5 0.75 1 P(B) 2 0.25
0.83 5 P(B) 1 0.5
0.33 5 P(B)
14. P(A or B) 5 P(A) 1 P(B) 2 P(A and 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}
Chapter 10, continued
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624Algebra 2Worked-Out Solution Key
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
10.5 Guided Practice (pp. 717–721)
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%.
Chapter 10, continued
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625Algebra 2
Worked-Out Solution Key
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)
6. P(A and B) 5 P(A) p P(B)
0.1 5 0.5 p P(B)
0.2 5 P(B)
7. P(A and B) 5 P(A) p P(B)
0.6 5 P(A) p 0.8
0.75 5 P(A)
8. P(A and B) 5 P(A) p P(B)
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
Chapter 10, continued
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626Algebra 2Worked-Out Solution Key
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
34. Answers will vary.
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
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 (0.50)(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%.
Chapter 10, continued
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627Algebra 2
Worked-Out Solution Key
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
Chapter 10, continued
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628Algebra 2Worked-Out Solution Key
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
Domain: all real numbers
51. f (x) p g(x) 5 (x2 1 2)(x 2 4) 5 x3 2 4x2 1 2x 2 8
Domain: all real numbers
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
Domain: all real numbers
54. g(f(x)) 5 g(x2 1 2) 5 (x2 1 2) 2 4 5 x2 2 2
Domain: all real numbers
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
Domain: all real numbers
56. g(g(x)) 5 g(x 2 4) 5 (x 2 4) 2 4 5 x 2 8
Domain: all real numbers
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
10.6 Guided Practice (pp. 725–727)
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
Chapter 10, continued
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629Algebra 2
Worked-Out Solution Key
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
Chapter 10, continued
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630Algebra 2Worked-Out Solution Key
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.
Chapter 10, continued
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631Algebra 2
Worked-Out Solution Key
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
The distribution is skewed because it is not symmetric about any vertical line. The most likely number of successes is 6.
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
The distribution is skewed because it is not symmetric about any vertical line. The most likely number of successes is 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.
39. Answers will vary.
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.
Chapter 10, continued
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632Algebra 2Worked-Out Solution Key
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
Chapter 10, continued
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633Algebra 2
Worked-Out Solution Key
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
Chapter 10, continued
n2ws-10b.indd 633 6/27/06 11:26:34 AM
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634Algebra 2Worked-Out Solution Key
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
Chapter 10, continued
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635Algebra 2
Worked-Out Solution Key
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
The most likely number of successes is 3.
2. n 5 14, p 5 0.58
The most likely number of successes is 8.
3. n 5 15, p 5 0.805
The most likely number of successes is 12.
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.
Chapter 10, continued
n2ws-10b.indd 635 6/27/06 11:26:49 AM
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636Algebra 2Worked-Out Solution Key
3. P(at least 2 have the same code)
5 1 2 P(none are the same)
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
P(C) 5 P(A and C) 1 P(B and C)
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
Chapter 10, continued
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637Algebra 2
Worked-Out Solution Key
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
}} Numbers from 1 to 30
5 15
} 30 5 1 } 2
16. P(multiple of 5) 5 Multiples of 5 from 1 to 30
}} Numbers from 1 to 30
5 6 } 30 5
1 } 5
17. P(factor of 60) 5 Factors of 60 from 1 to 30
}} Numbers from 1 to 30
5 11
} 30
18. P(prime) 5 Prime numbers from 1 to 30
}}} 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.
20. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
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
Chapter 10, continued
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638Algebra 2Worked-Out Solution Key
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
17. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
5 0.3 1 0.6 2 0.1
5 0.8
18. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
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.
Chapter 10, continued
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639Algebra 2
Worked-Out Solution Key
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
11. P(A or B) 5 P(A) 1 P(B) 2 P(A and B)
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
Chapter 10, continued
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640Algebra 2Worked-Out Solution Key
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.
Chapter 10, continued
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