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Solutions Key

Probability and Statistics11

CHAPTER

ARE YOU READY? PAGE 791

1. B 2. E

3. A 4. C

5.

6. 1 - 14___20

= 20___20

- 14___20

= 6___20

= 3___10

7.3__8

+ 5__6

= 9___24

+ 20___24

= 29___24

8.8___

15- 2__

5= 8___

15- 6___

15

= 2___15

9.1___

12+ 1___

10= 5___

60+ 6___

60

= 11___60

10.1__2

· 3__7

= 3___14

11. 2 1__3

· 1__4

= 7__3

· 1__4

= 7___12

12.4__5

÷ 1__2

= 4__5

· 2__1

= 8__5

13. 5 1__3

÷ 1__4

= 16___3

÷ 1__4

= 16___3

· 4__1

= 64___3

14. 7% of 150 = x

0.07(150) = x

x = 10.5

15. 90% of x = 45 0.9x = 45

x = 50

16. Price increased by 12% of 24 = 0.12(24) = $2.88

17. The amount of water to be changed is 20% of 65 = 0.20(65) = 13 gal.

18. 2, 4, 4, 6, 9 mean: 5 median: 4 mode: 4

19. 1, 1, 1, 2, 2, 2 mean: 1.5 median: 1.5 mode: 1, 2

20. 1, 2, 3, 4, 5, 6 mean: 3.5 median: 3.5 mode: none

21. 3, 14, 14, 18, 18, 18, 20 mean: 15 median: 18 mode: 18

11-1 PERMUTATIONS AND

COMBINATIONS, PAGES 794–800

CHECK IT OUT!

1a. start plot end 6 × 4 × 5 = 120 There are 120 adventures.

b. letter letter letter letter digit 52 × 52 × 52 × 52 × 10 = 73,116,160 There are 73,116,160 possible passwords.

2a. 8 P3 = 8!_______(8 - 3)!

= 8!__5!

= 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 ____________________ 5 · 4 · 3 · 2 · 1

= 8 · 7 · 6 = 336 There are 336 ways to award the costumes.

b. 5 P2 = 5!_______(5 - 2)!

= 5!__3!

= 5 · 4 · 3 · 2 · 1 ____________ 3 · 2 · 1

= 5 · 4 = 20 There are 20 ways a 2-digit number can be formed.

3. The order does not matter. It is a combination.

8 C2 = 8!________2!(8 - 2)!

= 8!____2!6!

= 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 ____________________ 2 · 1(6 · 5 · 4 · 3 · 2 · 1)

= 8 · 7____2 · 1

= 28

There are 28 ways to select 2 swimmers from 8.

THINK AND DISCUSS

1. Possible answer: selecting a 9-player batting order from 20 players; selecting 3 magazine subscriptions from a list of 20

2. 1; possible answer: there is only 1 way to choose the entire group from a group.

3. Possible answer: the number of ways to select 4 items from 3; you can’t select more than the total number of items.

4.

EXERCISES

GUIDED PRACTICE

1. important; permutation 2. blouse jacket skirt

3 × 3 × 2 = 18 There are 18 different outfits.

3. digit letter 9 × 25 = 225 There are 225 different codes.

4. 7 P2 = 7!_______(7 - 2)!

= 7!__5!

= 7 · 6 = 42 There are 42 ways to schedule the 2 activities.

405 Holt McDougal Algebra 2

5. 12 P3 = 12!________(12 - 3)!

= 12!___9!

= 12 · 11 · 10 = 1320

There are 1320 ways to listen to 3 songs.

6. 6 P3 = 6!_______

(6 - 3)!=

6!__3!

= 6 · 5 · 4 = 120

There are 120 ways the prizes can be awarded.

7. 21 C4 = 21!_________4!(21 - 4)!

= 21!_____4!17!

= 21 · 20 · 19 · 18

______________ 4 · 3 · 2 · 1

= 5985

There are 5985 ways to send 4 students to the

library.

8. 5 C3 = 5!________

3!(5 - 3)!=

5!____3!2!

= 5 · 4____2 · 1

= 10

There are 10 ways to choose 3 boxes of cereal.

PRACTICE AND PROBLEM SOLVING

9. lake cabin

4 × 3 = 12

There are 12 routes from the lake to the cabins.

10. posters markers

3 × 4 = 12

There are 12 different posters.

11. 9 P2 = 9!_______

(9 - 2)!=

9!__7!

= 9 · 8 = 72

There are 72 ways to choose a manager and an

assistant.

12. 26 P3 = 26!________

(26 - 3)!=

26!___23!

= 26 · 25 · 24 = 15,600

There are 15,600 possible identification codes.

13. 5 P2 = 5!_______

(5 - 2)!=

5!__3!

= 5 · 4 = 20

There are 20 ways to assign 2 planes to the

runways.

14. 6 C3 = 6!________

3!(6 - 3)!=

6!____3!3!

= 6 · 5 · 4_______3 · 2 · 1

= 20

There are 20 choices of 3 hamburger toppings.

15. 49 C7 - 49 C6

= 49!_________

7!(49 - 7)!-

49!_________6!(49 - 6)!

= 49!_____

7!42!-

49!_____6!43!

= 49 · 48 · 47 · 46 · 45 · 44 · 43

_________________________ 7 · 6 · 5 · 4 · 3 · 2 · 1

- 49 · 48 · 47 · 46 · 45 · 44

_____________________ 6 · 5 · 4 · 3 · 2 · 1

= 85,900,584 - 13,983,816

= 71,916,768

There are 71,916,768 more ways to select the

numbers.

16. 6 P6 = 6!_______

(6 - 6)!=

6!__0!

= 6 · 5 · 4 · 3 · 2 · 1 = 720

17. 5 C5 = 5!________

5!(5 - 5)!=

5!____5!0!

= 1

18. 9 P1 = 9!_______

(9 - 1)!=

9!__8!

= 9

19. 6 C1 = 6!________

1!(6 - 1)!=

6!____1!5!

= 6

20.2!__6!

= 1__________ 6 · 5 · 4 · 3

= 1____360

21.4!3!____2!

= (4 · 3 · 2 · 1)(3 · 2 · 1)

__________________ 2 · 1

= 4 · 3 · 2 · 1(3) = 72

22.9!__7!

= 9 · 8 = 72

23.8! - 5!_______(8 - 5)!

= 8! - 5!______

3!=

8!__3!

- 5!__3!

= (8 · 7 · 6 · 5 · 4) - (5 · 4)

= 6720 - 20 = 6700

24. 6 C2 = 6!________

2!(6 - 2)!=

6!____2!4!

= 6 · 5____2 · 1

= 15

25. 7 C4 = 7!________4!(7 - 4)!

= 7!____4!3!

= 7 · 6 · 5_______3 · 2 · 1

= 35

26. 7 P3 = 7!_______(7 - 3)!

= 7!__4!

= 7 · 6 · 5 = 210

7 C4 = 7!________4!(7 - 4)!

= 7!____4!3!

= 7 · 6 · 5_______3 · 2 · 1

= 35

Therefore, 7 P3 > 7 C4 .

27. 7 P4 = 7!_______(7 - 4)!

= 7!__3!

= 7 · 6 · 5 · 4 = 840

7 C3 = 7!________3!(7 - 3)!

= 7!____3!4!

= 7 · 6 · 5_______3 · 2 · 1

= 35

Therefore, 7 P4 > 7C3 .

28. 7 C3 = 7!________3!(7 - 3)!

= 7!____3!4!

= 7 · 6 · 5_______3 · 2 · 1

= 35

7 C4 = 7!________4!(7 - 4)!

= 7!____4!3!

= 7 · 6 · 5_______3 · 2 · 1

= 35

Therefore, 7 C3 = 7 C4 .

29. 10 C10 = 10!___________

10!(10 - 10)!=

10!_____10!0!

= 1

10 P10 = 10!_________

(10 - 10)!=

10!___0!

= 3,628,800

Therefore, 10 C10 < 10 P10 .


30.n! 4! 3! 2! 1

n(n - 1)! 4(3!) = 24 3(2)! = 6 2(1)! = 2 1(0)! = 1

n(n - 1)! = n!

(1)(1 - 1)! = 1!

1(0)! = 1

0! = 1

31. The Es in GEESE are identical. The order of the Es

is not important.

32. Number of sequences in a peal is 8! = 40,320.

It would take 0.25(40,320) = 10,080 s = 2.8 h to

ring a complete peal.

33a. President A A A A A A A A A A A A

Vice President B B B C C C D D D E E E

Secretary C D E B D E B C E B C D

b. President B B B B B B B B B B B B

Vice President A A A C C C D D D E E E

Secretary C D E A D E A C E A C D

A president, a vice president, and a secretary can

be chosen in 60 ways.

c.5!_______

(5 - 3)!= 60

d.5!________

3!(5 - 3)!= 10

The answer 60 is a number of permutations, and

the answer 10 is a number of combinations.

34. Possible answer:

n P r____ n C r

=

n!______(n - r)!________

n!_______r!(n - r)!

= n!______(n - r)!

× r !(n - r)!________

n!= r !;

6 P3____ 6 C3

= 3! = 6; 6 C3 = 6 P3____3!

;

the number of combinations of n items taken r at

a time, is the number of permutations of the items

divided by the number of ways to order the r items.

35. 9 C2 = 9!________

2!(9 - 2)!=

9!____2!7!

= 9 · 8____2 · 1

= 36;

9 C7 = 9!________

7!(9 - 7)!=

9!____7!2!

= 9 · 8____2 · 1

= 36;

10 C6 = 10!_________

6!(10 - 6)!=

10!____6!4!

= 10 · 9 · 8 · 7

___________ 4 · 3 · 2 · 1

= 210;

10 C4 = 10!_________

4!(10 - 4)!=

10!____4!6!

= 10 · 9 · 8 · 7

___________ 4 · 3 · 2 · 1

= 210;

n!________r !(n - r)!

is the same as n !________(n - r)! r !

.

36a. Jen can arrange the dice in 5! = 120 ways.

b. 5 C3 = 5!________

3!(5 - 3)!=

5!____3!2!

= 5 · 4____2 · 1

= 10

37. A; order is important.

38. Choosing 3 times from 9 digits: there are 9 possible

choices the first time, 8 the second, and 7 the third.

The total number of permutations is

9 × 8 × 7 = 504.

TEST PREP

39. D

14 C5 = 14!_________5!(14 - 5)!

= 14!____5!9!

40. J

9 C4 = 9!________

4!(9 - 4)!=

9!____4!5!

9 C5 = 9!________

5!(9 - 5)!=

9!____5!4!

41. 15 C4 = 15!_________

4!(15 - 4)!=

15!_____4!11!

= 15 · 14 · 13 · 12

______________ 4 · 3 · 2 · 1

= 1365

There are 1365 ways Rene can choose her

electives.

CHALLENGE AND EXTEND

42a. 4 points: 4!________2!(4 - 2)!

= 4!____2!2!

= 4 · 3____2 · 1

= 6

5 points: 5!________

2!(5 - 2)!=

5!____2!3!

= 5 · 4____2 · 1

= 10

6 points: 6!________

2!(6 - 2)!=

6!____2!4!

= 6 · 5____2 · 1

= 15

n points: n C2

b. 20 points: 20 C2 = 20!_________

2!(20 - 2)!=

20!_____2!18!

= 20 · 19______2 · 1

= 190

43. Select 12 jurors out of 30 potential jurors, 30 C12.

Select 2 alternate jurors out of the remaining

18 potential jurors, 18 C2 . Use the Fundamental

Counting Principle to combine the number of ways

the jurors can be selected, (30 C12 ) ( 18 C2 ).

SPIRAL REVIEW

44. p(x) = 0.15x; P(x) = 0.45x

P(x) = 3p(x) is a vertical stretch.

45. 17___n= 11___

77 1309 = 11n1309_____

11= 11n____

11 119 = n

46. 2.9___3.7

= x_____23.31

67.599 = 3.7x67.599______

3.7=

3.7x____3.7

18.27 = x


47. 2.2___n

= 1.6___9.5

20.9 = 1.6n

20.9____1.6

= 1.6n____1.6

13.0625 = n

48. x___36

= 98___18

18x = 352818x____18

= 3528_____18

x = 196

49. A = 6, B = 3, C = -9

B2 - 4AC = 3

2 - 4(6)(-9)

= 225

Because B2 - 4AC > 0, the equation represents a

hyperbola.

50. A = 8, B = 0, C = 8

B2 - 4AC = 0

2 - 4(8)(8)

= -256

Because B2 - 4AC < 0 and A = C, the equation

represents a circle.

11-2 THEORETICAL AND EXPERIMENTAL

PROBABILITY, PAGES 802–809

CHECK IT OUT!

1a. There are 36 possible outcomes, and 5 outcomes

with the sum of 6.

P(sum is 6) = 5___

36

b. There are 36 possible outcomes, and 0 outcomes

with a difference of 6.

P(difference is 6) = 0___36

= 0

c. There are 36 possible outcomes, and 15 outcomes

where the red cube is greater.

P(red cube is greater) = 15___36

= 5___12

2. There are 100 possible outcomes.

The number of possible outcomes where both

numbers are less than 9 is(first number < 9) (second number < 9) = 8 × 8 = 64.

P(both numbers less than 9) = 64____

100=

16___25

The probability that both numbers are less

than 9 is 16___25

.

3. Order is not important. It is a combination.

The number of possible outcomes is

8 C2 = 8!____

2!6!=

8 · 7____2 · 1

= 28.

There is only 1 way to play both of the retailer’s ads.

P(play both of the retailer’s ads) = 1___28

4. Area of large triangle is At = 1__2

(15)(15) = 112.5.

Area of small triangle is As = 1__2

(4)(4) = 8.

As___At

= 8_____

112.5=

16____225

The probability that the point is in the small triangle

is 16____225

.

5a. P(diamond) = 9___

26

b. P(not club) = 1 - P(club)

= 1 - 7___26

= 19___26

THINK AND DISCUSS

1. No, the probability of an event cannot exceed 1.

2. sum of 5 and sum of 9

3. experimental: 8___20

= 2__5

; theoretical: 1__2

4.

EXERCISES

GUIDED PRACTICE

1. theoretical probability

2. There are 8 possible outcomes, and 4 outcomes

where the quarter shows heads.

P(quarter shows heads) = 4__8= 1__

2


where the penny and the nickel both show heads.

P(penny and nickel show heads) = 2__8= 1__

4


where 1 coin shows heads.

P(one coin shows heads) = 3__8


where all coins land the same way.

P(all coins land the same way) = 2__8= 1__

4

6. P(does not end in 5) = 1 - P(ends in 5)

= 1 - 10____100

= 9___10

The probability that a random 2-digit number does

not end in 5 is 9___10

.

7. P(not in Dec or Jan) = 1 - P(in Dec or Jan)

= 1 - 31 + 31_______

365=

303____365

The probability that a date is not in December or

January is 303____365

.

8. Order is important. It is a permutation.


4 P4 = 4!__0!= 24.

There is only 1 way to place all the letters in the

correct envelope.

P(all letters are in correct envelopes) = 1___24




12 C3 = 12!____3!9!

= 12 · 11 · 10

__________3 · 2 · 1

= 220.

There is only 1 way to choose all 3 green balloons.

P(3 green balloons) = 1____220

10. Area of large circle is At = π (6)2 = 36π

Area of middle circle is Am = π (4)2 = 16π

Area of small circle is Au = π (2)2 = 4π

Shaded area is As = 16π - 4π = 12π

As___At

= 12π

____36π

= 1__3

The probability that a point is in the shaded area

is 1__3

.

11.Au___At

= 4π

____36π

= 1__9

The probability that a point is in the smallest

circle is 1__9

.

12. P (red) = 5___

20= 1__

4

13. P (red or blue) = 5 + 7_____

20=

3__5


14. There are 15 possible outcomes, and 5 outcomes of

a white marble.

P(white marble) = 5___

15= 1__

3


of a red or white marble.

P(red or white marble) = 12___15

= 4__5


The number of possible outcomes where both

numbers are greater than 2 is(first number > 2) (second number > 2) = 6 × 6 = 36

P(both numbers greater than 2) = 36___64

= 9___16

.

The probability that both numbers are greater

than 2 is 9___

16.


8 C3 = 8!____

3!5!=

8 · 7 · 6_______3 · 2 · 1

= 56

There is only 1 way to choose the 3 strongest

swimmers.

P(3 strongest swimmers) = 1___56


7 C2 = 7!____2!5!

= 7 · 6____2 · 1

= 21

There is only 1 way to choose books 1 and 2.

P(books 1 and 2) = 1___21

19. 2 ft = 24 in., and 4 ft = 48 in.

Area of platform is At = 24(48) = 1152

Area of hole is As = π (3)2 = 9π

As___At

= 9π

_____1152

= π

____128

≈ 1___42

The probability of the bag landing in the hole is 1___42

.

20. P(red card) = 16___28

= 4__7

The experimental probability that a card is red is 4__7

.

21. never; the theoretical probability of tossing tails on

a fair coin is 1__2

. Tossing the coin 25 times, implies

that tails will appear 12.5 times in order to be equal

to the theoretical probability.

22. P(state that does not border Mississippi)= 1 - P(state that borders Mississippi)

= 1 - 4___49

= 45___49

The probability that the winner will be from a state

that does not border Mississippi is 45___49

.

23a. Area of circle is Ac = π r 2

Area of square is As = (2r)2 = 4 r

2

Ac___As

= π r

2____4 r

2=

π

__4

b. Possible answer: 7852______

10,000≈

π

__4

,

and so, π ≈ 4 × 7852______

10,000≈ 3.141.

24. Possible answer: A roll of a die shows less than 20.

25a. P(made throws 1-25) = 17___25

= 0.68

P(made throws 26-50) = 21___25

= 0.84

P(made throws 51-75) = 19___25

= 0.76

P(made throws 76-100) = 16___25

= 0.64

b. P(throws made) = 73____

100= 0.73

c. The greater the number of experiments, the

closer the experimental probability will be to the

theoretical probability.

26a. P(two 4s) = P(4 and 4) = 1___36

The probability that Mei will have 5 of a kind is 1___36

.

b. P(one 4) = P(4 and not 4) + P(not 4 and 4)

= 5 + 5_____

36=

5___18

The probability that Mei will have 4 of a kind is 5___18

.

c. P(zero 4s) = P(not 4 and not 4) = 25___36

The probability that Mei will have three 4s is 25___36

.

d.1___36

+ 10___36

+ 25___36

= 1

27. Length −−

BD= 4 - 2 = 2; Length

−−

AF= 6 - 1 = 5

Length −−

BD________Length

−−

AF

= 2__5

Probability that a point lies on −−

BD is 2__5

.

28. P(temperature > 90°F in April)

= 5___30

= 1__6

≈ 0.167


29. June;

P(temperature ≤ 90°F)= 1 - P(temperature > 90°F)

= 1 - 26___30

= 4___30

≈ 0.13

30. it will be slightly greater:

P(temperature ≤ 90°F)= 1 - P(temperature > 90°F)

= 1 - 11___31

= 20___31

≈ 0.645 vs. = 1 - 11___30

= 19___30

≈ 0.633

31. No; yes; a theoretical probability of 1 means all

possible outcomes are favorable outcomes, but a

theoretical probability of 0.99 means there is at least

1 unfavorable outcome.

32. Let r represent the radius of the outer circle.

r = √ �� x 2 + x

2

r = √ �� 2x2

r = √ � 2x

Aouter = π ( √ � 2x) 2 = 2π x2

Ainner = π x2

Ainner_____Aouter

= π x

2_____2π x

2= 1__

2

The probability that a random point in the large

circle

is within the inner circle is 1__2

.

33.1__2

; each toss is independent.

34. College: female high school players have a better

chance, since 4100_______

456,900>

4500_______549,500

;

pro: male high school players have a better chance,

since 32_______

456,900< 44_______

549,500.

35. Possible answer: Theoretical probability is based on

all possible outcomes, while experimental probability

is based on sample results. The theoretical

probability that a rolled number cube will show a 4

is 1__6

. The experimental probability would be 3___

13 if is

rolled 13 times and shows a 4 three times.

TEST PREP

36. A

P(heads) = 1 - P(tails)

= 1 - 14___25

= 11___25

= 0.44

37. G

Alarge � = 8(16) = 128

Asmall � = 5(14) = 70

Ashaded_______A large �

= 128 - 70________

128

= 58____

128≈ 45%

38. C

2 · 2 · 2 · 2 = 8

39. H

P(sum is 5) = 4___36

= 1__9

40. Length −−

AD= 24 - 4 = 20; Length

−−

BC= 12 - 8 = 4

Length −−

BC________Length

−−

AD

= 4___20

= 1__5

The probability that a point will lie between points

B and C is 1__5

.


41. Possible answer: After a large number of trials,

experimental probability approaches theoretical

probability.


P(no one gets the right trumpet) = 8___

24=

3__8

43. There were 100 experiments.

SPIRAL REVIEW

44. x = - b___2a

= - (-0.85)_______2(0.25)

= 0.85____0.5

= 1.7

f(1.7) = 0.25 (1.7)2 - 0.85(1.7) + 1 = 0.2775

The minimum is 0.2775.

45. x = - b___2a

= - 20_____2(-2)

= 20___4

= 5

f(5) = -2(5)2 + 20(5) - 34 = 16

The maximum is 16.

46. Since the directrix is vertical and to the left of the

vertex, the parabola opens to the right and has the

form x = 1___4p

y 2 , where p > 0.

The distance between the directrix and the vertex

is 3, and so, p = 3, and 4p = 12.

equation: x = 1___12

y2

47. Since the directrix is horizontal and above the

vertex, the parabola opens down and has the

form y = 1___4p

x 2 , where p < 0.

The distance between the directrix and the vertex

is -5, and so, p = -5, and 4p = -20.

equation: y = - 1___20

x2

48a. 11 C5 = 11!_________5!(11 - 5)!

= 11!____5!6!

= 11 · 10 · 9 · 8 · 7

______________ 5 · 4 · 3 · 2 · 1

= 462

She can choose 5 players in 462 ways.

b. 11 P5 = 11!________(11 - 5)!

= 11!___6!

= 11 · 10 · 9 · 8 · 7 = 55,440

She can choose 5 players to play different

positions in 55,440 ways.


11-3 INDEPENDENT AND DEPENDENT

EVENTS, PAGES 811–818

CHECK IT OUT!

1a. Rolling a 6 once does not affect the probability of rolling a 6 again. The events are independent.P(6 and 6) = P(6) · P(6)

= 1__6· 1__

6= 1___

36

b. Tossing heads once does not affect the probability of tossing heads or tails again. The events are independent.P(H and H and T) = P(H) · P(H) · P(T)

= 1__2· 1__

2· 1__

2= 1__

8

2. The events are dependent because P(red > 4) is different when the sum is 9.

P(red > 4) = 2__6= 1__

3

P(sum > 9 | red > 4) = 5___12

P(sum > 9 AND red > 4)= P(red > 4) · P(sum > 9 | red > 4)

= 1__3· 5___

12= 5___

36; P(sum is 9) changes when it is

known that the red cube is greater than 4.

3a. P(other | Travis) = 5____350

≈ 0.014

b. P(Harris) = 1058_____3125

P(Bush | Harris) = 581_____1058

P(Harris and Bush | Harris) = 1058_____3125

· 581_____1058

≈ 0.186

4a. Replacing the first bead means that the occurrence of the first selection does not affect the probability of the second selection. The events are independent.P(white and red) = P(white) · P(red)

= 15____100

= 3___20

b. Not replacing the first bead means that there will be fewer beads to chose from, affecting the probability of the second selection. The events are dependent.P(white and red) = P(white) · P(red | white)

= 3___10

· 5__9= 1__

6

c. Not replacing the beads means that there will be fewer beads to chose from, affecting the probability of the second and third selections. So the events are dependent.P(not red and not red and not red)= P(not red) · P(not red | not red)· P(not red | not red and not red)

= 5___10

· 4__9· 3__

8= 1___

12

THINK AND DISCUSS

1. Possible answer: a coin landing heads up on one flip and landing heads up on the next flip

2. For independent events A, B, and C,P(A, then B, then C) = P(A) · P(B) · P(C);3 coin flips: P(H, then H, then H)

3.

EXERCISES

GUIDED PRACTICE

1. independent 2. Rolling a 1 once does not affect the probability of

rolling a 1 again. The events are independent.P(1 and 1) = P(1) · P(1)

= 1__6· 1__

6= 1___

36

3. Tossing heads once does not affect the probability of tossing heads again. The events are independent.P(H and H and H) = P(H) · P(H) · P(H)

= 1__2· 1__

2· 1__

2= 1__

8

4. The probability that the product is less than 20

decreases from 7__9

if the blue cube shows a 4.

P(blue 4) = 6___36

= 1__6

P(product < 20 | blue 4) = 4__6= 2__

3P(blue 4 and product < 20)= P(blue 4) · P(product < 20 | blue 4)

= 1__6· 2__

3= 1__

9

5. The probability that the yellow cube shows a

multiple of 3 increases from 1__3

if the product is 6.

P(yellow multiple of 3 | product is 6) = 2__4= 1__

2

6. P(not defective | shipped) = 942____952

= 471____476

7. P(Shipped AND Defective)

= (Defective AND Shipped)

_____________________ Total

= 10_____1000

= 1____100

8. Replacing the first checker means that the occurrence of the first selection does not affect the probability of the second selection. The events are independent.P(black and black) = P(black) · P(black)

= 10___20

· 10___20

= 1__4


9. Not replacing the first checker means that there will be fewer checkers to choose from, affecting the probability of the second selection, so the events are dependent.P(black and black) = P(black) · P(black | black)

= 10___20

· 9___19

= 9___38


10. The choice of activity of the first friend does not affect the probability of the choice of activity of the second friend. The events are independent.The first student will choose one activity. Then, the next student will choose an acitivity. Since there are 4 activities and his friend is in one, the probability of

them being in the same activiity is 1__4

· 1__4

= 1___16

.

11. Rolling an even number does not affect the probability of rolling a 6. The events are independent.P(even and 6) = P(even) · P(6)

= 1__2

· 1__6

= 1___12

12. The probability that the product is greater than 24

increases from 1__9

if the yellow cube is greater than 5

to 1__3

.

P(yellow > 5) = 1__6

P(product > 24 | yellow > 5) = 2__6

= 1__3

P(yellow > 5 and product > 24)= P(yellow > 5) · P(product > 24 | yellow > 5)

= 1__6

· 1__3

= 1___18

13. The probability that the product is 8 decreases

from 1___18

if the blue cube is less than 3.

P(blue < 3) = 12___36

= 1__3

P(product = 8 | blue < 3) = 1___12

P(blue < 3 and product is 8)= P(blue < 3) · P(product is 8 | blue < 3)

= 1__3

· 1___12

= 1___36

14a. P(Cuba | 1990) = 10,645______16,997

≈ 0.63

b. P(Spain) = 4471______65,846

P(2000 | Spain) = 1264_____4471

P(Spain and 2000 | Spain)

= 4471______65,846

· 1264_____4471

≈ 0.019

c. P(1995 | Ghana) = 3152______11,962

≈ 0.26

15. P(employed | advanced degree) = 0.104_____0.145

≈ 0.72

16. P(not a high school grad) = 1.894______13.697

P(not employed | not a high school grad) = 0.834_____1.894

P(not a high school grad and not employed)

= 1.894______13.697

· 0.834_____1.894

= 0.834______13.697

= 0.06

17. Not replacing the first slip means that there will be fewer slips to choose from, affecting the probability of the second selection. The events are dependent.P(even and even) = P(even) · P(even | even)

= 4__9

· 3__8

= 1__6

18. Replacing the first slip means that the occurrence of the first selection does not affect the probability of the second selection. The events are independent.P(even and even) = P(even) · P(even)

= 4__9

· 4__9

= 16___81

19. The tossing of heads on a coin does not affect the probability of rolling a 6 on a number cube.

The events are independent.

20. Drawing a 4 and not replacing it affects the probability of drawing an ace.

The events are dependent.

21. Rolling a 1 does not affect the probability of rolling a 4 on the same number cube.


22. Hitting the bull’s-eye the first time does not affect the probability of hitting the bull’s-eye again.


23a. P(won | second serve in) = 34___56

≈ 0.61

b. P(double fault | lost) = 3___56

≈ 0.05

24. P(present | present)= 0.9 · 0.95= 0.855

25a. P(not 5 and not 5) = P(not 5) · P(not 5)

= 5__6

· 5__6

= 25___36

P(first reroll no 5s and second reroll no 5s)= P(not 5 and not 5) · P(not 5 and not 5)

= 25___36

· 25___36

= 625_____1296

b. P(5 and 5) = P(5) · P(5)

= 1__6

· 1__6

= 1___36

c. P(5 | 5) = 1__6

26. P(girl) ≈ 147____270

≈ 0.54

27. P(girl | senior) ≈ 71____118

≈ 0.6

28. P(senior | male) ≈ 47____123

≈ 0.38


29. P(yellow and “Happy Birthday!”)= P(yellow) · P(“Happy Birthday!”)

= 80____100

· 50____100

= 40____100

There are 40 yellow balloons marked “Happy Birthday!” in the box.

30a.Scheduled Flights (thousands)

January to July

2003 2004 2005 Total

On Time 3102 3197 3237 9536

Delayed 598 846 877 2321

Canceled 61 68 82 211

Total 3761 4111 4169 12,068

b. P(canceled | 2004) = 68____4111

≈ 0.017

c. P(2005 | on time) = 3237_____9536

≈ 0.339

31. The events are not dependent. If the coin is fair, P(H) = P(T) = 0.5 for any toss.

TEST PREP

32. A. It cannot be Saturday again next year.

33. F 6 · P(doubles and doubles and doubles)= P(doubles) · P(doubles) · P(doubles)

= 1__6

· 1__6

· 1__6

= 1____216

P(three 5’s in a row)= P(five) · P(five) · P(five)

= ( 1__6

· 1__6

· 1__6) = 1____

216

34a. P(D | A) = 0.2; P(D | B) = 0.2; P(D | C) = 0.2

b. Independent; P(D) and P(E) do not change regardless of whether A, B, or C occurs first.

c. Possible answer: A ball has a 0. − 3 probability of

rolling into pipe A, B, or C. From any pipe, the probability of rolling to location D is 0.2 and to location E is 0.8.


35. 7; P(sum of 7) = 6___36

= 1__6

;

after a roll, P(sum of 7 | 1st roll = 1, 2, 3, 4, 5, 6) = 1__6

.

36a. Let x represent the size of the smallest group. To find the probability that 2 people share a birthday, subtract the complement from 1.1__2

≤ P(2 people share a birthday)

1__2

≤ 1 - P(no one shares a birthday)

1__2

≤ 1 - 365____365

· 364____365

· … · 365 - x_______365

1__2

≤ 1 - ( 1____365)

x

( 365!_________(365 - x)!)

From trial and error, x = 23.

b. The probability of a person not having a birthday

on February 29, in a four-year span, is 1460_____1461

.

Since the probability of one person’s birthday does not affect the probability of the next person’s birthday, the events are independent.

P(no one born on February 29) = ( 1460_____1461)

150 ≈ 0.9

c. Let x represent the size of the smallest group of people.1__2

≤ P(1 person born on February 29)

1__2

≤ 1 - P(no one born on February 29)

1__2

≤ 1 - ( 1460_____1461)

x

1__2

≤ ( 1460_____1461)

x

x ≥ 1012.34 The smallest group is 1013 people.

37. P(lower | woman) = 1 - P(upper | woman)

= 1 - 35___90

= 11___18

;

no, P(lower) ≠ P(lower | woman)

38a. Per 10,000 People Tested

Have

Strep

Do Not

Have StrepTotal

Test Positive 198 98 296

Test Negative 2 9702 9704

Total 200 9800 10,000

b. P(have strep | test positive) = 198____296

= 99____148

SPIRAL REVIEW

39a. p(d) = 18.3g; p(j) = 32.5g

b.

c. vertical stretch by a factor of ≈ 1.78

40. The graph of the first equation is a hyperbola. The graph of the second equation is a line. There may be as many as two points of intersection.

Solve each equation for y.

y = ± √ 1__2

x2 - 3

y = 2 The points of intersection are when x ≈ ±3.7 and y = 2.


41. The graph of both equations is a hyperbola. There

may be as many as four points of intersection.


y = ± √ �� 2x2 - 9

y = ± √ �� 1__6

x2 + 11___

3 The points of intersection

are when

x ≈ ± 2.6 and y ≈ ± 2.2.

42. The graph of the first equation is a circle. The graph

of the second equation is a parabola. There may be

as many as two points of intersection.


y = ± √ �� 16 - x2

y = - 3__2

- 5__2

x2

The points of intersection

are when

x ≈ ±1.0 and y ≈ -3.9.

43. P(sum is 12) = 1___36

44. P(sum < 5) = 6___36

= 1__6

45. P(at least one is odd)= 1 - P(even and even)

= 1 - 3__6

· 3__6

= 1 - 1__4

= 3__4

46. P(at least one < 3)= 1 - P(both ≥ 3)

= 1 - 4__6

· 4__6

= 1 - 4__9

= 5__9

11-4 COMPOUND EVENTS,

PAGES 819–825

CHECK IT OUT!

1a. Each student can only vote once.

b. P(votes for Kline � voted for Vila)= P(votes for Kline) + P(voted for Vila)= 20% + 55% = 75%

2a. P(king � heart)= P(king) + P(heart) - P(king heart)

= 4___52

- 13___52

- 1___52

= 4___13

b. P(red � face)= P(red) + P(face) - P(red face)

= 26___52

+ 12___52

- 6___52

= 8___

13

3. 61 - 28 = 33 people got a hair styling and a

manicure.

P(hair styling � manicure)= P(hair styling) + P(manicure)

- P(hair styling manicure)

= 96____160

+ 61____160

- 33____

160= 124____

160=

31___40

The probability that a customer had a hair styling or

a manicure is 31___40

.

4. P(all choose different) = 62 P5_____62

5

= 62 · 61 · 60 · 59 · 58

_________________ 62 · 62 · 62 · 62 · 62

≈ 0.8476

P(at least 2 choose same)= 1 - P(all choose different)= 1 - 0.8476 ≈ 0.152393394

The probability that at least 2 customers bought the

same style is 0.1524 or 15.24%.

THINK AND DISCUSS

1. If events A and B are mutually exclusive,

P(A B) = 0.

So, P(A � B) = P(A) + P(B) - 0 = P(A) + P(B).

2. February 29 occurs only once every 4 years, and

March 13 occurs once every year. You are more

likely to share a birthday with someone if you were

born on March 13.

3.

EXERCISES

GUIDED PRACTICE

1. inclusive events

2. A marble is either black or red.

3. P(red � blue)= P(red) + P(blue)

= 13___25

+ 2___25

= 3__5

4. The car cannot turn both left and right;

P(left � right)= P(left) + P(right)= 0.1 + 0.2 = 0.3

5. P(greater than 5 � odd)= P(greater than 5) + P(odd)

- P(greater than 5 odd)

= 5___

10+

5___10

- 2___10

= 4__5

6. P(8 � less than 5)= P(8) + P(less than 5)

= 1___10

+ 4___10

= 1__2

7. Pat least 1 even

= 1 - Podd odd

= 1 - 5___10

· 4__9

= 7__9


8. 400 ÷ 2 = 200 students have a college degree and

were married.

P(degree � married)= P(degree) + P(married) - P(degree � married)

= 400____650

+ 310____650

- 200____650

= 51___65

9. 400 ÷ 2 = 200 students have a college degree and

were not married.

650 - 310 = 340 students were not married.

P(degree � not married)= P(degree) + P(not married) - P(degree � not married)

= 400____650

+ 340____650

- 200____650

= 54___65

10. 650 - 400 = 250 students do not have a college

degree.

310 - 200 = 110 students were not married and do

not have a college degree.

P(no degree � married)= P(no degree) + P(married)- P(no degree � married)

= 250____650

+ 310____650

- 110____650

= 9___

13

11. P(all choose different) = 8 P6____8

6

= 8 · 7

_______________ 8 · 8 · 8 · 8 · 8 · 8

≈ 0.08

P(at least 2 choose same)= 1 - P(all choose different)= 1 - 0.0769 = 0.92

The probability that at least 2 employees purchased

the same drink is 0.92 or 92%.


12. The jump rope is either red or green.

13. P(red � green)= P(red) + P(green)

= 1__6+ 1__

3= 1__

2

14. P(E � G)= P(E) + P(G)

= 1___16

+ 1___16

= 1__8

15. P(E � vowel)= P(E) + P(vowel) - P(E � vowel)

= 1___16

+ 4___16

- 1___16

= 1__4

16. 98 × 1__7= 14 teachers teach math.

P(woman � math)= P(woman) + P(math) - P(woman � math)

= 42___98

+ 14___98

- 8___98

= 24___49

17. 98 - 42 = 56 teachers are men.

14 - 8 = 6 math teachers are men.

P(man � math)= P(man) + P(math) - P(man � math)

= 56___95

+ 14___98

- 6___98

= 32___49

18. 56 - 6 = 50 teachers are men and don’t teach

math.

98 - 14 = 84 teachers do not teach math.

P(man � not math)= P(man) + P(not math) - P(man � not math)

= 56___98

+ 84___98

- 50___98

= 45___49

19. P(no heart) = 39___52

= 0.75

Replacing the card after it is drawn means that the

draw of the first card does not affect the draw of the

second card. The events are independent.

P(at least one heart)= 1 - P(no heart)= 1 - 0.75

13 ≈ 0.976

20. No; possible answer: If event A is rolling a 3 on a

number cube and event B is rolling a 4 on a number

cube, then, the outcomes 1, 2, 5, and 6 are common

to both A′ and B′.

21. P(NBA � CSI)= P(NBA) + P(CSI)= 0.22 + 0.15 = 0.37

experimental because it is based on a small sample

22. the percent of schools that offer both music and

dance classes

23. The minimum will occur when there is the largest

possible intersection between the two events.

largest intersection is 19%

P(music � drama)= P(music) + P(drama) - P(music � drama)= 87 + 19 - 19 = 87%

The minimum probability that music or drama are

offered is 87%.

The maximum will occur when there is smallest

possible intersection between the two events.

smallest intersection is 6%

P(music � drama)= P(music) + P(drama) - P(music � drama)= 87 + 19 - 6 = 100%

The maximum probability that music or drama are

offered is 100%.

24a. P(purple) = 2.25

2_____9

2

=2.25____81

=1___

36

b. P(red) = 3

2 - 1.5

2________9

2

= 6.75____81

= 1___12

c. P(red � blue)= P(red) + P(blue) - P(red � blue)

= 1__9+

16___81

- 1___36

= 91____

324

d. P(yellow)= 1 - P(not yellow)

= 1 - 91____234

= 233____324

25a. Possible answer: The probability that a person is

born color-blind or male will be greater, because it

includes more successful outcomes, such as

color-blind females and non-color-blind males.


b. P(male � color-blind) = 0.08 · 52 = 4.16%

P(color-blind) = 0.08 · 52 + 0.005 · 48 = 4.4

P(male � color-blind)= P(male) + P(color-blind) - P(male � color-blind)= 52 + 4.4 - 4.16 = 52.24%

26a. P(3 and 5 � 4 and 4)= P(3 and 5) + P(4 and 4)

= 2___36

+ 1___36

= 1___12

b. P(one 3) + P(two 3)

= 1 - P(no 3s)

= 1 - 25___36

= 11___36

Since two 3’s is also a small straight, 11___36

- 1___36

= 10___36

= 5___

18.

27. P(under 18 � owner’s property)= P(under 18) + P(owner’s property)

- P(under 18 � owner’s property)

0.95 = 0.8 + 0.64 - P(under 18 � owner’s property)P(under 18 � owner’s property) = 0.49

28a. P(all male � all female)= P(all male) + P(all female)

= 10 C4_____ 24 C4

+ 14 C4_____ 24 C4

≈ 0.11

b. At any given point the committee must have at

least one man or one woman. The probability is 1.

29a. P(vowel) = 42____100

= 0.42

b. P(Y) = 2____100

= 0.02

c. P(vowel or Y) = 44____100

= 0.44;

it is the sum of the probabilities.

30. Possible answer: There are a total of 4 outcomes:

HH, HT, TH, TT

. Three of these have at least one

heads. The probability is 3__4

.

The event “at least one heads” is the complement of

the event “no heads.” The probability is

1 - ( 1__2

· 1__2) = 1 - 1__

4=

3__4

.

TEST PREP

31. D

P(5 and 2 and 7)

= 1___10

· 1___10

· 1___10

= 1_____1000

32. F

P(tails) = 1__2

33. D

P(5 � greater than 3)= P(5) + P(greater than 3) - P(5 � greater than 3)

= 1__6

+ 3__6

- 1__6

= 1__2

34. 1; the complement of an event contains all

unfavorable outcomes. The probability of an

event or its complement is the probability of all

outcomes, 1.


35. P(at least 2 people share a birthday)= 1 - P(no one shares a birthday)

= 1 - 365 P10______365

10

= 1 - ( 1____365)

10 (

365!__________ (365 - 10)!) ≈ 0.12

36. P(ferry � train)= P(ferry) + P(train) - P(ferry � train)

= 47____162

+ 80____162

- 27____162

= 50___81

37. P(ferry � rental car)= P(ferry) + P(rental car) - P(ferry � rental car)

= 47____162

+ 94____162

- 24____162

= 13___18

38. P(train � ferry � train � rental car)= P(train � ferry) + P(train � rental car)

- P(train � ferry � train � rental car)

= 27____162

+ 19____162

- 11____162

= 35____162

39. P(B � C)= P(B) + P(C) - P(B � C)= 0.3 + 0.7 - 0.1 = 0.9

40. P(A � B � C)= P(A) + P(B) + P(C) - P(B � C) - P(A � C)

- P(A � B) + P(A � B � C)

= 0.5 + 0.3 + 0.7 - 0.1 - 0.3 - 0.2 + 0.1 = 1

41. P(A � C)= P(A) + P(C) - P(A � C)= 0.5 + 0.7 - 0.3 = 0.9

P(B � (A � C))= P(B) + P(A � C) - P(B � (A � C))= 0.3 + 0.9 - 1 = 0.2

SPIRAL REVIEW

42. The data can be modeled by

y = 0.5 x3 + 1.5 x

2 - 3x - 4.

43. The data can be modeled by

y = -1.5x3 - 6 x

2 + 16.5x + 45.

44. The function is composed of a constant piece and a

linear piece. The domain is divided at x = -1.

x f(x) = 2 f(x) = 2x + 4

-2 2 0

-1 2 2

0 2 4

No circle is required at (-1, 2) because the function

is connected at that point.


45. The function is composed of two parabolas. The domain is divided at x = 1.

x f(x) = 1 - x2

f(x) = x2 - 1

0 1 -1

1 0 0

2 -3 3

No circle is required at (0, 0) because the function is connected at that point.

46. P(heads and heads)= P(heads) · P(heads)

= 1__2

· 1__2

= 1__4

47. P(heads and heads and tails and tails)= P(heads) · P(heads) · P(tails) · P(tails)

= 1__2

· 1__2

· 1__2

· 1__2

= 1___16

48. P(sum > 10 and first cube is a 6)= P(sum > 10) · P(first cube is a 6 | sum > 10)

= 3___36

· 2__3

= 1___18

READY TO GO ON? PAGE 827

1. 10 P5 = 10!________(10 - 5)!

= 10!___5!

= 10 · 9 · 8 · 7 · 6= 30,240

2. 8 C4 = 8!________4!(8 - 4)!

= 8!____4!4!

= 8 · 7 · 6 · 5 __________ 4 · 3 · 2 · 1

= 70

3. 6 P5 = 6!_______(6 - 5)!

= 6!__1!

= 6 · 5 · 4 · 3 · 2= 720

4. P(iced tea) = 3___18

= 1__6

5. P(out of ink and out of ink)= P(out of ink) · P(out of ink | out of ink)

= 2__9

· 1__8

= 1___36

6. Area of large triangle is At = 1__2

(15)(4) = 30

Area of shaded region is As = 1__2

(4)(4) = 8

As___At

= 8___30

= 4___15

The probability that the point is in the shaded

region is 4___15

.

7. P(not rolling a 2) = 1 - P(rolling a 2)

= 1 - 12___50

= 19___25

8. The result of a toss does not affect the probability of the next toss.P(tails and tails and tails and tails and heads)= P(tails) · P(tails) · P(tails) · P(tails) · P(heads)

= 1__2

· 1__2

· 1__2

· 1__2

· 1__2

= 1___32

9. P(sum ≥ 10) changes after a red 6 has occurred.

P(sum ≥ 10 and red 6)

P(sum ≥ 10) = 1__6

P(red 6 | sum ≥ 10) = 1__2

10. P(11th grade | geometry) = 33____127

11. Not replacing the red checker after it is selected affects the probability of the next selection. The events are dependent.P(red and black)= P(red) · P(black | red)

= 15___25

· 10___24

= 1__4

12. P(even � 1)

= P(even) + P(1)

= 15___30

+ 1___30

= 8___15

13. P(even � multiple of 7)= P(even) + P(multiple of 7)

- P(even � multiple of 7)

= 15___30

+ 4___30

- 2___30

= 17___30

14. 85 - 60 = 25 part-time employees. 85 - 40 = 45 not married employees.P(part time � not married) = 1 - P(full time � married)

= 1 - P(full time) + P(married)

- P(full time � married)

= 1 -

60___85

+ 40___85

- 30___85

�

= 15___

85P(part time � not married)= P(part time) + P(not married)

- P(part time � not married)

= 25___85

+ 45___85

- 15___85

= 11___17

11-5 MEASURES OF CENTRAL

TENDENCY AND VARIATION,

PAGES 828–836

CHECK IT OUT!

1a. mean: 26___4

= 6.5

median: 6 + 8_____

2= 7

mode: no mode

b. mean: 21___5

= 4.2

median: 5 mode: 2 and 6

2. expected value = 0(0.75) + 1(0.15) + 2(0.08) + 3(0.02)= 0.37

The expected number of accidents in one week is 0.37.


3. 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 17, 18, 18, 19, 22, 23

minimum: 11 first quartile: 13 maximum: 23 third quartile: 18 median: 14 IQR: 5

4. − x = 14___10

= 1.4

0 3 1 1 0

x - − x -1.4 1.6 -0.4 -0.4 -1.4

(x - − x)

2 1.96 2.56 0.16 0.16 1.96

5 1 0 3 0

x - − x 3.6 -0.4 -1.4 1.6 -1.4

(x - − x)

2 12.96 0.16 1.96 2.56 1.96

σ 2 = 26.4____

10= 2.64; σ = √ �� 2.64 ≈ 1.6

5. The mean is about 5.4 and the standard deviation is about 4.5.

Two standard deviations is about 2(4.5) = 9. Values 9 units below the mean are negative and would not make sense in this problem. Values greater than 9 units away from the mean are outliers. So, 19, the number of scored runs, is an outlier.

The mean increases from ≈ 4.3 to ≈ 5.4, and the standard deviation increases from ≈ 2.3 to ≈ 4.5.

THINK AND DISCUSS

1. Possible answer: The mean increases by the constant.

2. Possible answer: The standard deviation is unchanged.

3. Possible answer: The standard deviation is multiplied by √ � 2.

4.

EXERCISES

GUIDED PRACTICE

1. variance 2. mean: 36___6

= 6

median: 6 + 7_____

2= 6.5

mode: 7

3. mean: 43___8

= 5.375

median: 6 + 6_____

2= 6

mode: 6

4. mean: 90___5

= 18

median: 18 mode: no mode

5. expected value= 0(0.9359) + 1(0.05) + 5(0.01) + 20(0.003)

+ 100(0.001) + 1000(0.0001)= 0.36

The expected value of the prize is $0.36. 6. 1, 2, 2, 3, 5, 8, 9, 11

minimum: 1 first quartile: 2 maximum: 11 third quartile: 8.5 median: 4 IQR: 6.5

7. 1, 2, 2, 2, 4, 4, 4, 7 minimum: 1 first quartile: 2 maximum: 7 third quartile: 4 median: 3 IQR: 2

8. 22, 27, 31, 33, 34 minimum: 22 first quartile: 24.5 maximum: 34 third quartile: 33.5 median: 31 IQR: 9

9. − x = 20___5

= 4

3 3 4 5 5

x - − x -1 -1 0 1 1

(x - − x)

2 1 1 0 1 1

σ 2 = 4__

5= 0.8 σ = √ �� 0.8 = 0.89

10. − x = 112____7

= 16

10 12 14 15 18 20 23

x - − x -6 -4 -2 -1 2 4 7

(x - − x)

2 36 16 4 1 4 16 49

σ 2 = 126____

7= 18 σ = √ �� 18 = 4.24


11. − x = 147____6

= 24.5

7 14 21 28 35 42

x - − x -17.5 -10.5 -3.5 3.5 10.5 17.5

(x - − x)

2 306.25 110.25 12.25 12.25 110.25 306.25

σ 2 = 857.5_____

6= 142.92 σ = √ �� 142.92 = 11.95

12. The mean is 46. −

6 and the standard deviation is about 8.4.

Three standard deviations is about 3(8.4) = 25.2. Values greater than 25.2 units away from the mean are outliers. So 19, the width of the desk, is an outlier.

The mean decreases from ≈ 49.2 to 46. −

6 , and the standard deviation increases from ≈ 0.72 to ≈ 8.4.


13. mean: 139____6

≈ 23.1 −

6

median: 16 + 25_______

2= 20.5

mode: no mode

14. mean: 169____8

= 21.125

median: 7 + 7_____

2= 7

mode: 7

15. mean: 75___5

= 15

median: 15 mode: no mode

16. expected value

= 0 ( 1__8) + 1 ( 3__

8) + 2 ( 3__8) + 3 ( 1__

8)= 1.5

The expected number of heads is 1.5.

17. 6, 12, 12, 15, 18, 29 minimum: 6 first quartile: 12 maximum: 29 third quartile: 18 median: 13.5 IQR: 6

18. 2, 2, 2, 2, 3, 8, 8, 42 minimum: 2 first quartile: 2 maximum: 42 third quartile: 8 median: 2.5 IQR: 6

19. 1, 2, 3, 3, 4 minimum: 1 first quartile: 1.5 maximum: 4 third quartile: 3.5 median: 3 IQR: 2

20. − x = 21___5

= 4.2

4 4 4 4 5

x - − x -0.2 -0.2 -0.2 -0.2 0.8

(x - − x)

2 0.04 0.04 0.04 0.04 0.64

σ 2 = 0.8___

5= 0.16 σ = √ �� 0.16 = 0.4

21. − x = 245____7

= 35

8 12 30 35 48 50 62

x - − x -27 -23 -5 0 13 15 27

(x - − x)

2 729 529 25 0 169 225 729

σ 2 = 2406_____

7≈ 343.71 σ = √ �� 343.71 ≈ 18.54

22. − x = 132____4

= 33

14 26 40 52

x - − x -19 -7 7 19

(x - − x)

2 361 49 49 361

σ 2 = 820____

4= 205 σ = √ �� 205 ≈ 14.32

23. The mean is about 22.8 and the standard deviation is about 11.5.

Three standard deviations is about 3(11.5) = 34.5. Values greater than 34.5 units away from the mean are outliers. So, 58 is an outlier.

The mean increases from ≈ 19.8 to ≈ 22.8, and the standard deviation increases from ≈ 5.6 to ≈ 11.5.

24. Possible answer: {3, 3, 9, 9}

25. The mean; 37° is an outlier and affects the mean greatly.

26. 25; Q3 + 1.5 (IQR) = 5 + 1.5(2)= 8;

25 > 8

27. 15; Q1 - 1.5 (IQR) = 79 - 1.5(11)= 62.5;

15 < 62.5

28. 92 and 1; Q 3 + 1.5 (IQR)= 36 + 1.5(3) = 40.5

92 > 40.5

Q1 - 1.5 (IQR)= 33 - 1.5(3) = 28.5

1 < 28.5

29. The time intervals in which the duration time is an outlier are three times the standard deviation away from the mean, < 0.3 min or > 6.9 min. There are no outliers for duration times.

30. The intervals in which the time between eruptions is an outlier are three times the standard deviation away from the mean, < 32.6 min or > 109.4 min. There are no outliers for time between eruptions.

31. Ruth; possible answer: 13

32. Ruth; possible answer: 6

33. Possible answer: Ruth: 36; Aaron: 18

34. Aaron; Ruth’s data are more spread out, and the set’s IQR is twice the IQR for Aaron’s set.

35. expected value = 500(0.001) - 1(0.999) = -0.499 The expected gain is -$0.499.


36. expected value = 100(0.10) - 2(0.30) + 0(0.60) = 9.40 The expected value is $9.40.

37. B; the student should have squared the differences.

38. Sometimes; possible answer: when 2 coins are tossed, the expected number of heads is 1, a possible outcome. When 3 coins are tossed, the expected number, 1.5, is not a possible outcome.

39a. Product 1 2 3 4 5 6 8 9 10

Probability 1___36

1___18

1___18

1___12

1___18

1__9

1___18

1___36

1___18

Product 12 15 16 18 20 24 25 30 36

Probability 1__9

1___18

1___36

1___18

1___18

1___18

1___36

1___18

1___36

expected value = 12 1__4

b. P(product greater than 12 1__4)

= P(15) + P(16) + P(18) + P(20) + P(24) + P(25)+ P(30) + P(36)

= 1___18

+ 1___36

+ 1___18

+ 1___18

+ 1___18

+ 1___36

+ 1___18

+ 1___36

= 13___36

c. P(product less than 12 1__4)

= 1 - P(product greater than 12 1__4)

= 1 - 13___36

= 23___36

d. No; possible answer: the expected value is a mean, not a median value.

40a. − x = 91___10

= 9.1 in.

9.4 17.0 7.3 7.0 16.1

x - − x 0.3 7.9 -1.8 -2.1 7

(x - − x)

2 0.09 62.41 3.24 4.41 49

5.4 6.9 8.5 4.2 9.2

x - − x -3.7 -2.2 -0.6 -4.9 -0.1

(x - − x)

2 13.69 4.84 0.36 24.01 0.01

σ 2 = 162.06______

10= 16.206 σ = √ �� 16.206 ≈ 4.0

The mean annual precipitation is 9.1 in., and the standard deviation is 4.0.

b. One standard deviation from the mean is precipitation less than 5.1 in or greater than 13.1 in.

In the years 1995, 1998, and 2002 the precipitation was more than one standard deviation away from the mean.

c. median: 7.3 + 8.5________

2= 7.9; IQR: 2.5

TEST PREP

41. D. Datas are close together.

42. H. Two different variances for the 2 sets.

43. C. Datas cannot have a mean of 50.


44a. The mean, median, and standard deviation grow by a factor of 5.

mean: 20; median: 15; standard deviation: 1.6 · 5 = 8

b. The mean and median increase by 5, while the standard deviation remains the same.

mean: 9; median: 8; standard deviation: 1.6

45. A deck of 1 card:

Cards in

Same Position1

Probability 1

Expected value = 1

A deck of 2 cards:

Cards in

Same Position2 0

Probability 1__2 1__

2

Expected value = 1

A deck of 3 cards:

Cards in

Same Position3 1 0

Probability 1__6

1__2 1__

3

Expected value = 1

A deck of 4 cards:

Cards in

Same Position4 2 1 0

Probability 1___24

1__4 1__

3 3__

8

Expected value = 1

Following the pattern, the expected number of cards that will be in the same position after a deck of cards is shuffled is 1.

SPIRAL REVIEW

46. Let m represent the magazines sold. 725 + 1.75m = 1425 1.75m = 700

m = 400 Li has sold 400 magazines.

47. (2 - x2) (2x

2 + 5x - 3)

= 2 (2x2) + 2 (5x) + 2 (-3) - x

2 (2x

2) - x2 (5x)

- x2 (-3)

= 4 x2 + 10x - 6 - 2 x

4 - 5 x

3 + 3 x

2

= -2x4 - 5 x

3 + 7 x

2 + 10x - 6

48. 4x y 2 (x2

y + 3 x2 - 2y)

= 4x y 2 (x2

y) + 4x y 2 (3x

2) + 4x y 2 (-2y)

= 4 x3 y

3 + 12 x

3 y

2 - 8x y

3

49. P(even � 1)

= P(even) + P(1)

= 1__2

+ 1__6

= 2__3

50. P(odd � 4)= P(odd) + P(4)

= 1__2

+ 1__6

= 2__3


51. P(divisible by 2 � divisible by 6)= P(divisible by 2) + P(divisible by 6)

- P(divisible by 2 � divisible by 6)

= 1__2

+ 1__6

- 1__6

= 1__2

11-6 BINOMIAL DISTRIBUTIONS,

PAGES 837–843

CHECK IT OUT!

1a. (x - y)5

= 5 C0 x5 (-y)

0 + 5 C1 x

4 (-y)

1 + 5 C2 x

3 (-y)

2

+ 5 C3 x2 (-y)

3 + 5 C4 x

1 (-y)

4 + 5 C5 x

0 (-y)

5

= x5 - 5 x

4y + 10 x

3 y

2 - 10 x

2 y

3 + 5x y

4 - y

5

b. (a + 2b) 3

= 3 C0 a3 (2b) 0 + 3 C1 a

2 (2b) 1 + 3 C2 a

1 (2b) 2

+ 3 C3 a0 (2b) 3

= 1 · a3 + 3 · 2 a

2b + 3 · 4a b

2 + 1 · 8 b

3

= a3 + 6 a

2b + 12a b

2 + 8 b

3

2a. The probability that a student will be assigned to

Counselor Jenkins is 1__3

.

P(2) = 3 C2 ( 1__3)

2 ( 2__3)

1

= 3 ( 1__9) (

2__3)

= 2__9

≈ 0.22

The probability that exactly 2 students will be

assigned to Counselor Jenkins is about 22%.

b. The probability that Ellen will guess the right

answer is 1__4

.

P(2) + P(3) + P(4) + P(5)

= 5 C2 ( 1__4)

2 ( 3__4)

3 + 5 C3 ( 1__

4)3 ( 3__4)

2 + 5 C4 ( 1__

4)4 ( 3__4)

1

+ 5 C5 ( 1__4)

5 ( 3__4)

0

= 10 ( 1___16) (

27___64) + 10 ( 1___

64) ( 9___16) + 5 ( 1____

256) ( 3__4) + ( 1_____

1024)= 47____

128≈ 0.37

The probability that Ellen will get at least 2 answers

correct by guessing is about 37%.

3a. P(at least 2 correct)

= 1 - P(0 or 1 corrrect)

= 1 - 20 C0 (0.25)0 (0.75)

20 + 20 C1 (0.25)

1 (0.75)

19

≈ 1 - 0.003 + 20(0.25)(0.004)

= 1 - 0.023

= 0.977

The probability that Wendy will get at least 2 correct

answers by guessing is about 98%.

b. P(23 or fewer)

= 1 - P(24 or 25)

= 1 - 25 C24 (0.98)24

(0.02)1 + 25 C25 (0.98)

25 (0.02)

0

≈ 1 - 25(0.6158)(0.02) + 0.6035

= 1 - 0.9114

= 0.0886

The probability that there are 23 or fewer acceptable

parts is about 9%.

THINK AND DISCUSS

1. Possible answer: 1; a binomial experiment has only

2 outcomes, and the probability of those outcomes

are p and q. So, p + q = 1.

2. n C r, pr, and q

n - r

3.

EXERCISES

GUIDED PRACTICE

1. 2

2. (x + 3) 4

= 4 C0 x4 3

0 + 4 C1 x

3 3

1 + 4 C2 x

2 3

2 + 4 C3 x

1 3

3 + 4 C4 x

0 3

4

= 1 · x4 + 4 · 3 x

3 + 6 · 9 x

2 + 4 · 27x + 1 · 81

= x4 + 12 x

3 + 54 x

2 + 108x + 81

3. (3x + 5) 3

= 3 C0 (3x)3 5

0 + 3 C1 (3x)

2 5

1 + 3 C2 (3x)

1 5

2

+ 3 C3 (3x)0 5

3

= 1 · 27 x3 + 3 · 45 x

2 + 3 · 75x + 1 · 125

= 27 x3 + 135 x

2 + 225x + 125

4. (p - 2) 6

= 6 C0 p6 (-2)

0 + 6 C1 p

5 (-2)

1 + 6 C2 p

4 (-2)

2

+ 6 C3 p3 (-2)

3 + 6 C4 p

2 (-2)

4 + 6 C5 p

1 (-2)

5

+ 6 C6 p0 (-2)

6

= 1 · p6 + 6 · (-2p

5) + 15 · 4 p4 + 20 · (-8p

3) + 15 · 16 p

2 + 6 · (-32p) + 1 · 64

= p6 - 12 p

5 + 60 p

4 - 160 p

3 + 240 p

2 - 192p + 64

5. (x + y)6

= 6 C0 x6 y

0 + 6 C1 x

5 y

1 + 6 C2 x

4 y

2 + 6 C3 x

3 y

3

+ 6 C4 x2 y

4 + 6 C5 x

1 y

5 + 6 C6 x

0 y

6

= x6 + 6 x

5y + 15 x

4 y

2 + 20 x

3 y

3 + 15 x

2 y

4 + 6x y

5

+ y6


6. P(4) = 6 C4 (0.30)4 (0.7)

2

= 15(0.0081)(0.49)

≈ 0.060

The probability that exactly 4 athletes will be chosen

is about 0.060.

P(4) + P(5) + P(6)

≈ 0.060 + 6 C5 (0.30)5 (0.7)

1 + 6 C6 (0.30)

6 (0.7)

0

≈ 0.060 + 0.0102 + 0.0007

≈ 0.070

The probability that at least 4 athletes are chosen is

about 0.070.

7. P(3) = 4 C3 ( 1__5)

3 ( 4__5)

1

= 4 ( 1____125) (

4__5)

= 16____

625≈ 0.026

The probability of getting exactly 3 coupons is about

0.026.

P(at least 2)

= 1 - P(0 or 1)

= 1 - 4 C0 ( 1__

5)0 ( 4__5)

4 + 4 C1 ( 1__

5)1 ( 4__5)

3 �

= 1 - 256____625

+ 4 ( 1__5) (

64____125)

�

= 1 - 512____625

= 113____625

≈ 0.181

The probability of getting at least 2 coupons is about

0.181.

8. P(at least 2 upside-down stamps)

= 1 - P(0 or 1 upside-down stamps)

= 1 - 30 C0 (0.02)0 (0.98)

30 + 30 C1 (0.02)

1 (0.98)

29

≈ 1 - 0.545 + 30(0.02)(0.557)

= 1 - 0.8792

≈ 0.121

The probability of getting at least 2 boxes with an

upside-down stamp is 0.121.


9. (y + 5) 4

= 4 C0 y4 5

0 + 4 C1 y

3 5

1 + 4 C2 y

2 5

2 + 4 C3 y

1 5

3

+ 4 C4 y0 5

4

= 1 · y4 + 4 · 5 y

3 + 6 · 25 y

2 + 4 · 125y + 1 · 625

= y4 + 20 y

3 + 150 y

2 + 500y + 625

10. (2m - 1) 3

= 3 C0 (2m)3 (-1)

0 + 3 C1 (2m)

2 (-1)

1

+ 3 C2 (2m)1 (-1)

2 + 3 C3 (2m)

0 (-1)

3

= 1 · 8 m3 + 3 · (-4m

2) + 3 · 2m + 1 · (-1)

= 8 m3 - 12 m

2 + 6m - 1

11. (4 + 3x)5

= 5 C0 45 (3x)

0 + 5 C1 4

4 (3x)

1 + 5 C2 4

3 (3x)

2

+ 5 C3 42 (3x)

3 + 5 C4 4

1 (3x)

4 + 5 C5 4

0 (3x)

5

= 1 · 1024 + 5 · 768x + 10 · 576 x2 + 10 · 432 x

3

+ 5 · 324 x4 + 1 · 243 x

5

= 1024 + 3840x + 5760 x2 + 4320 x

3 + 1620 x

4

+ 243 x5

12. (2a + 3c)3

= 3 C0 (2a)3 (3c)

0 + 3 C1 (2a)

2 (3c)

1 + 3 C2 (2a)

1 (3c)

2

+ 3 C3 (2a)0 (3c)

3

= 1 · 8 a3 + 3 · 12 a

2c + 3 · 18a c

2 + 1 · 27 c

3

= 8 a3 + 36 a

2c + 54a c

2 + 27 c

3

13. P(6) + P(7) + P(8)

= 8 C6 (0.83)6 (0.17)

2 + 8 C7 (0.83)

7 (0.17)

1

+ 8 C8 (0.83)8 (0.17)

1

≈ 28(0.327)(0.029) + 8(0.271)(0.17) + 0.225

≈ 0.86

The probability that at least 6 students agree with

the statement is about 0.86.

14. P(2) = 5 C2 (0.15)2 (0.85)

3

≈ 10(0.023)(0.614)

≈ 0.14

The probability that exactly 2 marbles are black is

about 0.14.

P(at least 2)

= 1 - P(0 or 1)

= 1 - 5 C0 (0.15)0 (0.85)

5 + 5 C1 (0.15)

1 (0.85)

4

≈ 1 - 0.444 + 5(0.15)(0.522)

= 1 - 0.8355

≈ 0.16

The probability that at least 2 marbles are black is

about 0.16.

15. P(2 girls and 1 boy) = 3 C2 ( 1__2)

2 ( 1__2)

1

= 3 ( 1__4) (

1__2)

= 3__8

= 0.375

P(3 girls) = 3 C3 ( 1__2)

3 ( 1__2)

0

= 1__8

= 0.125

16. P(at least 4)

= 1 - P(0, 1, 2, or 3)

= 1 - 15 C0 (0.25)0 (0.75)

15 + 15 C1 (0.25)

1 (0.75)

14

+ 15 C2 (0.25)2 (0.75)

13 + 15 C3 (0.25)

3 (0.75)

12

≈ 1 - 0.461

≈ 0.54

17. (x - y)5

= 5 C0 x5 (-y)

0 + 5 C1 x

4 (-y)

1 + 5 C2 x

3 (-y)

2

+ 5 C3 x2 (-y)

3 + 5 C4 x

1 (-y)

4 + 5 C5 x

1 (-y)

5

= x5 - 5 x

4y + 10 x

3 y

2 - 10 x

2 y

3 + 5x y

4 - y

5


18. (c + 6) 3

= 3 C0 c3 6

0 + 3 C1 c

2 6

1 + 3 C2 c

1 6

2 + 3 C3 c

0 6

3

= 1 · c3 + 3 · 6 c

2 + 3 · 36c + 1 · 216

= c3 + 18 c

2 + 108c + 216

19. (4k - 1) 4

= 4 C0 (4k)4 (-1)

0 + 4 C1 (4k)

3 (-1)

1 + 4 C2 (4k)

2 (-1)

2

+ 4 C3 (4k)1 (-1)

3 + 4 C4 (4k)

0 (-1)

4

= 1 · 256 k4 + 4 · (-64k

3) + 6 · 16 k2 + 4 · (-4k)

+ 1 · 1

= 256 k4 - 256 k

3 + 96 k

2 - 16k + 1

20. (p + q)7

= 7 C0 p7 q

0 + 7 C1 p

6 q

1 + 7 C2 p

5 q

2 + 7 C3 p

4 q

3

+ 7 C4 p3 q

4 + 7 C5 p

2 q

5 + 7 C6 p

1 q

6 + 7 C7 p

0 q

7

= p7 + 7 p

6q + 21 p

5 q

2 + 35 p

4 q

3 + 35 p

3 q

4

+ 21 p2 q

5 + 7p q

6 + q

7

21. P(2) = 3 C2 (0.8)2 (0.2)

1

= 3(0.64)(0.2)

= 0.384

22. P(1) = 5 C1 (0.5)1 (0.5)

4

= 5(0.5)(0.0625)

= 0.15625

23. P(2) = 4 C2 ( 1__3)

2 ( 2__3)

2

= 6 ( 1__9) (

4__9)

= 8___

27≈ 0.30

24. To seat all of the passengers who arrive, only 20 or

less of the passengers who bought tickets need to

show up.

P(20 or less)

= 1 - P(21 or 22)

= 1 - 22 C21 (0.91)21

(0.09)1 + 22 C22 (0.91)

22 (0.09)

0

≈ 1 - 0.40

= 0.60

The probability that all of the passengers who arrive

will have a seat is about 0.60.

25. P(4 males) = 4 C4 ( 1__2)

4 ( 1__2)

0

= 1___16

= 0.0625

P(at least 3 males)

= P(3 males) + P(4 males)

= 4 C3 ( 1__2)

3 ( 1__2)

1 + 1___

16

= 4 ( 1__8) (

1__2) + 1___

16

= 5___16

≈ 0.3125

26. P(more than 7 heads)

=P(8) + P(9) + P(10)

= 10 C8 ( 1__2)

8 ( 1__2)

2 + 10 C9 ( 1__

2)9 ( 1__2)

1 + 10 C10 ( 1__

2)10

( 1__2)

0

= 45_____

1024+

5____512

+ 1_____1024

= 7____128

≈ 0.055

27. P(at least 2 heads)

= 1 - P(0 or 1 heads)

= 1 -

10 C0 ( 1__

2)0 ( 1__2)

10 + 10 C1 ( 1__

2)1 ( 1__2)

9

= 1 - 11_____1024

= 1013_____1024

≈ 0.989

28. P(5 heads)

= 10 C5 ( 1__2)

5 ( 1__2)

5

= 63____256

≈ 0.25

29. P(0 or 1 defective)

= 8 C0 (0.05)0 (0.95)

8 + 8 C1 (0.05)

1 (0.95)

7

≈ 0.94

The probability of no more than 1 defective part in a

box of 8 is about 0.94.

30a. Possible answer: randBin(5, 0.8, 5)

b. P(at least 4)

= P(4) + P(5)

= 5 C4 (0.8)4 (0.2)

1 + 5 C5 (0.8)

5 (0.2)

0

≈ 0.74

c. Possible answer: 0.6 < 0.74

31.

The bar heights increase nearly exponentially from 0

successes to 7 successes, maximize at 8, and drop

at 10. The expected value is 8.

32. 3 of one and 1 of the other:

2 · ( 4 C3 ( 1__2)

3 ( 1__2)

1) = 2 · 4___16

= 0.5

2 of each: 6___16

= .375

33a. P(rain) = 82____365

≈ 0.22

b. P(3 rainy days) = 7 C3 (0.22)3 (0.78)

4

≈ 0.14


c. P(at least 3 rainy days)= 1 - P(0, 1, or 2 rainy days)

= 1 - 7 C0 (0.22)0 (0.78)7 + 7 C1 (0.22)1 (0.78)6

+ 7 C2 (0.22)2 (0.78)5

≈ 0.19

34. The trials are dependent. For a binomial experiment the trials must be independent.

35. P(delayed at least 3 times)= P(3) + P(4)

= 4 C3 (0.2046)3 (0.7954)1 + 4 C4 (0.2046)4 (0.7954)0

≈ 0.03

36a. P(home run)

= 6 C1 ( 1__2)

1 ( 1__2)

5

= 3___32

≈ 0.09375

b. P(out)

= 6 C0 ( 1__2)

0 ( 1__2)

6 + 6 C2 ( 1__

2)2 ( 1__2)

4 + 6 C3 ( 1__

2)3 ( 1__2)

3

= 1___64

+ 15___64

+ 5___16

= 9___16

≈ 0.5625

c. P(hit)

= 6 C2 ( 1__2)

2 ( 1__2)

4 + 6 C5 ( 1__

2)5 ( 1__2)

1 + 6 C6 ( 1__

2)6 ( 1__2)

0 + 3___

32

= 15___64

+ 3___32

+ 1___64

+ 3___32

= 7___16

≈ 0.4375

d. They are complements.

37. P(fewer than 3)

= 5 C0 (0.45)0 (0.55)5 + 5 C1 (0.45)1 (0.55)4

+ 5 C2 (0.45)2 (0.55)3

≈ 0.59

38. Possible answer: in 20 coin flips, the probability of getting fewer than 18 heads

39. P(2 or fewer)≈ 0.017 + 0.08 + 0.195≈ 0.3

40. Possible answer: ≈ 0.33; P(0 successes) ≈ 0.017. So, solve x10 = 0.017 to find the complement, ≈ 0.67, and subtract from 1.

TEST PREP

41. B. Binomial trials are independent.

42. H

2 C1 (0.4)1 (0.6)1

= 0.48

43. B. It matches the Binomial trial formula.

44. P(0 or 1 impefect parts)

= 10 C0 (0.04)0 (0.94)10 + 10 C0 (0.04)0 (0.94)10

≈ 0.94 = 94%

45. P(3 or more)= 1 - P(0, 1, or 2)

= 1 - 10 C0 (0.188)0 (0.812)10 + 10 C1 (0.188)1 (0.812)9

+ 10 C2 (0.188)2 (0.812)8

≈ 0.29


46a. 65; number of people × probability left-handed

b. standard deviation = √ npq

= √ 650(0.1)(0.9) ≈ 7.6485

So, n is approximately between 57.3515 and 72.6485

So, { n | 58 ≤ n ≤ 72 }.

47a. P(at least one 1 in 6 rolls)= 1 - P(zero 1s)

= 1 - 6 C0 ( 1__6)

0 ( 5__6)

6

≈ 0.67

b. P(at least two 1s in 12 rolls)= 1 - P(zero or one 1)

= 1 - 12 C0 ( 1__

6)0 ( 5__6)

12 + 12 C1 ( 1__

6)1 ( 5__6)

11 �

≈ 0.62

48. P(at least 4)= 1 - P(at most 3)

enter: 1 - binomcdf(20, 0.4, 3) ≈ 0.984

49. n C r + n C r + 1

= n!________r !(n - r )!

+ n!_________________ (r + 1)![n - (r + 1)]!

= n!________r !(n - r )!

+ n!________________ (r + 1)!(n - r - 1)!

= n!_________________ r !(n - r ) (n - r - 1)!

+ n!_________________ (r + 1)r ! (n - r - 1)!

= ( r + 1____r + 1)

n!_________________ r !(n - r )(n - r - 1)!

+

( n - r_____n - r )

n!________________ (r + 1)r !(n - r - 1)!

= n!(r + 1)

_____________________ (r + 1)!(n - r )(n - r - 1)!

+ n!(n - r)

_____________ (r + 1)r !(n - r )!

= n!(r + 1 + n - r)

_____________ (r + 1)!(n - r )!

= n!(n + 1)

____________ (r + 1)!(n - r )!

= (n + 1)!____________

(r + 1)!(n - r )!= n + 1 Cr + 1

50a. P(1) = 2 C1 p1 (1 - p)1

The equation is: 2p(1 - p) = 0.4

2p - 2 p2 = 0.4

-2(p2 - p + 0.2) = 0.

p = 1 ± √ 1 - 4(0.2)

______________2

≈ 0.72 or ≈ 0.28

So, p ≈ 0.72 or ≈ 0.28

b. P(2) = 2 C2 (0.72)2 (0.28)0

≈ 0.52 or ≈ 0.076 if p ≈ 0.28.


SPIRAL REVIEW

51. f(-3) = - (-3)2 + 2(-3) - 4 = -19

f(0) = - (0)2 + 2(0) - 4 = -4

f(2) = - (2)2 + 2(2) - 4 = -4

52. f(-3) = [-(-3)]2 - 3(-3) + 1 = 19

f(0) = [-(0)]2 - 3(0) + 1 = 1

f(0) = [-(2)]2 - 3(2) + 1 = -1

53. first difference: 1.2 1.2 1.2 1.2

No; y is a linear function of x.

54. first difference: 12 14 16 18

No; y is a quadratic function of x.

55. mean: 67___5

= 13.4

median: 15

mode: 18

56. mean: 45___6

= 7.5

median: 6 + 7_____

2= 6.5

mode: 6

57. mean: 150____6

= 25

median: 24

mode: 24

58. mean: 43___5

= 8.6

median: 8

mode: 5

READY TO GO ON? PAGE 845

1. mean: 3.4; median: 3; mode: 2

2. expected value

= 0(0.82) + 1(0.11) + 2(0.04) + 3(0.02) + 4(0.01)

= 0.29

3. 17, 18, 21, 22, 23, 25, 28, 45

minimum: 17 first quartile: 19.5

maximum: 45 third quartile: 26.5

median: 22.5 IQR: 7

4. mean: 23; standard deviation: ≈ 6.9856997

The lengths 16.01 in. to 29.99 in. are 1 standard

deviation away from the mean.

5. mean: 55; standard deviation: ≈ 54.3

6. Three standard deviations is about 3(54.3) = 162.9.

Negative values don’t make sense for this question,

and values greater than 217.9 are outliers.

Outlier: 280; the mean increases from 43.2 to 55,

and the standard deviation increases from ≈ 17.3 to

≈ 54.3.

7. (m - 2n)3

= 3 C0 m3 (-2n)

0 + 3 C1 m

2 (-2n)

1 + 3 C2 m

1 (-2n)

2

+ 3 C3 m0 (-2n)

3

= 1 · m3 + 3 · (-2m

2n) + 3 · 4m n

2 + 1 · (-8n

3)= m

3 - 6 m

2n + 12mn

2 - 8 n

3

8. P(5) = 10 C5 (0.25)5 (0.75)

5

≈ 0.058

9. P(at least 3)

= 1 - P(0, 1 or 2)

= 1 - 10 C0 (0.25)0 (0.75)

10 + 10 C1 (0.25)

1 (0.75)

9

+ 10 C2 (0.25)2 (0.75)

8

≈ 0.47

10. P(5) = 5 C5 ( 1__3)

5 ( 2__3)

0

= 1____243

≈ 0.004

11. P(1) = 5 C1 ( 1__3)

1 ( 2__3)

4

= 80____

243≈ 0.33

12. P(0) = 5 C0 ( 1__3)

0 ( 2__3)

5

= 32____

243≈ 0.13

13. P(at least 1)

= 1 - P(0)

= 1 - 32____

243

= 211____243

≈ 0.87

STUDY GUIDE: REVIEW, PAGES 848–851

1. dependent events 2. expected value

3. permutation

LESSON 11-1

4. 7 · 10 · 10 · 10 · 10 · 10 · 10 = 7,000,000

There are 7,000,000 possible different 7-digit

telephone numbers.

5. 12 C5 = 12!_________5!(12 - 5)!

= 12 · 11 · 10 · 9 · 8

_______________ 5 · 4 · 3 · 2 · 1

= 95,040______

120= 792

There are 792 ways to choose groups of 5.

6. 14 C6 = 14!________(14 - 6)!

= 14 · 13 · 12 · 11 · 10 · 9= 2,162,160

There are 2,162,160 ways to visit 6 companies.

7. 10 C7 = 10!________

(10 - 7)!= 10 · 9 · 8 · 7 · 6 · 5 · 4= 604,800

There are 604,800 ways to arrange 7 people into

the van.

8. 6 C3 = 6!________

3!(6 - 3)!

= 6 · 5 · 4_______3 · 2 · 1

= 120____6

= 20

There are 20 ways to choose 3 entrées.


LESSON 11-2

9. P(sum is 8) = 5___36

10. P(difference is 1) = 10___36

= 5___18

11. P(sum is even) = 18___36

= 1__2

12. P(product < 30) = 33___36

= 11___12

13. P(4 lowest grades) = 1_____ 10 C4

= 1____210

14. P(same 5 digits) = 10 _________________ 10 · 10 · 10 · 10 · 10

= 1______10,000

15. P(shaded) = shaded area ___________total area

= 1__2

(5)(8)______8(12)

= 20___96

= 5___24

16. P(not in circle) = area outside circle _______________ total area

= area of square - area of circle

_________________________ total area

= 10(10) - π (5)

2

_____________10(10)

= 100 - 25π

_________100

≈ 0.21

17. P(H, H) = 10___50

= 1__5

18. P(at least 1 T )= 1 - P(H, H)

= 1 - 1__5

= 4__5

19. P(no H) = 14___50

= 7___25

20. P(1 T) = 26___50

= 13___25

21. P(H, H) = 1__2

· 1__2

= 1__4

22. P(at least 1 T)

= 1 - P(H, H)

= 1 - 1__4

= 3__4

23. P(no H) = 1__2

· 1__2

= 1__4

24. P(1 T) = 2 · 1__4

= 1__2

LESSON 11-3

25. The result of any roll does not affect the probability of any other outcome.

P(3 doubles) = 6___36

· 6___36

· 6___36

= 1____216

26. Replacing the first pen means the occurrence of the first selection does not affect the probability of the second selection.

P(red, then blue) = 10___25

· 15___25

= 6___25

27. P(single | 30-35) = 22___42

= 11___21

28. P(66+ | married) = 4___52

= 1___13

29. P(married | 18-50) = 26___62

= 13___31

30. P(single and 18-34)= P(single) · P(18-34 | single)

= 47___99

· 14___47

= 14___99

LESSON 11-4

31. Each coupon offers only 1 discount.

32. P(10% � 15%) = P(10%) + P(15%)

= 1__3

+ 1__2

= 5__6

33. P(red � 5) = P(red) + P(5) - P(red � 5)

= 26___52

+ 4___52

- 2___52

= 7___13

34. P(club � heart) = P(club) + P(heart)

= 13___52

+ 13___52

= 1__2

35. P(passed � male)= P(passed) + P(male) - P(passed � male)

= 170____300

- 120____300

- 80____300

= 7___10

LESSON 11-5

36. mean: 5.4; median: 6; mode: 8

37. mean: 13. −

3 ; median: 13; mode: 12, 13, 15

38. expected value = 0(0.65) + 1(0.22) + 2(0.1) + 3(0.03)= 0.51

39. 29, 33, 33, 48, 50, 52, 65, 71, 83 minimum: 29 first quartile: 33 maximum: 83 third quartile: 68 median: 50 IQR: 35

40. − x = 7.5;

5 7 4 11 8

x - − x -2.5 -0.5 -3.5 3.5 0.5

(x - −

x)2 6.25 0.25 12.25 12.25 0.25

10 8 6 9 7

x - − x 2.5 0.5 -1.5 1.5 -0.5

(x - −

x)2 6.25 0.25 2.25 2.25 0.25

σ 2 = 43___

10= 4.3 σ ≈ 2.1

The number of wins within 1 standard deviation from the mean are between 5.4 and 9.6.

41. Yes; 3 standard deviations above the mean is 94.5. Her score is more than 3 standard deviations above the mean.

42. The mean decreases from 75.5 to 69.3, and the standard deviation increases from ≈ 21.5 to ≈ 25.1.


LESSON 11-6

43. (5 + 2x)3

= 3 C0 53 (2x)

0 + 3 C1 5

2 (2x)

1 + 3 C2 5

1 (2x)

2

+ 3 C3 50 (2x)

3

= 1 · 125 + 3 · 50x + 3 · 20 x2 + 1 · 8 x

3

= 125 + 150x + 60 x2 + 8 x

3

44. (x - 2y)4

= 4 C0 x4 (-2y)

0 + 4 C1 x

3 (-2y)

1 + 4 C2 x

2 (-2y)

2

+ 4 C3 x1 (-2y)

3 + 4 C4 x

0 (-2y)

4

= 1 · x4 + 4 · (-2x

3y) + 6 · 4 x

2 y

2 + 4 · (-8x y

3) + 1 · 16 y

4

= x4 - 8 x

3y + 24 x

2 y

2 - 32x y

3 + 16 y

4

45. expected value = 75 · 0.65

= 48.75

standard deviation = √ �� 75(0.65)(0.35)

≈ 4.13

46. P(3) = 8 C3 ( 1__6)

3 ( 5__6)

5

= 56 ( 1____216) (

31 25_____7776) ≈ 0.10

P(at least 2)

= 1 - P(0 or 1)

= 1 - 8 C0 ( 1__

6)0 ( 5__6)

8 + 8 C1 ( 1__

6)1 ( 5__6)

7

≈ 0.40

CHAPTER TEST, PAGE 852

1. 6 · 4 · 8 = 192

The mannequin can be dressed in 192 ways.

2. 8 P3 = 8!_______

(8 - 3)!= 8 · 7 · 6 = 336

There are 336 ways to award first, second, and third

places.

3. 30 C3 = 30!_________

3!(30 - 3)!

= 30 · 29 · 28

__________3 · 2 · 1

= 4060

4. P(4 jacks, queens, or kings)

= 3 · 4!_____ 52 P4

= 3_______

270,725≈ 0.000011

5. P(T, T) = 6___

20=

3___10

6. Replacing the first letter means that the occurrence

of the first selection does not affect the probability of

the second selection. The events are independent.

P(D, then J) = 1___26

· 1___26

= 1____676

7. Not replacing the vowel means that there will be

fewer vowels to chose from, affecting the probability

of the second and third selections. The events are

dependent.

P(vowel, then vowel, then vowel)

= 5___

26· 4___

25· 3___

24= 1____

260

8. P(C � even) = 1__9

+ 2__9

= 1__3

9. P(odd � multiple of 3)= P(odd) + P(multiple of 3) - P(odd multiple of 3)

= 3__9

+ 3__9

- 2__9

= 4__9

10. expected value

= 0 ( 7___20) + 1 ( 5___

20) + 2 ( 4___20) + 3 ( 3___

20) + 4 ( 1___20)

= 13___10

= 1.3

11. minimum: 0 first quartile: 0.5

maximum: 63 third quartile: 30

median: 2 IQR: 29.5

12. No; mean ≈ 15.6; standard deviation ≈ 20.5;

3 standard deviations above the mean ≈ 77.1,

and 77.1 > 63.


13. 12; the mean decreases from 104.4 to 96.7, and the

standard deviation increases from ≈ 10.6 to ≈ 27.5.

14. (3x + y)4

= 4 C0 (3x)4 y

0 + 4 C1 (3x)

3 y

1 + 4 C2 (3x)

2 y

2

+ 4 C3 (3x)1 y

3 + 4 C4 (3x)

0 y

4

= 1 · 81 x4 + 4 · 27 x

3y + 6 · 9 x

2 y

2 + 4 · 3x y

3 + 1 · y

4

= 81x4 + 108 x

3y + 54 x

2 y

2 + 12x y

3 + y

4

15. P(2) = 10 C2 (0.15)2 (0.85)

8

≈ 0.28

16. P(at least 2)

= 1 - P(0 or 1)

= 1 - 10 C0 (0.15)0 (0.85)

10 + 10 C1 (0.15)

1 (0.85)

9

≈ 0.46


chapter solutions key 11 probability and statistics · solutions key 11 probability and statistics...

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