MODULE Probability 18 Solutions Key -...

© Houghton Mifflin Harcourt Publishing Company 215 Holt McDougal Analytic Geometry

ProbabilitySolutions Key

18-1 GEOMETRIC PROBABILITY

CHECK IT OUT!

1. P ( −−

BD ) = P ( −−

BC ) + P ( −−

CD ) = 3 ___ 12

+ 5 ___ 12

= 8 ___ 12

= 2 __ 3

2. P(not red) = P(green or yellow)= P(green) + P(yellow)

= 25 ___ 60

+ 5 ___ 60

= 30 ___ 60

= 1 __ 2

3. P = 80 + 100 ________ 360

= 180 ____ 360

= 1 __ 2

4. Area of the triangle (which contains the circle) is ≈ 187 m 2 .Area of the trapezoid is 75 m 2 .Area of the rectangle is 900 m 2

P ≈ 900 - (187 + 75)

_______________ 900

= 638 ____ 900

≈ 0.71

THINK AND DISCUSS

1. In a geometric model, there are an infinite number of outcomes in each event.

2. Subtract 1 _ 2 and 1 _

3 from 1 to find what part of the

spinner is yellow.

3.

Lengthmodel:

Angle measuremodel:

Areamodel:

GeometricProbability

A B C

5 52

4

90°

135°

135° 21

EXERCISES

GUIDED PRACTICE

1. Possible answer: a spinner

2. P ( −−

XZ ) = P ( −−

XY ) + P ( −−

YZ ) = 5 ___ 10

+ 3 ___ 10

= 8 ___ 10

= 4 __ 5

3. P (not −−

XY ) = P ( −−−

WX ) + P ( −−

YZ ) = 2 ___ 10

+ 3 ___ 10

= 5 ___ 10

= 1 __ 2

4. P ( −−−

WX or −−

YZ ) = P ( −−−

WX ) + P ( −−

YZ ) = 1 __ 2

5. P ( −−−

WY ) = P ( −−−

WX ) + P ( −−

XY ) = 2 ___ 10

+ 5 ___ 10

= 7 ___ 10

6. P = 1.5 min _______ 10 min

= 0.15

7. P(wait < 3 min, once) = 3 + 1.5 _______ 10

= 0.45

In 20 times, expect to wait < 3 min, 0.45(20) = 9 times.

8. P = 45 ____ 360

= 1 __ 8

9. P = 45 + 90 _______ 360

= 135 ____ 360

= 3 __ 8

10. P = 360 - 120 _________ 360

= 240 ____ 360

= 2 __ 3

11. P = 60 + 90 _______ 360

= 150 ____ 360

= 5 ___ 12

12. Area of the triangle is A = 1 _ 2 (10)(10) = 50 ft 2

Area of the rectangle is A = (48)(24) = 1152 ft 2

P = 50 _____ 1152

≈ 0.04

13. Area of the trapezoid is A = 1 _ 2 (18 + 12)(6) = 90 ft 2

P = 90 _____ 1152

≈ 0.08

14. Area of the square is A = (10 ) 2 = 100 ft 2 P = 100 _____

1152 ≈ 0.09

15. The combined area of the smaller shapes is A = 50 + 90 + 100 = 240 ft 2 .

P = 1152 - 240 __________ 1152

≈ 0.79

PRACTICE AND PROBLEM SOLVING

16. HM = 16.4 + 21.9 + 15.3 + 14.8 = 68.4

P ( −−

JK ) = 21.9 ____ 68.4

≈ 0.32

17. P (not −−

LM ) = 68.4 - 14.8 __________ 68.4

≈ 0.78

18. P ( −−

HJ or −−

KL ) = P ( −−

HJ ) + P ( −−

KL )

= 16.4 ____ 68.4

+ 15.3 ____ 68.4

≈ 0.46

19. P (not −−

JK or −−

LM ) = P ( −−

HJ or −−

KL ) ≈ 0.46

20. P = 0.75 min ________ 15 min

= 0.05

21. P = 15 - (5 + 0.75)

_____________ 15

≈ 0.62

22. Assume that if the report has started, you must wait to hear the whole next one.P = 1 ___

15

In 50 times, expect to wait < 1 min, 1 ___ 15

(50) ≈ 3 times.

23. P(red) = 180 ____ 360

= 1 __ 2

24. P(yellow or blue) = 45 + 45 _______ 360

= 90 ____ 360

= 1 __ 4

25. P(not green) = 360 - 90 ________ 360

= 270 ____ 360

= 3 __ 4

26. P(red or green) = 180 + 90 ________ 360

= 270 ____ 360

= 3 __ 4

18MODULE

G_MGAESK893952_M18.indd 215G_MGAESK893952_M18.indd 215 4/9/12 6:25:16 PM4/9/12 6:25:16 PM


27. Area of the triangle is

A = 1 __ 2 ( 30 _____

√ � 3 /2 ) (30) = 1 __

2 (20 √ � 3 ) (30) = 300 √ � 3 m 2

Area of the rectangle is A = (20 √ � 3 ) (30) = 600 √ � 3 m 2

P = 300 √ � 3

______ 600 √ � 3

= 1 __ 2 or 0.5

28. Area of the square is A = (10 √ � 2 ) 2 = 200 m 2

P = 200 ______ 600 √ � 3

≈ 0.19

29. Area of the circle is A = π(10 ) 2 = 100π m 2 P = 100π - 200 __________

600 √ � 3 ≈ 0.11

30. The circle and square are inside the triangle, so the remaining area is A = 600 √ � 3 - 300 √ � 3 = 300 √ � 3 m 2

P = 300 √ � 3

______ 600 √ � 3

= 1 __ 2 or 0.5

31. The value in A is incorrect because the sectors have different angle measures, so they are not equally likely outcomes.

32. Area of rectangle: A = (15 - 2)(8 - 1) = 91 units 2 Area of triangle: A = (5)(4) - 1 _

2 (1)(4) - 1 _

2 (4)(2) - 1 _

2 (5)(2)

= 20 - 2 - 4 - 5 = 9 units 2 P = 9 ___

91 ≈ 0.10

33. Only half of � P lies inside ABCD. The area of the semicircle is A = 1 _

2 π(3 ) 2 = 4.5π units 2 .

P = 91 - 4.5π

_________ 91

≈ 0.84

34. Area of outer square: A = (10 ) 2 = 100 units 2 Area of parallelogram: A = (2)(3) = 6 units 2 P = 6 ____

100 = 0.06

35. Area of circle: A = π(2 ) 2 = 4π units 2 P = 4π

____ 100

≈ 0.13

36. Area of triangle is A = 1 _ 2 (3)(3) = 4.5 units 2

P = 4.5 + 4π

________ 100

≈ 0.17

37. P = 100 - (4.5 + 6 + 4π)

__________________ 100

≈ 0.77

38a. Area of central region is A = π(6.1 ) 2 cm 2 Area of target is A = π(61 ) 2 cm 2

P = π(6.1 ) 2

______ π(61 ) 2

= ( 6.1 ___ 61

) 2 = (0.1 ) 2 = 0.01

b. Inner radius of blue rings: r = 4(6.1) cmOuter radius of black rings: r = 8(6.1) cmArea of blue and black rings: A = π(8(6.1) ) 2 - π(4(6.1) ) 2

= π(6.1 ) 2 (64 - 16) = (48)π(6.1 ) 2 cm 2

P = 48π(6.1 ) 2

________ π(61 ) 2

= 48 ____ 100

= 0.48

c. Area of 5 inner rings: A = π(5(6.1) ) 2 = 25π(6.1 ) 2 cm 2

P = 25π(6.1 ) 2

________ π(61 ) 2

= 25 ____ 100

= 0.25

d. The probabilities might be different because an archer would be aiming for the center, not a random point.

39. Possible answer: The point lies on −−

AC .

40. Possible answer: The point lies in the red or yellow region.

41. Possible answer: The point lies in the blue or green triangle.

42a. Area of blue parallelogram: A = (2)(1) = 2 units 2 Area of tangram: A = (4 ) 2 = 16 units 2

P = 2 __ 16

= 1 _ 8

b. Area of purple triangle: A = 1 _ 2 (2)(2) = 2 units 2

P = 2 __ 16

= 1 _ 8

c. Area of large yellow triangle: A = 1 _

2 (4)(2) = 4 units 2

P = 4 __ 16

= 1 _ 4

d. No, because areas are the same.

43. P = 4 _ 8 = 1 _

2 ; it does not matter which regions are

shaded because they all have the same area.

44a. Area of each balloon: A = π(1.5 ) 2 = 2.25π in. 2 Area of board: A = (50)(30) = 1500 in. 2 For 40 balloons,

P = 40(2.25π)

_________ 1500

≈ 0.19

b. For n balloons, if probability is ≥ 0.25,

P = n(2.25π)

________ 1500

≥ 0.25

n ≥ 1500 _____ 2.25π

(0.25) ≈ 53.1

n ≥ 54 balloons

TEST PREP

45. AP =

2(1.5) _____

6(3.5) = 3 ___

21 ≈ 0.14

46. GP (

−− AB ) = 18 _______

18 + 24 = 18 ___

42 ≈ 0.43

47. DArea of triangle: A = 1 _

2 (10)(20) = 100 m 2

Area of circle: A = π(20 ) 2 = 400π m 2 Area of square: A = (25 ) 2 = 625 m 2 Area of field: A = 100(70) = 7000 m 2

P = 7000 - (100 + 400π + 625)

_______________________ 7000

≈ 0.717



48a. Let P(r), P(b), P(g) be the probabilities of each color. From the given info:P(r) = 2P(b), P(g) = P(b)Substitute into this eqation: P(r) + P(b) + P(g) = 12P(b) + P(b) + P(b) = 1 4P(b) = 1 P(b) = 1 _

4

P(g) = P(b) = 1 _ 4 or 0.25

b. 3; the probability of landing on green is 0.25, so the number of green regions is 0.25(12) = 3.

CHALLENGE AND EXTEND

49. Area of each red region: A = 1 2 - 4 ( 1 _

4 π(0.5 ) 2 ) = 1 - 0.25π

P = 1 - 0.25π

_________ 1 = 4 - π

_____

4 ≈ 0.21

50. P = s 2 _______ (18)(24)

= s 2 ____ 432

= 1 __ 3

s 2 = 144 s = 12 ft The square will be 12 ft by 12 ft.

P = s 2 ____ 432

= 3 __ 4

s 2 = 324 s = 18 ft

The square will be 18 ft by 18 ft.

51. Possible answer: The probabilities must add to 1, so P(yellow) + P(blue) + P(red) = 1. I would make the regions different sizes, and I would want each region to be worth more points the smaller it is. The point value for red is 6 times the point value for yellow, so I would make 6 · P(red) = P(yellow).

The point value for blue is 3 times the point value for yellow, so I would make 3 · P(blue) = P(yellow).

Then P(yellow) + 1 __ 3 P(yellow) + 1 __

6 P(yellow) = 1.

This means P(yellow) = 2 __ 3 , P(blue) = 2 __

9 ,

and P(red) = 1 __ 9 .

The angle measure for the yellow region would be 240°, for the blue region would be 80°, and for the red region would be 40°.

FOCUS ON MATHEMATICAL PRACTICES

52a. Since the entire chip must land in the inner circle, the center of the chip must be at least

r chip = d chip

_ 2 = 3 _

2 =1.5 in. from the boundary of the

inner circle. Therefore the center of the chip must land within

r circle - 1.5 = d circle

_ 2 - 1.5 = 12 _

2 - 1.5 = 4.5 in.

of the center of the target.

b. P = π(4.5) 2

_______ π ( 36 _

2 )

2 = ( 4.5 ___

18 )

2 = 1 ___

16 = 0.0625

53a. P( _

AB ) = P( _

AC ) - P( _

BC ) = 0.8 - 0.5 = 0.3

Note that point B must be between points

A and C, else P( _

AB ) would be

P( _

AC ) + P( _

BC ) = 0.8 + 0.5 = 1.3,

which does not make sense in this context.

b. P( _

CD ) = P( _

AD ) - P( _

AC ) = 1 - 0.8 = 0.2

18-2 PERMUTATIONS AND COMBINATIONS

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 P 3 = 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 P 2 = 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 C 2 = 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. Fundamental

Counting Principle Permutation

Formula

Examples 5 shirts × 3 skirts × 4 pairs of shoes

= 5 × 3 × 4 outfits

Permutations of 8 items taken 3 at a time

3 items chosen from 8

Combination

! _____ ( - )!

! ______ !( - )! 1

� 2

� … �

8! ______ (8 - 3)!

= 8! __ 5!

= 336 8! _______ 3!(8 - 3)!

= 8! ____

3!5! = 56



EXERCISESGUIDED 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 P 2 = 7! _______ (7 - 2)!

= 7! __ 5!

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

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

= 12! ___ 9!

= 12 · 11 · 10 = 1320 There are 1320 ways to listen to 3 songs.

6. 6 P 3 = 6! _______ (6 - 3)!

= 6! __ 3!

= 6 · 5 · 4 = 120 There are 120 ways the prizes can be awarded.

7. 21 C 4 = 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 C 3 = 5! ________ 3!(5 - 3)!

= 5! ____ 3!2!

= 5 · 4 ____ 2 · 1

= 10

There are 10 ways to choose 3 boxes of cereal.


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 P 2 = 9! _______ (9 - 2)!

= 9! __ 7!

= 9 · 8 = 72 There are 72 ways to choose a manager and an

assistant.

12. 26 P 3 = 26! ________ (26 - 3)!

= 26! ___ 23!

= 26 · 25 · 24 = 15,600 There are 15,600 possible identification codes.

13. 5 P 2 = 5! _______ (5 - 2)!

= 5! __ 3!

= 5 · 4 = 20 There are 20 ways to assign 2 planes to the

runways.

14. 6 C 3 = 6! ________ 3!(6 - 3)!

= 6! ____ 3!3!

= 6 · 5 · 4 _______ 3 · 2 · 1

= 20

There are 20 choices of 3 hamburger toppings.

15. 49 C 7 - 49 C 6

= 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 P 6 = 6! _______ (6 - 6)!

= 6! __ 0!

= 6 · 5 · 4 · 3 · 2 · 1 = 720

17. 5 C 5 = 5! ________ 5!(5 - 5)!

= 5! ____ 5!0!

= 1

18. 9 P 1 = 9! _______ (9 - 1)!

= 9! __ 8!

= 9

19. 6 C 1 = 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 C 2 = 6! ________ 2!(6 - 2)!

= 6! ____ 2!4!

= 6 · 5 ____ 2 · 1

= 15

25. 7 C 4 = 7! ________ 4!(7 - 4)!

= 7! ____ 4!3!

= 7 · 6 · 5 _______ 3 · 2 · 1

= 35

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

= 7! __ 4!

= 7 · 6 · 5 = 210

7 C 4 = 7! ________ 4!(7 - 4)!

= 7! ____ 4!3!

= 7 · 6 · 5 _______ 3 · 2 · 1

= 35

Therefore, 7 P 3 > 7 C 4 .

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

= 7! __ 3!

= 7 · 6 · 5 · 4 = 840

7 C 3 = 7! ________ 3!(7 - 3)!

= 7! ____ 3!4!

= 7 · 6 · 5 _______ 3 · 2 · 1

= 35

Therefore, 7 P 4 > 7 C 3 .



28. 7 C 3 = 7! ________ 3!(7 - 3)!

= 7! ____ 3!4!

= 7 · 6 · 5 _______ 3 · 2 · 1

= 35

7 C 4 = 7! ________ 4!(7 - 4)!

= 7! ____ 4!3!

= 7 · 6 · 5 _______ 3 · 2 · 1

= 35

Therefore, 7 C 3 = 7 C 4 .

29. 10 C 10 = 10! ___________ 10!(10 - 10)!

= 10! _____ 10!0!

= 1

10 P 10 = 10! _________ (10 - 10)!

= 10! ___ 0!

= 3,628,800

Therefore, 10 C 10 < 10 P 10 .

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 P 3

____ 6 C 3

= 3! = 6; 6 C 3 = 6 P 3

____ 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 C 2 = 9! ________ 2!(9 - 2)!

= 9! ____ 2!7!

= 9 · 8 ____ 2 · 1

= 36;

9 C 7 = 9! ________ 7!(9 - 7)!

= 9! ____ 7!2!

= 9 · 8 ____ 2 · 1

= 36;

10 C 6 = 10! _________ 6!(10 - 6)!

= 10! ____ 6!4!

= 10 · 9 · 8 · 7 ___________ 4 · 3 · 2 · 1

= 210;

10 C 4 = 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 C 3 = 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 C 5 = 14! _________

5!(14 - 5)! = 14! ____

5!9!

40. J 9 C 4 = 9! ________

4!(9 - 4)! = 9! ____

4!5!

9 C 5 = 9! ________ 5!(9 - 5)!

= 9! ____ 5!4!

41. 15 C 4 = 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.


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 C 2

b. 20 points: 20 C 2 = 20! _________ 2!(20 - 2)!

= 20! _____ 2!18!

= 20 · 19 ______ 2 · 1

= 190

43. Select 12 jurors out of 30 potential jurors, 30 C 12 . Select 2 alternate jurors out of the remaining 18 potential jurors, 18 C 2 . Use the Fundamental Counting Principle to combine the number of ways the jurors can be selected, ( 30 C 12 ) ( 18 C 2 ).




44. 6 P 4 = 6! _______

(6 - 4)! = 6 · 5 · 4 · 3 · 2 · 1 ______________

2 · 1 = 6 · 5 · 4 · 3

= 360

6 C 4 = 6! ________

4!(6 - 4)! = 6 · 5 · 4 · 3 · 2 · 1 ______________

4 · 3 · 2 · 1 · 2 · 1 = 6 · 5 ____

2 · 1 = 15

15 ____ 360

≈ 0.0417

The number of combinations is about 4.17% of the number of permutations.

45. Possible answer:

5 P 1 = 5! _______

(5 - 1)! = 5 · 4 · 3 · 2 · 1 ____________

4 · 3 · 2 · 1 = 5 and

5 C 1 = 5! ________

1!(5 - 1)! = 5 · 4 · 3 · 2 · 1 ____________

1 · 4 · 3 · 2 · 1 = 5, so in this

case x P y = x C y .

18-3 THEORETICAL AND EXPERIMENTAL PROBABILITY

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 C 2 = 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 A t = 1 __ 2 (15)(15) = 112.5.

Area of small triangle is A s = 1 __ 2 (4)(4) = 8.

A s ___ A t

= 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. ExperimentalThe experimental probability of rolling a 5on a number cube that was rolled 30 timesand landed on 5 7 times is

TheoreticalThere are 6 equally-likely outcomes when

rolling a number cube, so each hasa theoretical probability of

ComplementThe complement of the experimentalprobability above is 1 - , or , which is the experimental probability of not rolling a 5.

GeometricIf a circle of radius 3 is inside a square

with side length 10 and any point insidethe square is equally likely, the probability

of a random point being in the circle is

Probability7 __ 30

π (3) 2 ____ 10 2 ≈ 0.28

7 __ 30

23 __ 30

1 __ 6

.


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

3. There are 8 possible outcomes, and 2 outcomes where the penny and the nickel both show heads.

P (penny and nickel show heads) = 2 __ 8 = 1 __

4

4. There are 8 possible outcomes, and 3 outcomes where 1 coin shows heads.

P (one coin shows heads) = 3 __ 8

5. There are 8 possible outcomes, and 2 outcomes 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. The number of possible outcomes is

4 P 4 = 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

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



The number of possible outcomes is

12 C 3 = 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 A t = π (6) 2 = 36π

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

Area of small circle is A u = π (2) 2 = 4π

Shaded area is A s = 16π - 4π = 12π

A s ___ A t

= 12π

____ 36π

= 1 __ 3

The probability that a point is in the shaded area is 1 __

3 .

11. A u

___ A t

= 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

15. There are 15 possible outcomes, and 12 outcomes of a red or white marble.

P (red or white marble) = 12 ___ 15

= 4 __ 5

16. There are 64 possible outcomes. 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 C 3 = 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 C 2 = 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 A t = 24(48) = 1152 Area of hole is A s = π (3) 2 = 9π

A s ___ A t

= 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 A c = π r 2 Area of square is A s = (2r) 2 = 4 r 2

A c ___ A s

= 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 = √ �� 2 x 2 r = √ � 2 x

A outer = π ( √ � 2 x) 2 = 2π x 2

A inner = π x 2

A inner _____ A outer

= 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 A large = 8(16) = 128 A small = 5(14) = 70

A shaded

_______ A large

= 128 - 70 ________ 128

= 58 ____ 128

≈ 45%

38. C2 · 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.

42. There are 24 possible outcomes. P (no one gets the right trumpet) = 8 ___

24 = 3 __

8

43. There were 100 experiments.

B

Y

B

R

R

G


44. Let r = radius of a small circle

Area of a small circle = πr 2 Area of 3 small circles = 3πr 2

Area of large circle = π(3r ) 2 = 9πr 2

Since the 3 small circles cover one third of the area of the target, and Yves has a 1 – 0.25 = 0.75 chance of hitting the target, he has a 0.75 ÷ 3 = 0.25 chance of hitting one of the small circles.

45. Possible answer: The numbers 5 and 6 came up about twice as often as the other numbers. It is unlikely that the cube is fair, because it would be rare to get these results by chance.


Conditional ProbabilitySolutions Key

19-1 INDEPENDENT AND DEPENDENT EVENTS

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.

Similarities Probabilities are multiplied using the

Fundamental Counting Principle. Probability within [0, 1] range

Differences Probability of a previous event does

not affect probability for independent events but does for dependent events.

Probability of Independent Events vs. Probability of Dependent Events


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 __

3

P (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

19MODULE



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.

0.1

0.9

0.88

0.12

0.05

0.95

AA

AP

PA

PP

A

P

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

HaveStrep

Do NotHave Strep

Total

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


39. Since the tiles are drawn without replacement, they are dependent events, but Topher made the calculation as if they were independent;

12 ___ 28

· 11 ___ 27

· 10 ___ 26

= 55 ____ 819

≈ 0.0672.

40. The probability that it is not cloudy in the morning is 1 – 0.4 = 0.6. If it is not cloudy, the probability that it rains is 1 – 0.92 = 0.08.

(0.06)(0.08) = 0.048

41. Possible answer: No; the probability that the number on the tile is greater than the number rolled.



19-2 TWO-WAY TABLES

CHECK IT OUT!

1. Find the total number of books sold: 28 + 52 + 94 + 36 = 210. Divide each value in the

table by 210 to find the joint relative frequencies, and add each row and column to find the marginal relative frequencies.

Fiction Nonfiction Total

Hardcover 0.133 0.248 0.381Paperback 0.448 0.171 0.619Total 0.581 0.419 1

2a. Find the total enrollment:

38 + 52 + 86 + 24 = 200. Divide each value in the table by 200, and add each row and column.

BalletYes No Total

TapYes 0.19 0.26 0.45No 0.43 0.12 0.55Total 0.62 0.38 1

b. Taking ballet: 0.62; of these, also not taking tap: 0.43.

0.43 ____ 0.62

= 0.69

3. Create a table of joint and marginal relative frequencies:

Pass Fail Total

Al’s Driving 0.28 0.16 0.44Drive Time 0.22 0.14 0.36Crash Course 0.1 0.1 0.2Total 0.6 0.4

P (passing at Al’s Driving) = 0.28 ____ 0.44

≈ 0.64

P (passing at Drive Time) = 0.22 ____ 0.36

≈ 0.61

P (passing at Crash Course) = 0.1 ___ 0.2

= 0.5

Al’s Driving School is best.

THINK AND DISCUSS

1. Joint relative frequencies are the values in each category divided by the total number of values, while marginal relative frequencies are found by adding the joint relative frequencies in each column and row.

2. To find a conditional relative frequency, divide a joint relative frequency in the two-way table by the marginal relative frequency in that row or column.

3. Relative Frequencies

Joint Marginal

Divide each value by the total number of values.

Add the joint relative frequencies in each row and column.

Divide each joint relative frequency in the two-way table by the marginal relative frequency in that row or column.

Conditional

EXCERCISESGUIDED PRACTICE

1. marginal 2. conditional

3. The joint relative frequencies are determined by dividing the value in each cell in the table by the total number of people surveyed. There were 50 people surveyed, so each value is divided by 50.

UnderClassmates

UpperClassmates Total

Morning 0.16 0.28 0.44Afternoon 0.36 0.2 0.56Total 0.52 0.48 1

The marginal relative frequencies, recorded in the row and column labeled Total, are found by adding the joint relative frequencies in each row and column.

4. The joint relative frequencies are determined by dividing the value in each cell in the table by the total number of people surveyed. There were 150 people surveyed, so each value is divided by 150.

Arlington Towson Parkville Total

Yes 0.27 0.23 0.27 0.77No 0.12 0.07 0.04 0.23Total 0.39 0.3 0.31 1


5a. The joint relative frequencies are determined by dividing the value in each cell in the table by the total number of sophomores who play an instrument or a sport. Data was collected on 203 people, so each value is divided by 203.

Play Sport

Yes No Total

Play Instrument

Yes 0.23 0.19 0.42No 0.25 0.33 0.58Total 0.48 0.52 1



b. There are 47 + 38 = 85 students that play an instrument. Of the students, 47 out of 85 also play a sport.

P (student that plays an instrument plays a sport) = 47 ÷ 85 = 0.55

c. There are 47 + 51 = 98 students that play a sport. Of the students, 47 of 98 also play an instrument. P(student that plays a sport plays an instrument) = 47 ÷ 98 = 0.48

6a. The joint relative frequencies are determined by dividing the value in each cell in the table by the total number of attempted sales. Data was collected on 36 attempted sales, so each value is divided by 36.

Successful Unsuccessful TotalBecky 0.17 0.17 0.33Raul 0.11 0.14 0.25Darrell 0.17 0.25 0.42Total 0.45 0.56 1

b. The probability of a salesman being successful is determined by dividing the total number of attempts by the total number of salesP(Becky being successful) = 6 ÷ 12 = 0.50P(Raul being successful) = 4 ÷ 9 = 0.44P(Darrell being successful) = 6 ÷ 15 = 0.40

c. Becky has the highest success rate.


7. The joint relative frequencies are determined by dividing the value in each cell in the table by the total number of shirts and sweatshirts sold. There were 60 items of clothing sold, so each value is divided by 60.

Students Adults TotalT-Shirts 0.267 0.383 0.65Sweatshirts 0.117 0.233 0.35Total 0.384 0.616 1


8. Possible answer: First find the total number of individuals in the survey or study by adding all the values in the two-way table. Next find the joint relative frequencies by dividing each cell in the two way table by the total number of individuals in the survey. Record these results in a new table. Finally, add the rows and columns in this new table to find the marginal relative frequencies.

9a. The joint relative frequencies are determined by dividing the value in each cell in the table by the total number of customers that were surveyed. There were 118 customers surveyed, so each value is divided by 118.

Satisfied Dissatisfied Total

Team 1 0.17 0.07 0.24Team 2 0.29 0.1 0.39Team 3 0.29 0.08 0.37Total 0.75 0.25 1

b. To determine the probability that a customer will be satisfied after working with a particular team, take the number of the customers that were satisfied after working with the team and divide by the total number of customers that worked with the team.P(Being satisfied with Team 1) = 20 ÷ 28 = 0.71P(Being satisfied with Team 2) = 34 ÷ 46 = 0.74P(Being satisfied with Team 3) = 34 ÷ 44 = 0.77

c. Team C has the highest rate of customer satisfaction.

10. The value is always equal to 1. It represents the portion of the people in the survey who are in the survey, so it must equal 1.

11. Maria has made an error. If you add up all the joint relative frequencies in her table the sum is 1.1. The sum should be 1. The sum of the joint relative frequencies in Brennan’s table is 1.

12. 0.48 is a little less than half. A little less than half of the 107 brownies and muffins is around 50.

13a. The joint relative frequencies are determined by dividing the value in each cell in the table by the total number of questions asked. There were 120 questions asked, so each value should be divided by 120.

Work less than 5 miles from home?

Yes No TotalUse new system?

Yes 0.2 0.27 0.47No 0.37 0.17 0.54Total 0.57 0.44 1

b. P(Works close to home would use the new system) = 24 ÷ 68 = 0.35

c. P(Would use the system lives far from home) = 32 ÷ 56 = 0.57



TEST PREP

14. B; there were 9 + 14 = 23 teachers polled.

15. Use a table to find the joint and marginal relative frequencies:

Yes No TotalJunior 0.223 0.313 0.536Senior 0.167 0.298 0.464Total 0.390 0.610 1

Inspecting the table shows that C is correct.

16. 0.25 + x = 0.3 x = 0.05


17. To find the marginal relative frequencies, add the rows and columns in the joint relative frequency table.

Yes No TotalChildren 0.125 0.1 0.225Teenagers 0.725 0.05 0.775Total 0.85 0.15 1

Marginal relative frequencies are in the row and column labeled Total.

18. P(teenager purchasing a ticket book) = 0.725 ÷ 0.775 = 0.94 = 94%

19. According to the table 12.5% of the fair attendees are children who will buy a ticket at the entrance. Because 12.5% of the 80 teenagers and children who attend the fair equals 10, 10 children will buy a ticket at the entrance.

20. According to the table, 10% of the fair attendees are children who did not buy a ticket at the entrance. According to the problem, 10% of the total fair goers is equal to 12. 10% of 120 is 12, so there are 120 children and teenagers attending the fair.

21. The total of the marginal relative frequencies is always 1, so 1 - 1 = 0.

22. The maximum is 1. It can’t be higher because that would represent more than 100% of the data.


23. Antoine found P(large | green) instead of

P(green | large); 2 _____ 2 + 4

= 2 __ 6 = 1 __

3 .

24. It is possible. One way to start is to find the overall percent of people who are women and like classical music (0.25 · 0.44 = 0.11), then find the total percent of men (1 – 0.44 = 0.56) and the overall percent of people who are men and like classical music (0.21 – 0.11 = 0.1). The remaining categories can easily be filled using subtraction:

Male Female TotalYes 0.1 0.11 0.21No 0.46 0.33 0.79

Total 0.56 0.44 1

19-3 COMPOUND EVENTS

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 P 5

_____ 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.

Multiplying Probabilities rolling two 2’s in a row

on a number cube

Compound Events rolling exactly one 5

on two rolls of a number cube

Inclusive Events drawing a king or a diamond

from a deck of cards

Adding Probabilities rolling a 2 or a 3

on a number cube

Mutually Exclusive Events rolling a 2 or a 3

on a number cube

Probabilities




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 P 6

____ 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 C 4

_____ 24 C 4

+ 14 C 4

_____ 24 C 4

≈ 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 P 10

______ 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


42. P(art � theater) = P(art) + P(theater) - P(art � theater) = 0.10 + 0.15 - 0.19 = 0.06

43. Possible answer: 1. List all 36 possible outcomes for two rolls and count how many have at least one 5. 2. Add the probability of the first roll being a 5 to the probability of the second roll being a 5, and then subtract the probability of rolling two 5s.

44. The greatest possible value for P(A � B) is P(A), if B occurs whenever A occurs. Therefore, the smallest possible value forP(A � B) = P(A) + P(B) - P(A � B) is 0.3 + 0.4 - 0.3 = 0.4.

45a. Since there is no overlapping area between pink and orange, no student chose that combination.

b. P(orange � blue) = P(orange) + P(blue) - P(blue � orange) =

6 + 5 _____ 30

+ 5 + 12 + 2 __________ 30

- 5 ___ 30

= 25 ___ 30

= 5 __ 6


G_MGAESK893952_M19.indd 231G_MGAESK893952_M19.indd 231 4/11/12 1:52:16 AM4/11/12 1:52:16 AM

MODULE Probability 18 Solutions Key -...

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