Unit 2 Chapter 15 Answers

8/10/2019 Unit 2 Chapter 15 Answers

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Unit 2 Answers: Chapter 15 Macmillan Publishers Limited 2013

Chapter 15 Probability

Try these 15.1

(a)1 2 3 4 5 6

1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12

S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}n (1) = 1

(b) S = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T), (5, T), (6, T)}n (S ) = 12

(c) S = {RG, RRG, RRRG, RRRRG}n(S ) = 4

Exercise 15A

1

D1 1 1P(H and K )2 4 8

= =

2


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(a) P(three heads) 0.4 0.4 0.4= (b) P(two tails and a head)

P(HTT) P(TTH) P(THT)= + + + (0.4 0.6 0.6) (3)=

= 0.432(c) P(at least one tail)

= 1 P(no tails)= 1 0.064 = 0.936

3 (a)

(b) (i) 3 3 9P(two blue balls)8 8 64

= =

(ii) nd P(blue ball in his 2 draw P(BB) P(WB)= + 9 5 364 8 8

= +

9 15 24 364 64 64 8

= + = =

(iii) nd stP(white on 2 | white on1 )

nd st

st

5 5P(white on 2 white on 1 ) 58 8

5 8P(white on1 )8

= = =

4 (a)

(b) (i) 2 4 1P(both red)10 12 15

= =

(ii) 8 8 4 2 8P(both green)10 12 5 3 15

= = =

(iii) P(one red and one green)


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2 8 8 410 12 10 12

= +

1 2 4 1 6 25 3 5 3 15 5

= + = =

(iv) P(at least one is red) = P(one red) + P(2 red)2 1 75 15 15

= + =

5 (a)

(b) (i) P(at least one blue) = P(one blue) + P(two blue)4 2 2 4 2 26 6 6 6 6 6

= + +

20 536 9

= =

(ii) 4 2 2 4 16 4P(one red and one blue)6 6 6 6 36 9

= + = =

(iii) 4 4 2 2 20 5P(two of the same colour) P(two red) P(two blue)6 6 6 6 36 9

= + = + = =

(iv)nd st

nd stst

P(blue on 2 red on 1 )P(blue on 2 draw| red on1 draw)P(red on1 )

=

4 22 16 6

4 6 36

= = =

6 P(X) = 0.4, P(Y) = 0.5(a) (i) P(X Y) P(X) P(Y), since the events are independent= 0.4 0.5= 0.20

(ii) P(X Y) = P(X) + P(Y) P(X Y)= 0.4 + 0.5 0.20= 0.70

(b) X and Y are not mutually exclusive since P(X Y) 0

7 1P ( | )5

A B = , 1P( )8

B = , 1P( )6

A =

(a) P( ) P( )P( | ) P( | )1P( )6

A B A B B A B A

A

= =


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3 1 5P( )7 4 28

F G = =

(b) P( )P( | )P( )F G

F GG

=

51 283 P( )G

=

5P( ) 328

G =

1528

=

(c) P( )P( | )P( )G F

G F F

=

52837

=

528

= 7

4

3

512

=

12 In favour Against

Male 145 255

Female 40 160

(a) 145 40 185 37P(in favour)600 600 120

+= = =

(b)

40P(in f avour female) 40 1600P(in favour | female) 200P(female) 200 5

600

= = = =

(c) 255 17P(male and against)600 40

= =

(d) 185 200 40 23P(in favour or female)600 600 600 40

= + =

(e) 400P(male)600

=

185P(in favour)600

=

145 29P(male and in favour)600 120

= =

400 185 37P(male) P(in favour)600 600 180

= =

Since P(male and in favour ) P(male) P(in favour) Events are not independent


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13 (a) 65 40 33 69P(male)260 130

+ += =

(b)

40P(male Nissan) 40 10260

P(male | Nissan) 76P(Nissan) 76 19260

= = = =

(c) 31P(female who purchased a Honda)260

=

(d) P(male who purchased a Toyota or a female who purchased a Honda) 65 31 24260 260 65

= + =

14 (a) P(< 10 000 per month) 40 45 35 30 150 15260 260 26

+ + += = =

(b) 50 5P(male and earns at least10000)260 26

= =

(c) P(female or earns less than 8000) 125 75 35 165 33260 260 260 260 52

= + = =

15 News Movies

Male 110 150Female 90 50

(a) P(male or prefer news) 260 200 110400 400 400

= +

350 7

400 8= =

(b) P(male or female who prefers watching movies)260 50 310 31400 400 400 40

= + = =

16 (a) (i) P(prefers watching comedy) 24 85 109250 250

+= =

(ii) P(female and prefers comedy) 85 17250 50

= =

(iii) P(prefers watching action or male) 141 120 96 33250 250 250 50

= + =

(b) P(female) 130250=

P(watching action) 141250

=

P(female and watching action) 45250

=

P(female) and P(action) 130 141 1833 45250 250 6250 250

= =

Not independentFor mutually exclusive P(A B) = 0 since P(female and watching action) 0 Not mutually exclusive


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Try these 15.2

(a) 43 792Total no. of arrangements = 5! = 120 No. of odd arrangements (with 3, 7 or 9 in the final position)

4 3 2 1 3 = 72

P(an odd arrangement)72

120=

35

=

(b) CALALOO

(i) Total no. of arrangements = 7! 6302! 2! 2!

=

No. of arrangements beginning and ending with O = 2 5! 1 302! 2! 2!

=

Required probability30 1630 21

= =

(ii) No. of arrangements with the Ls together 6! 1802!2!

= =

Required probability180 2630 7

= =

(iii) No. of arrangement with Os and As next to each other4! 4!

722! 2! 2!= =

Required probability72 4

630 35= =

Exercise 15B

1 (a) Number of arrangements 10! 907 2002! 2!

= =

(b) Number of arrangements beginning with O9!1 9! 90720

2! 2!= =

P(arrangement begins with O) = 90 720 1907 200 10

=

2 (a) Number of arrangements = 6! = 720(b) Number of arrangements beginning with C and ending with E 1 4 3 2 11 24= =

Required probability 24 1720 30

= =

3 (a) Total Number of arrangements 8! 201602!

= =

Number of arrangement with 2 Ss together = 7! = 5040


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P(2 Ss together) 5040 120 160 4

= =

(b) Number of arrangements with 2 Ss separate = 20 160 5040= 15 120

P(2 Ss separate) 15 120 320 160 4

= =

4 Number of arrangements starting and ending with E 3 7 6 5 4 3 2 1 2 50403!

= =

Total number of arrangements 9! 604803!

= =

P(start and end with E) 5040 160 480 12

= =

5 (a) 7! 25202!

=

(b) Number of arrangements starting and ending with I2 5 4 3 2 1 1 120

2!= =


= =

6 (a) 6! 3602!

=

(b) Number of odd numbers 5 4 3 2 1 3 1802!

= =

P(odd number) 180 1

360 2= =

7 (a) P(all the men stand next to each other) 7! 5! 111! 66

= =

(b) P(all the women stand next to each other)6! 6! 1

11! 77

= =

(c) P(no two women stand next to each other)W M W M W M W M W M W

6 5 5 4 4 3 3 2 2 1 1 111! 462

= =

8 SCANDALOUS has 10 letters including 2 As and 2 Ss(a) Total Number of arrangements 10! 907 200

2!2!= =

Number of arrangements with all vowels together7! 4! 15 1202!2!

= =

Required probability 15 120 1907 200 60

= =

(b) Number of arrangements starting and ending with A2 8 7 6 5 4 3 2 1 1 20 160

2! 2!

= =


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= =

(c) SS AA CN DLOUTreat the two As as one unit and the two Ss as one unitAlso treat the two As together with the two Ss as one unit Number of arrangements with As and Ss together = 7! 2!

Required probability 7! 2! 1907 200 90

= =

(d) Number of arrangements ending with A9! 2

181 4402! 2!

= =

Number of arrangements ending with D 9! 1 90 7202!2!

= =

Number of arrangements ending with A or D = 181 440 + 90 720 = 272 160

Required probability272 160 3907 200 10= =

9 SSS TTT II AC

Total Number of arrangements 10!3!3!2!

= = 50 400

(a) Number of arrangements beginning and ending with S3 8! 2

33603!3!2!

= =

Required probability 3360 150 400 15

= =

(b) Number of arrangements with all the Ts together8!

33603!2!= =


= =

(c) Number of arrangements beginning with I 2 9!3! 3! 2!

=

= 10 080


= =

10 (a) Total number of arrangements = 3 3 3 = 27 Number of odd arrangements greater than 300 = 2 3 3 = 18

Probability 18 227 3

= =

(b) Total No.of arrangements with no repetition = 3! = 6 Numbers greater than 300 must start with 3 or 5, an odd number must end in 3 or 5,therefore the middle number is 23 2 55 2 3


= =

Exercise 15C


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1 5M 2W 1B

(a) Total number of choices = 8C 4 = 70

Number of choices with the boy = 1C 1 7C 3 = 35

Probability that selection includes the boy 35 170 2

= =

(b) Number of choices with exactly 3 men = 5C 3 3 C 1 = 30

Probability that the selection includes exactly 3 men 30 370 7

= =

2 R N D M A O

Number of ways of choosing 3 consonantsP(all three are consonants) total number of ways of choosing 3 letters=

43

63

4 120 5

C

C = = =

3 T R P Z M A E I U

P(all vowels are chosen)4

49

4

1126

C

C = =

4

(a) Total number of choices = 8C 2 = 28

Number of ways of choosing two boys = 4C 2 = 6

P(choosing two boys) 6 328 14

= =

(b) Number of ways of choosing a boy and a girl = 4C 1 4C 1 = 16

P(choosing a boy and a girl) 16 428 7

= =

5 9 boys and 4 girls

or

9 boys, 3 girls and Amanda

Possibilities with at least one girl are 1G 3B, 2G 2B, 3G 1B, 4G

Number of choices with at least one girl and Amanda


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= 1C 1 9C 3 + 1C 1 3C 1 9C 2 + 1C 1 3C 2 9C 1 + 1C 1 3C 3

= 84 + 108 + 27 + 1

= 220

Total number of ways of choosing 4 people = 13C 4 = 715


= =

6 Number of children chosen = 3

Number of ways of choosing 3 children out of 13 = 13C 3 = 286

Since we need more boys than girls, possibilities are 2B 1G, 3B

Number of ways of choosing more boys than girls =7C 2

6C 1 +

7C 3 = 126 + 35 = 161

Required probability 161 0.563286

= =

7

(a) P(choosing one wicket keeper)

2 181 1020

11

C C

C

=

87 516 0.521167 690

= =

(b) P(choosing the two most experienced players)2 18

2 920

11

48 620 0.289167 960

C C

C = = =

8

(a) P(all 4 boys chosen)5

48

4

5 170 14

C

C = = =

(b) P(all three girls chosen)3 5

3 18

4

5 170 14

C C

C = = =

(c) P(exactly 2 boys and exactly 2 girls)5 3

2 2

8 4

30 3

70 7

C C

C

= = =


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(d) P(more boys than girls)5 3 5

3 1 48

4

35 170 2

C C C

C

+= = =

9

(a) P(all red)2

28

2

128

C

C = =

(b) P(exactly one green ball)3 5

1 18

2

1528

C C

C = =

(c) P(all balls of the same colour)2 3 3

2 2 28

2

C C C

C

+ +=

7 128 4

= =

10

(a) P(3S, 2M, 1D)3 3 4

3 2 110

5

12 1252 21

C C C

C = = =

(b) P(4D, 1S)4 3

4 110

5

3 1252 84

C C

C = = =

(c)3 4

3 2

10 5

6 1P(3M, 2D)252 42

C C

C = = =

11

Number to be chosen = 5

(a) P(4 bowlers and 1 batsman)4 6

4 113

5

6 0.004661287

C C

C

= = =


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Required probability

252533

31 3133

= =

13


(a) P(VP included)1 12

1 513

6

792 61716 13

C C

C = = =

(b) P(exactly 5 parents)8 5

5 113

6

280 701716 429

C C

C = = =

(c) P(4 parents or less)5 8 5 8 5 8 5 8

5 1 4 2 3 3 2 413

6

C C C C C C C C

C

+ + +=

1408 328 140 560 7001716 1716 39

+ + += = =

14

Number chosen = 2

(a) P(both bruised)3

212

2

3 166 22

C

C = = =

(b) P(at least one bruised)3 9 3

1 1 2

12 2

30 5

66 11

C C C

C

+= = =

(c) P(two bruised | at least one bruised)

P(two bruised at least one bruised) p(at least one bruised)

=

1122

5 1011

= =


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Review Exercise 15

1 P(choosing four apples) 12 11 10 9 90 4514 13 12 11 182 91

= = =

2 P(2 most experienced)

2 182 9

2011

1138

C C

C = =

3 P(straight) = 0.55P(right) = 0.10P(left) = 0.35(a) P(all three turn left) = 0.35 0.35 0.35

= 0.0429(b) P(all go in different directions)

= P(LRS) + P(LSR) + P(RLS) + P(RSL) + P(SLR) + P(SRL)= 6(0.35 0.10 0.55)

= 0.1155(c) P(two turn right and one turns left) = P(RRL) + P(RLR) + P(LRR)

= (0.1) (0.1) (0.35) 3= 0.0105

4 2 black, 1 red, 4 yellow

(a)

(b)(i) P(one red) 1 2 1 4 2 1 4 17 6 7 6 7 6 7 6

= + + +

2 4 2 442 42 42 42

= + + +

12 242 7

= =

(ii) P(two yellow) 4 3 12 27 6 42 7

= = =

(iii) P(one black and one yellow)2 4 4 2 16 87 6 7 6 42 21

= + = =

5 P( A ) = 0.4


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P( B ) = 0.9P( A B ) = 0.6(a) P( A ) P( B ) = 0.4 0.9 = 0.36

P( A B ) = 0.6Since P( A B ) P( A ) P( B ) A and B are not independent

(b) P(A B) = 0.6Since mutually exclusive events cannot occur together A and B are not mutually exclusive

6 (a)

(b) P(Ian does not arrive on time)= 0.2 0.4 + 0.3 0.1 + 0.5 0.2

= 0.08 + 0.03 + 0.10= 0.217 P( X | Y ) = 0.8, P(Y | X ) = 0.5, P( X Y ) = 0.24

(a) P( ) 0.24P( | ) 0.8P( ) P( ) X Y

X Y Y Y

= =

0.24P( ) 0.30.8

Y = =

(b) P( ) 0.24P( | ) 0.5 P( ) 0.48P( ) P( )Y X

Y X X X X

= = =

P( ) P( ) 0.48 0.3 0.144 P( ) X Y X Y = =

X

andY

are not independent.(c) P( X Y )= P( X ) P( X Y )= 0.48 0.24= 0.24

8 (a) Positive difference between scores on two dice1 2 3 4 5 6

1 0 1 2 3 4 52 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 0 1 2

5 4 3 2 1 0 1


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6 5 4 3 2 1 0

6 1P( )36 6

X = =

(b) Product of the scores on two dice

1 2 3 4 5 6

1 1 2 3 4 5 6

2 2 4 6 8 10 12

3 3 6 9 12 15 18

4 4 8 12 16 20 24

5 5 10 15 20 25 30

6 6 12 18 24 30 36

17P( )36

Y =

(c) 2 1P( )36 18

X Y = =

Since P( X Y ) 0 X and Y are not mutually exclusive

9 1 1P( ) , P( )9 24

C C D= =

(a) Since C and D are independentP(C D ) = P(C ) P( D )

1 1 P( )24 9

D=

9 3P( )24 8

D = =

(b) P( ) P( ) P( ) P( )C D C D C D = + 1 3 1

9 8 24= +

49

=

10 1 1P( ) , P( )3 6

X Y = =

(a) P( ) 0, since and are mutually exclusive X Y X Y =

(b) 1 1 3 1P( ) P( ) P( ) P( ) 03 6 6 2

X Y X Y X Y = + = + = =

(c) 1P( ) P( ) P( )6

Y X Y X Y = =

11 P(

A

) = 0.6P( A B ) = 0.8


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P( A B ) = 0.3(a) P( A B ) = P( A ) + P( B ) P( A B )

0.8 = 0.6 + P( B ) 0.3P( ) 0.5 B =

(b) P( ) P( ) P( ) B A A A B = = 0.6 0.3 = 0.3

(c) P( ) P( ) A B A B = = 1 0.8 = 0.2

12 (a) (i) P(smoker) 95 19400 80

= =

(ii) P(UTT graduate or smoker)= P(UTT graduate) + P(smoker) P(UTT graduate smoker)

155 95 25 9400 400 400 16

= + =

(b) P(smoker) 95400=

P(UTT graduate) 155400

=

P(smoker UTT graduate) 25 1400 16

= =

P(smoker) P(UTT graduate) 95 155 589 1400 400 6400 16

= =

Events are not independentEvents are not mutually exclusive since P(UTT graduate smoker) 0

13 TT W I H O U

(a) Total number of four letter codes 7! 25202!

= =

Number of codes without the Ts = 5 4 3 2 = 120

P(neither of the Ts) 120 12520 21

= =

(b) Number of codes with both Ts 5 24!1 1202!

C = =

P(code has both Ts) 120 12520 21

= =

14


(a) P(four boys)8

414

4

70 0.06991001

C

C = = =

(b) P(two boys and two girls)8 6

2 214

2

420 0.41961001

C C

C = = =

(c) P(more boys than girls) = P(four boys) + P(three boys and one girl)8 8 6

4 3 114 4

70 3360.40561001

C C C

C

+ += = =

15 C C N L D O U E


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(a) Number of ways of arranging all letters without restrictions 8! 20 1602!

= =

Number of arrangements with all consonants together 4! 5! 14402!

= =


= =

(b) Number of arrangements beginning with a consonant and ending with a vowel5 6! 3

54002!

= =


= =

(c) Number of arrangements with all vowels together6! 3! 2160

2!

= =

Required probability 2160 3 .20160 28

= =

16 (a) Total number of arrangements 7! 12602!2!

= =

Number of arrangements beginning and ending with a 72 5! 1

602!2!

= =


= =

(b) Number of odd numbers6! 4

7202!2!= =

P(odd arrangement) 720 41260 7

= =

(c) P(starts and ends with 7 | odd no)P(starts and ends with 7 odd)

P(odd)

=

1121

4 127

= =

17

(a) P(both bruised)4

220

2

6 3190 95

C

C = = =

(b) P(both red and at least one bruised)12 3 3

1 1 220

2

39190

C C C

C

+= =

(c) P(at least one bruised | both red)


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P(at least one bruised both red)P(both red)

=

32

3202215 15

2 220

2

3 1105 35

C

C C C C

C

= = = =

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Unit 2 Chapter 15 Answers