math t volume 2 chapter 6

8/8/2019 math T volume 2 chapter 6

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Ace Ahead Mathematics T Volume 2

Exam Practice 6

1. Let T: Teacher

M: Male student

F: Female student(a) Number of arrangements

__T __ __ __ __ __ 5!

__ __ __ __ __ __T 5!

Therefore, the total number of

arrangements

= 5! 2 = 240

(b)T F

1 F

2

Number of arrangements = 4!

M1M

2M

3

Number of arrangements = 3!

Therefore, the total number of arrangements

= 4! 3! = 144

2. Out of the 15 cards numbered from 1 to 15,

there are 8 cards with odd numbers and 7 cards

with even numbers.

(a) Number of ways of choosing cards with 5

even numbers and 4 odd numbers

= 7C58C

4

= 1470

(b) Number of ways of arranging 4 cards with

even numbers followed by 4 cards with

odd numbers

= 7P48P

4

= 1 411 200

3. The number of four-digit numbers (from 1000

to 9999) whose digits are all different

= 9 9 8 7

= 4536

The number of integers from 1000 to 9999

= 9999 1000 + 1 = 9000

Hence, the number of four-digit numbers from

1000 to 9999 which do not have four different

digits

= 9000 4536

= 4464

4. (a) Number of arrangements for all the letters

of MISSISSIPPI (with 1M, 4 Is, 4 Ss

and 2 Ps) without restrictions

11!=

4!4!2!

= 34 650

(b) If the word must begin and end with the

letter P, then it is left with 1 M, 4 Is and

4 Ss at the centre to be arranged.

9!Number of arrangements = = 630

4!4!

(c) If all theIs must be grouped together,

then there are 8 objects with 1 M, 4

similar Ss, 2 similar Ps and 1 group ofIs.

8!Number of arrangements = = 8404!2!

(d) If the word must begin and end with the

letter P and at the same time all the Is

must be grouped together, then it is left

with 1M, 4 Ss and 1 group ofIs at the

centre to be arranged.

6!Number of arrangements = = 30

4!

5. (a) Number of arrangements if the numbers

begin with 0 but do not end with 0

= 1 10 10 9 = 900

(b) Number of arrangements if the numbers

do not begin with 0 but end with 0

= 9 10 10 1 = 900

(c) Number of arrangements if the numbers

begin and end with 0

= 1 10 10 1 = 100

P(the number begins with 0 or ends with 0)

900 + 900 + 100=

10 00019

= 100

6. If there are no restrictions, then the number of

word codes which can be formed6!

= = 3602!

(a) If the two Ps are side-by-side, then the

number of word codes which can be

formed

= 5! = 120

P(the two Ps are side-by-side)

120 1= =

360 3

Hence, P(the word codes formed do not

have two Ps side-by-side)

1 2= 1 =

3 3(b) U P

Number of ways to arrange R, P,L,Eis 4!

U R

4!Number of ways to arrange P, P,L,Eis

2!

U L

4!Number of ways to arrange P,R, P,Eis

2!

M1 M

2 M

3

Sum of parts

(a), (b) and (c)

ACE Ahd Mth STPM V2 E6 (WAns) 4th.indd 1ACE Ahd Mth STPM V2 E6 (WAns) 4th.indd 1 1/7/2009 10:28:05 AM1/7/2009 10:28:05 AM


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E P

Number of ways to arrange U,R, P,L is 4!

E R

4! Number of ways to arrange P, U, P,L is

2!

E L

4!Number of ways to arrange P, U,R, P is

2!

Therefore, the number of word codes

which can be formed if they begin with a

vowel and end with a consonant

4!= 4 + (2 4!) = 962!Hence, P(the word codes begin with a

vowel and end with a consonant)

96 4= =

360 15

7.

Based on the Venn diagram,

P(AB) = P[(AB') (A'B)] + P(AB)

P(AB) is also given by

P(AB) = P(A) + P(B) P(AB)

Hence, P[(AB') P(A'B)] + P(AB)

= P(A) + P(B) P(AB)

17 7 7 + P(AB) = + P(AB)20 20 10

1

2P(A

B) = 51

P(AB) = 10

7 7 49P(A) P(B) = = .

20 10 200

Since P(A) P(B) P(AB), thenA andB

are not independent.

8.

1(a) P(EF) = P(C) =

5

(b) P(EF)

= P(E) + P(F) P(EF)

= P(A B C) + P(CD) P(C)

= P(A) + P(B) + P(C) + P(C) + P(D) P(C)

= P(A) + P(B) + P(C) + P(D)

1 1 1 1= + + +

5 5 5 5

4=

5

9. X: Event that the cartridges are supplied by

supplier X.

Y: Event that the cartridges are supplied by

supplier Y.

D : Event that a cartridge is defective.

(a) P(defective)

= P(XD) + P(YD)

= (0.4 0.05) + (0.6 0.03)

= 0.038

(b) 1000 cartridges are bought. This meansthat there are 38 defective cartridges and

962 good condition cartridges bought.

P(at least one of the two cartridges chosen

is defective)

= 1 P(both the cartridges chosen are in

good condition)

962 961= 1

1000 999

= 0.0746

10. (a) P(B'/A) = 0.7

P(B'A) = 0.7

P(A)P(A) P(AB)

= 0.7P(A)

0.4 P(AB) = 0.7

0.4

0.4 P(AB) = 0.28

P(AB) = 0.12

(b) P(B/A') = 0.5

P(B A') = 0.5

P(A')

P(B) P(AB) = 0.5

1 P(A)

P(B) 0.12 = 0.5

1 0.4P(B) = 0.42

Hence, P(AB) = P(A) + P(B) P(AB)

= 0.4 + 0.42 0.12

= 0.70

11. (a) P(A B) = P(A) + P(B) P(A B)

3 2 1= +

5 3 223

= 30

A BS

A BS

E:

F:

A B

C D

S

0.05

0.95

0.03

0.97

0.4

0.6

X

Y

XD

XD

YD

YD

Outcomes

D

D

D

D

A B' A B A' B

A BS



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P(A' B') = P(A B)'

= 1 P(A B)

23= 1

30

7=

30

(b) P(A' B') = P(A B)'

= 1 P(A B)1

= 1 2

1=

2

212. P(B/A) =

3P(A B) 2 =

P(A) 3

P(A B) 2 =

1 33

2P(A B) =

91

P(A/B) = 2

P(A B) 1 =

P(B) 2

29 1

= P(B) 2

4P(B) =

9

P(A B) = P(A) + P(B) P(A B)

1 4 2= +

3 9 9

5= 9

13. H: Event that a networking breakdown is due

to hardware problem.

K: Event that a networking breakdown is due

to software problem.

The given problem can be represented by the

following Venn diagram.

(a) P(hardware problem)

= P(H) = 0.15 + 0.1 = 0.25

(b) P(hardware problem or software problem)

= P(H K) = 0.15 + 0.1 + 0.5 = 0.75

(c) P(not hardware problem and software

problem at the same time)

= P(H K)'= 1 P(H K) = 1 0.1 = 0.9

(d) P(not hardware problem or software

problem or both)

= P(HK)'= 1 P(HK) = 1 0.75 = 0.25

14.Age

(years)

Probability of

TotalSkilled

worker

Unskilled

worker

20 24 0.08 0.32 0.40

25 29 0.13 0.37 0.50

30 34 0.09 0.01 0.10

Total 0.30 0.70 1.00

(a) P(skilled worker chosen from the age

group of 25 29)

0.13=

0.13 + 0.37

= 0.26

(b) P(unskilled worker chosen from the age

group of 25 29)

0.37=

0.13 + 0.37

= 0.74

Hence, the expected number of unskilledworkers in the age group of 25 29

= 0.74 1 000 000

= 740 000

(c) P(at least one of the two workers chosen

from the age groups of 20 34 is

unskilled worker)

= 1 P(both of them are skilled workers)

= 1 (0.3 0.3)

= 0.91

(d) P(skilled worker chosen from the age

groups of 25 34)

0.13 + 0.09=

0.13 + 0.37 + 0.09 + 0.010.22

= 0.6

11=

30

15. (a)

Qualification

GenderGraduate

Non-

graduateTotal

Male 25 10 35

Female 20 45 65

Total 45 55 100

(b) P(exactly one teacher out of two teachers

receives the RM300 service allowance)

= P(exactly one teacher out of two

teachers is a graduate)

= P(one graduate and one non-graduate)

45 55 55 45= + 100 99 100 99

1= 2

0.25

0.50.10.15

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(c) P(exactly one teacher receives the RM300

service allowance if a male teacher and

a female teacher are chosen)

= P(exactly one graduate and one non-

graduate if a male teacher and a

female teacher are chosen)

= P(male graduate and female non-

graduate) + P(male non-graduate and

female graduate)

25 45 10 20= + 35 65 35 65

53= 91

16. A: Event that the taxi booked is from company

A.

B: Event that the taxi booked is from company

B.

C: Event that the taxi booked is from company

C.

L: Event that a taxi booked is late.

(a) P(the taxi is from companyA and it is not

late)

= P(AL)

= 0.5 0.98

= 0.490

(b) P(the taxi booked is late)

= P(A L) + P(B L) + P(C L)

= (0.5 0.02) + (0.3 0.04) + (0.2 0.05)

= 0.032

P(CL)P(C/L) =

P(L)

0.2 0.05=

0.032

= 0.3125

17. P(F) = 0.60

P(B) = 0.25

P(F B) = 0.15

(a) Since P(F B) = 0.15 0, FandB are

not mutually exclusive.

P(F) P(B) = 0.60 0.25 = 0.15

Since P(F B) = P(F) P(B) = 0.15, F

andB are independent.

P(F B) 0.15(b) P(F/B) = = = 0.60

P(B) 0.25

18 T:Event that Janice takes tuition.

P:Event that Janice passes the Pure

Mathematics paper.

(a) P(Janice passes the Pure Mathematics

paper)

P(TP) + P(TP)

= (0.7 0.9) + (0.3 0.6)

= 0.81

(b) P(T/P)

P(T P)=

P(P)

0.9 0.7=

0.817=

9

19. X: Event that a tyre is produced by machine X.

Y: Event that a tyre is produced by machine Y.

Z: Event that a tyre is produced by machine Z.

D: Event that a tyre produced is defective.

P(X/D)

P(XD)=

P(D)P(XD)

= P(XD) + P(YD) + P(ZD)

0.25 0.05=

(0.25 0.05) + (0.35 0.04) + (0.40 0.02)

0.0125=

0.0345

= 0.362

20. (a) P(at least one of the two ping-pong balls

drawn is white)

= 1 P(both the ping-pong balls drawn

are orange)

90 89= 1

100 9921

= 110

0.9

0.1

0.6

0.4

0.7

0.3

TP

Outcomes

TP

TP

P

P

P

P TP

T

T

From (a).

0.05

0.040.25

0.40

X

Z

XD

XD

YD

YD

Outcomes

0.02

Y

ZD

ZD

0.35

D

D

D

D

D

D

0.02

0.98

0.04

0.96

0.5

0.2

A

C

AL

AL

BL

BL

Outcomes

0.05

0.95

B

CL

CL

L

L

L

L

L

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(b) A: Event that both the ping-pong balls

drawn are white.

B: Event that at least one of the two ping-

pong balls drawn is white.

P(A/B)

P(A B)=

P(B)

P(both the ping-pong balls drawn are white)=

21110

10 9 100 99

= 21110

1=

21

21. X= {1, 2, 3, 4, 5, 6}

Y= {1, 2, 3, 4, 6, 7, 8, 9}

(a) A = {(1, 3), (2, 5), (3, 7), (4, 9)}

n(A

) 4 4 2P(A) = = = = n(S) 6 9 54 27

(b) B = {(2, 5), (4, 2)}

n(B) 2 1P(B) = = =

n(S) 54 27

(c) A B = {(2, 5)}

n(A B) 1P(A B) = =

n(S) 54

1

P(A B) 54 1P(A/B) = = =

1 2P(B)

27

(d) A B = {(1, 3), (2, 5), (3, 7), (4, 2), (4, 9)}

P(A' B') = (A B)'= 1 P(A B) n(A B) 5 49

= 1 = 1 = n(S) 54 54

49

P(A'B') 54 49P(A'/B') = = =

1 52P(B') 1

27

22. Have

seen the

advertisement

(A)

Have not

seen the

advertisement

(A)

Total

Purchase

(B) 30 10 40

Do not

purchase

(B)

30 30 60

Total 60 40 100

A: Event that a visitor has seen the

advertisement.

B: Event that a visitor makes purchases.

P(a visitor who has not seen the advertisement

makes a purchase)

= P(B/A')

P(BA')=

P(A')

10100

= 40100

1=

4

23. (a)

Spade

(S)

Heart

(H)

Diamond

(D)

Club

(C)

Total

King

(K)

2 1 3 3 9

Queen

(Q)

3 5 3 2 13

Jack

(J)

1 3 2 2 8

Total 6 9 8 7 30

(i) P(KD) = P(K) + P(D) P(KD)

9 8 3= +

30 30 30

7=

15

(ii) P(J'D') = P(JD)'

= 1 P(JD)= 1 [P(J) + P(D) P(JD)]

8 8 2= 1 + 30 30 30

7 8= 1 =

15 15

P(H K)(iii) P(H/K) =

P(K)

130

= 9

30

1= 9

(b) Number of permutations of one Spade,

one Heart, one Diamond and one

Club = 4!

P(one Spade, one Heart, one Diamond

and one Club are drawn)

6 9 8 7= 4!30 29 28 27

16=

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25. C: Event that Azmi chooses a correct icon.

C: Event that Azmi chooses an incorrect icon.

1 1 1(a) P(A B) = P(CC) = =

4 7 28

1 1 3 1(b) P(B) = P(CC) + P(CC) = +4 7 4 9

5=

42

(c) P(A B) = P(A) + P(B) P(A B)

1 5 1 71

= + = 4 24 28 168

1

P(A B) 28 3(d) P(A/B) = = =

P(B) 5 1042

26. A: Event that Ling recalls the first digit

correctly.

B: Event that Ling recalls the second digit

correctly.

P(BA) 0.62(a) P(B/A) = = = 0.886

P(A) 0.70

(b) P(AB') = P(A) P(AB) = 0.70 0.62

= 0.080

(c) P(A'B) = P(B) P(AB) = 0.76 0.62

= 0.140

P(B'A')(d) P(B'/A') =

P(A')

P(B A)'=

P(A')1 P(A B)

= 1 P(A)

1 (0.08 + 0.62 + 0.14)=

1 0.700.16

= 0.30

= 0.533

24.

P(HM) = P(BRY) + P(BYR)

2 5 3 2 3 5= + 5 8 7 5 8 7

3= [shown]

14

(a) (i) P(H/M)P(HM)

= P(M)

314

= P(WRY) + P(WYR) + P(BRY) + P(BYR)

314

= 3 4 3 3 3 4 + 5 7 6 5 7 6

2 5 3 2 3 5 + + 5 8 7 5 8 73

14=

3970

5=

13

(ii) P(HM) = P(H) + P(M) P(HM)

2 39 3= +

5 70 14

26=

35

(b) Colour of

objectsWhite Black Red Yellow

Numberof objects

3 2 9 6

Number of permutations of one White,

one Black, one Red and one Yellow

objects = 4!

P(one White, one Black, one Red and

one Yellow objects are obtained)

3 2 9 6= 4!20 20 20 20

243=

5000

WRR

Outcomes

WYR

WRY

R

Y

R

Y WYY

BRR

BYR

BRY

R

Y

R

Y BYY

R

Y

R

Y 4R

2YW

BBag

3R

3Y

4R

3Y

5R

2Y

4R

3Y

5R

3Y

3W

2B

Black

box

37White

box

47

36

36

26

46

37

47

27

573

8

58

35

2

5

CC

Outcomes

CC

CC

C

C

C CC

C

C

C

3

4

1

4

8

9

1

9

6

7

1

7

A B

0.16

0.08 0.62 0.14

S

math t volume 2 chapter 6

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