27[A Math CD]

1 Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. P(picking an English book)= 66 + 8= 37Answer: B

2. Let the number of boys be x.P(choosing a boy) = xtotal number of students 47 =

x42 x = 4 427 = 24Answer: C

3. P(picking a red ball) = 18x 29 =

18x x = 9 182 = 81Answer: D

4. P(drawing a 30-sen stamp)= 84 + 7 + 8 + 5= 13Answer: A

5. P(choosing a fresh egg) = 65h 57 =

65h

h = 65 75 = 91Answer: C

6. P(pickingablackball)=1 25 = 35Answer: C

7. P(pickingabrownenvelope)=1 14 = 3414 represents 6 envelopes.34 represents 6 3 = 18 envelopes.Hence, the number of brown envelopes is 18.Answer: C

8. P(picking a brown plastic chair)=1 514

37=1 514

614= 314Answer: A

9. P(thematchisadraw)=1 411 211

= 511Answer: B

10. P(picking a student taking part in the camp)=1 25= 35Total number of students in the Science stream= 45 53= 75Answer: C

CHAPTER

27 Probability IICHAPTER

2 Mathematics SPM Chapter 27

Penerbitan Pelangi Sdn. Bhd.

Paper 2

1. Let B be boy and G be girl.1st pupil 2nd pupil Outcomes

B1B2 (B1, B2)G (B1, G)

B2B1 (B2, B1)G (B2, G)

GB1 (G, B1)B2 (G, B2)

Hence, the possible outcomes= {(B1, B2), (B1, G), (B2, B1), (B2, G), (G, B1), (G, B2)}

2. 1st card 2nd card Outcomes

S E (S, E)T (S, T)

E S (E, S)T (E, T)

T S (T, S)E (T, E) Hence, the possible outcomes = {(S, E), (S, T), (E, S), (E, T), (T, S), (T, E)}

3.

M A T H

Box Q

Box P12

3

Hence, the possible outcomes = {(M, 1), (M, 2), (M, 3), (A, 1), (A, 2), (A, 3),

(T, 1), (T, 2), (T, 3), (H, 1), (H, 2), (H, 3)}

4. Let R be red ball and G be green ball. 1st ball 2nd ball Outcomes

RG1 (R, G1)G2 (R, G2)

G1R (G1, R)G2 (G1, G2)

G2R (G2, R)G1 (G2, G1) n(S) = 6

P(both of the balls are of different colours)= 46= 23

5. n(S) = 12A = {cards with even numbers and greater than 6} = {8, 10, 12}n(A) = 3P(the card is an even number and greater than 6)= 312= 14

6. n(S) = 10 B = {marbles with prime numbers or odd numbers} = {1, 2, 3, 5, 7, 9}n(B) = 6P(the number on the marble is a prime or odd number)= 610= 35

Paper 1

1. Total number of pens= 16 74= 28P(picking a red pen)= 428= 17P(picking a pen which is not blue)= P(picking a red pen or a black pen)= 17 +

47

= 57Answer: D

2. P(choosing a girl) = 58 2020 + y =

58

160 = 100 + 5y 5y = 60 y = 12Answer: B



3. Let x be the number of red marbles.P(choosing a red marble) = 14 x12 + 9 + x =

14

4x = 12 + 9 + x 3x = 21 x = 7P(choosing non-black marble)= 9 + 712 + 9 + 7= 1628= 47Answer: C

4. Let the number of black markers taken out be x.Before taking out the black markers: Number of blue markers= 40 25= 16After taking out the black markers:

P(picking a blue marker) = 16 + x40 16 + x40 =

58 8(16 + x) = 5 40 128 + 8x = 200 8x=200128 x = 728 = 9Answer: C

5. Let x be the number of red apples. P(picking a red apple) = 58 x9 + x =

58

8x = 45 + 5x 3x = 45 x = 15Answer: B

6. P(choosing a boy from the Science stream)= 2020 + 30 + 18 + 16= 2084= 521Answer: A

7. Let xbethenumberoffictionbooks.P(pickingafictionbook)= 513 x48 + x =

513

13x = 240 + 5x 8x = 240 x = 30Answer: C

8. The number of girls= 60 35 10= 15P(choosing a girl)= 1550= 310Answer: C

9. Multiples of 2 are 2, 4, 8, 14 and 16P(choosing a multiple of 2) = 511Answer: B

10. 4 pens represent 29 .

6 pens represent 6 29 14 =

39P(choosing a red pen)=1 29

39= 49Answer: C



Paper 2

1. Box A Box B Outcomes

X

P (X, P)3 (X, 3)4 (X, 4)Q (X, Q)

YP (Y, P)3 (Y, 3)4 (Y, 4)Q (Y, Q)

2P (2, P)3 (2, 3)4 (2, 4)Q (2, Q)

Sample space, S = {(X, P), (X, 3), (X, 4), (X, Q), (Y, P), (Y, 3), (Y, 4), (Y, Q), (2, P), (2, 3), (2, 4), (2, Q)}n(S) = 12

(a) Let G = Event of getting both cards labelled with a letter

= {(X, P), (X, Q), (Y, P), (Y, Q)}

P(G) = n(G)n(S) = 412 = 13

(b) Let H = Event of getting at least one card labelled with a number

= {(X, 3), (X, 4), (Y, 3), (Y, 4), (2, P), (2, 3), (2, 4), (2, Q)}

P(H) = n(H)n(S) = 812 = 23

2. (a) (i) Sample space, S = {(Alif, Hana), (Ezul, Hana), (Ken, Hana),

(Alif, Ezul), (Alif, Ken), (Ezul, Ken)} (ii) Let A = Event that two boys are chosen = {(Alif, Ezul), (Alif, Ken),

(Ezul, Ken)} P(A) = n(A)n(S) = 36 = 12

(b) (i)OutcomesGirlsBoys

Hana (Alif, Hana)

Ling (Alif, Ling)

(Alif, Raha)

(Ezul, Hana)

(Ezul, Ling)

(Ezul, Raha)

(Ken, Hana)

(Ken, Ling)

(Ken, Raha)

(Siva, Hana)

(Siva, Ling)

(Siva, Raha)

Raha

Hana

Ling

Raha

Hana

Ling

Raha

Hana

Ling

Raha

Alif

Ezul

Ken

Siva

Sample space, S = {(Alif, Hana), (Alif, Ling), (Alif, Raha),

(Ezul, Hana), (Ezul, Ling), (Ezul, Raha), (Ken, Hana), (Ken, Ling), (Ken, Raha), (Siva, Hana), (Siva, Ling), (Siva, Raha)}

(ii) Let B = Event that both librarians from class Q are chosen

= {(Siva, Ling), (Siva, Raha)} P(B) = n(B)n(S) = 212 = 16



3. (a) First spin Second spin Outcomes

E

E EE

EF

EG

FF

FG

FH

GF

GH

HE

HG

EH

FE

GE

GG

HF

HH

F

G

H

E

F

G

H

E

F

G

H

E

F

G

H

F

G

H

(b) (i) Let A = Event that the pointer stops at the same sector twice = {(EE), (FF), (GG), (HH)} P(A) = n(A)n(S) = 416 = 14

(ii) Let B = Event that the pointer stops at the sector F at least once = {(EF), (FE), (FF), (FG), (FH), (GF), (HF)} P(B) = n(B)n(S) = 716

4. (a) Box X Box Y Outcomes

3 (2, 3)

(2, 4)

(2, 7)

(3, 3)

(3, 4)

(3, 7)

(5, 3)

(5, 4)

(5, 7)

(6, 3)

(6, 4)

(6, 7)

(9, 3)

(9, 4)

(9, 7)

42

3

5

6

9

7

3

4

7

3

4

7

3

4

7

3

4

7

Sample space, S = {(2, 3), (2, 4), (2, 7), (3, 3), (3, 4), (3, 7), (5, 3), (5, 4), (5, 7), (6, 3), (6, 4), (6, 7), (9, 3), (9, 4), (9, 7)}

(b) (i) Let A = Event of getting both cards labelled with odd numbers

= {(3, 3), (3, 7), (5, 3), (5, 7), (9, 3), (9, 7)} P(A) = n(A)n(S) = 615 = 25

(ii) Let B = Event of getting one card labelled with an even number and the other card labelled with a prime number = {(2, 3), (2, 4), (2, 7), (3, 4), (5, 4), (6, 3), (6, 7)} P(B) = n(B)n(S) = 715



5. (a) 1st card 2nd card Outcomes

12 123 134 145 15

21 213 234 245 25

31 312 324 345 35

4

1 412 423 435 45

5

1 512 523 534 54

Sample space, S= {12, 13, 14, 15, 21, 23, 24, 25, 31, 32, 34,

35, 41, 42, 43, 45, 51, 52, 53, 54}

(b) (i) Let A = Event that the number begins with 2

= {21, 23, 24, 25} P(A) = n(A)n(S) = 420 = 15(ii) Let B = Event that the number consists of

two even digits or two odd digits = {13, 15, 24, 31, 35, 42, 51, 53} P(B) = n(B)n(S) = 820 = 25

Paper 1

1. P(drawing a non-RM5 note) = 7 + 47 + 2 + 4= 1113Answer: B

2. P(choosing a non-excellent result student)= 14 + 610 + 14 + 6= 2030= 23Answer: D

3. The numbers with factor 4 are 8, 12, 4, 16 and 24. P(choosing a number with a factor 4)= 515= 13Answer: C

4. The numbers whose sum of digits is equal to 5 are 50, 32, 14, 23 and 41.P(choosing a number whose sum of digits is equal to 5)= 512Answer: D

5. P(drawing a blue card)= 4 + 84 + 8 + 7 + 11= 1230= 25Answer: B



6. P(choosing a letter that is for company A)= 6 + 146 + 14 + 12 + 18= 2050= 25Answer: A

7. The number of Science students = 24 + 6= 30P(picking a Science student)= 3024 + 76= 30100= 310Answer: C

8. Let the number of black beads be x. Then, the number of white beads is 3x. x + 3x = 48 4x = 48 x = 484 = 12P(picking a black bead)= 12 + 448 + 4

= 1652= 413Answer: A

9. P(picking a bear toy or orang utan toy)= 4 + 84 + 6 + 8= 1218= 23Answer: D

10. Let the number of Science books be x. P(choosing a Science book) = x60 + x x60 + x =

14 4x = 60 + x 4xx = 60 3x = 60 x = 603 = 20Answer: B

11. P(picking a blue ball)=1 27

17= 4727 represents 18 balls.47 represents 18 2 = 36 balls.Hence, the number of blue balls is 36. Answer: C

12. Let the total number of students who play tennis be x. P(student plays badminton) = 186 + 18 + x

186 + 18 + x = 38

1824 + x = 38

18 83 = 24 + x x=4824 = 24Answer: C

13. Let the number of blue balls be x. P(picking a red ball) = 3030 + x 3030 + x =

35 5 30 = 3(30 + x) 150 = 90 + 3x 15090=3x x = 603 = 20Answer: B



14. 14 represents 8 students.44 represents 8 4 = 32 students.

P(choosing a student who wears spectacles) = 8 + 432 + 4

= 1236= 13Answer: D

15. P(picking a green card) =1 49

13= 2949 represents 48 cards.

29 represents 482 = 24 cards.

Hence, the number of green cards is 24.Answer: D

16. Number of students attending Mathematics tuition= 47 56= 32Number of students attending Science tuition=5632= 24P(choosing a student who attend Science tuition)= 2456 + 8= 2464= 38Answer: C

17. Number of green diskettes= 1114 84= 66Number of red diskettes=8466= 18P(picking a red diskette)= 18 + 1284 + 12= 3096= 516Answer: D

18. P(picking a yellow marble)=1 27

14= 132827 represents 16 marbles.828 represents 16 marbles.

1328 represents 1328 16

828 = 26 marbles.

Hence, the number of yellow marbles is 26.Answer: B

19. Let the number of girls who walk to school be x. P(choosing a student who walks to school) = 10 + x 25 + 15

25 = 10 + x 40

2 40 = 5(10 + x) 80 = 50 + 5x 5x=8050 x = 305 = 6 Answer: D

20. P(choosing a grade-A egg) = (4015)+n 40 + n 45 =

25 + n 40 + n 4(40 + n) = 5(25 + n) 160 + 4n = 125 + 5n 5n4n=160125 n = 35Answer: A



21. Number of mango drinks = 80 12= 40Number of lychee drinks = 80 15= 16Number of orange drinks =804016= 24Answer: C

22. P(choosing a model P motorcycle)= 2472= 13

P(choosing a model R motorcycle)=1 518

13= 718Answer: D

Paper 2

1. (a)

1

1

2

2 3 4 5 6

3

4

5

6

Black dice

Whi

te d

ice

Event A

Event B

(b) (i) Let A = Event that the total score is 6 P(A) = 536

(ii) Let B = Event that 3 on black dice and 6 on white dice

P(B) = 136

2. Let R1, R2 and R3 be red sweets and Y1 and Y2 be yellow sweets.

Rosli Aliman Outcomes

R1

R2 (R1, R2)R3 (R1, R3)Y1 (R1, Y1)Y2 (R1, Y2)

R2

R1 (R2, R1)R3 (R2, R3)Y1 (R2, Y1)Y2 (R2, Y2)

R3

R1 (R3, R1)R2 (R3, R2)Y1 (R3, Y1)Y2 (R3, Y2)

Y1

R1 (Y1, R1)R2 (Y1, R2)R3 (Y1, R3)Y2 (Y1, Y2)

Y2

R1 (Y2, R1)R2 (Y2, R2)R3 (Y2, R3)Y1 (Y2, Y1)

Sample space, S = {(R1, R2), (R1, R3), (R1, Y1), (R1, Y2), (R2, R1), (R2, R3), (R2, Y1), (R2, Y2), (R3, R1), (R3, R2), (R3, Y1), (R3, Y2), (Y1, R1), (Y1, R2), (Y1, R3), (Y1, Y2), (Y2, R1), (Y2, R2), (Y2, R3), (Y2, Y1)} n(S) = 20(a) Let A = Event that both get red sweets = {(R1, R2), (R1, R3), (R2, R1), (R2, R3),

(R3, R1), (R3, R2)} P(A) = n(A)n(S) = 620 = 310(b) Let B = Event that both get sweets of different

colours = {(R1, Y1), (R1, Y2), (R2, Y1), (R2, Y2), (R3, Y1), (R3, Y2), (Y1, R1), (Y1, R2), (Y1, R3), (Y2, R1),

(Y2, R2), (Y2, R3)} P(B) = n(B)n(S) = 1220 = 35

10

Mathematics SPM Chapter 27



T1T1 (T1, T1)A (T1, A)T2 (T1, T2)

T2T1 (T2, T1)A (T2, A)T2 (T2, T2)

A T1 (A, T1)T2 (A, T2)

Sample space, S = { (T1, T1), (T1, A), (T1, T2), (T2, T1), (T2, A), (T2, T2), (A, T1), (A, T2)}

(b) (i) Let X = Event that both cards are labelled T = {(T1, T1), (T1, T2), (T2, T1), (T2, T2)}

P(X) = n(X)n(S) = 48 = 12(ii) Let Y = Event that one card is labelled T = {(T1, A), (T2, A), (A, T1), (A, T2)} P(Y) = n(Y)n(S) = 48 = 12

4. (a) Let R1, R2 and R3 be red envelopes and W1, W2, W3 and W4 be white envelopes.

Drawer P Drawer Q Outcomes

R1

R2 (R1, R2)R3 (R1, R3)W2 (R1, W2)W3 (R1, W3)W4 (R1, W4)

W1

R2 (W1, R2)R3 (W1, R3)W2 (W1, W2)W3 (W1, W3)W4 (W1, W4)

Sample space, S = {(R1, R2), (R1, R3), (R1, W2), (R1, W3), (R1, W4), (W1, R2), (W1, R3), (W1, W2), (W1, W3), (W1, W4)}

(b) (i) Let A = Event that both envelopes are of the same colour

= {(R1, R2), (R1, R3), (W1, W2), (W1, W3), (W1, W4)} P(A) = n(A)n(S) = 510 = 12

(ii) Let B = Event that at least one envelope is red

= {(R1, R2), (R1, R3), (R1, W2), (R1, W3), (R1, W4), (W1, R2), (W1, R3)}

P(B) = n(B)n(S) = 710

5. Let J1, J2, J3, J4 and J5 be visitors from Johor and K1, K2 and K3 be visitors from Kedah.

Group A Group B Outcomes

J1

J3 (J1, J3)J4 (J1, J4)J5 (J1, J5)K2 (J1, K2)K3 (J1, K3)

J2

J3 (J2, J3)J4 (J2, J4)J5 (J2, J5)K2 (J2, K2)K3 (J2, K3)

K1

J3 (K1, J3)J4 (K1, J4)J5 (K1, J5)K2 (K1, K2)K3 (K1, K3)

Sample space, S = {(J1, J3), (J1, J4), (J1, J5), (J1, K2), (J1, K3), (J2, J3), (J2, J4), (J2, J5), (J2, K2), (J2, K3), (K1, J3), (K1, J4), (K1, J5), (K1, K2), (K1, K3)} n(S) = 15

11



(a) Let A = Event that both visitors are from Johor = {(J1, J3), (J1, J4), (J1, J5), (J2, J3),

(J2, J4), (J2, J5)} P(A) = n(A)n(S) = 615 = 25(b) Let B = Event that one visitor is from Johor

and one visitor is from Kedah = {(J1, K2), (J1, K3), (J2, K2), (J2, K3),

(K1, J3), (K1, J4), (K1, J5)} P(B) = n(B)n(S) = 715

6. (a) Box X Box Y Outcomes

L3 (L, 3)5 (L, 5)6 (L, 6)

O3 (O, 3)5 (O, 5)6 (O, 6)

V3 (V, 3)5 (V, 5)6 (V, 6)

E3 (E, 3)5 (E, 5)6 (E, 6)

Sample space, S = { (L, 3), (L, 5), (L, 6), (O, 3), (O, 5), (O, 6), (V, 3), (V, 5), (V, 6), (E, 3), (E, 5), (E, 6)}

(b) (i) Let A = Event of getting one marble labelled with L and one marble labelled with a prime number

= {(L, 3), (L, 5)} P(A) = n(A)n(S) = 212 = 16

(ii) Let B = Event of getting one marble labelled with L or one marble labelled with a prime number

= { (L, 3), (L, 5), (L, 6), (O, 3), (O, 5), (V, 3), (V, 5), (E, 3), (E, 5)}

P(B) = n(B)n(S) = 912 = 34


W

R WRI WIT WTE WE

R

W RWI RIT RTE RE

I

W IWR IRT ITE IE

T

W TWR TRI TIE TE

E

W EWR ERI EIT ET

Sample space, S = { WR, WI, WT, WE, RW, RI, RT, RE, IW, IR, IT, IE, TW, TR, TI, TE, EW, ER, EI, ET}

12



(b) (i) Let A = Event that the code ends with T = {WT, RT, IT, ET}

P(A) = n(A)n(S) = 420 = 15(ii) Let B = Event that the code consists of two

consonants or two vowels = {WR, WT, RW, RT, TW, TR, IE,

EI} P(B) = n(B)n(S) = 820 = 25

8. (a) Box P Box Q Outcomes

2 M (2, M)N (2, N)

4 M (4, M)N (4, N)

7 M (7, M)N (7, N)

8 M (8, M)N (8, N)Sample space, S = { (2, M), (2, N), (4, M), (4, N), (7, M), (7, N), (8, M), (8, N)}

(b) (i) Let A = Event of getting a card with a prime number and the card labelled N

= {(2, N), (7, N)} P(A) = n(A)n(S) = 28 = 14(ii) Let B = Event of getting a card with a

number divisible by 2 or the card labelled M

= {(2, M), (2, N), (4, M), (4, N), (8, M), (8, N), (7, M)}

P(B) = n(B)n(S) = 78


82 (8, 2)3 (8, 3)7 (8, 7)

28 (2, 8)3 (2, 3)7 (2, 7)

38 (3, 8)2 (3, 2)7 (3, 7)

78 (7, 8)2 (7, 2)3 (7, 3)

Sample space, S = { (8, 2), (8, 3), (8, 7), (2, 8), (2, 3), (2, 7), (3, 8), (3, 2), (3, 7), (7, 8), (7, 2), (7, 3)}

(b) (i) Let A = Event of getting 2 and 8 = {(8, 2), (2, 8)}

P(A) = n(A)n(S) = 212 = 16

(ii) Let B = Event that the sum of the numbers is less than 10

= {(2, 3), (2, 7), (3, 2), (7, 2)} P(B) = n(B)n(S) = 412 = 13

13



10. (a) Let B1 and B2 be boys and G1, G2, G3 and G4 be girls.

1st student 2nd student Outcomes

B1

B2 (B1, B2)G1 (B1, G1)G2 (B1, G2)G3 (B1, G3)G4 (B1, G4)

B2

B1 (B2, B1)G1 (B2, G1)G2 (B2, G2)G3 (B2, G3)G4 (B2, G4)

G1

B1 (G1, B1)B2 (G1, B2)G2 (G1, G2)G3 (G1, G3)G4 (G1, G4)

G2


G3


G4


Sample space, S = {(B1, B2), (B1, G1), (B1, G2), (B1, G3), (B1, G4), (B2, B1), (B2, G1), (B2, G2), (B2, G3), (B2, G4), (G1, B1), (G1, B2), (G1, G2), (G1, G3), (G1, G4), (G2, B1), (G2, B2), (G2, G1), (G2, G3), (G2, G4), (G3, B1), (G3, B2), (G3, G1), (G3, G2), (G3, G4), (G4, B1), (G4, B2), (G4, G1), (G4, G2), (G4, G3)}

n(S) = 30

Let A = Event that both students are girls = {(G1, G2), (G1, G3), (G1, G4), (G2, G1),

(G2, G3), (G2, G4), (G3, G1), (G3, G2), (G3, G4), (G4, G1), (G4, G2), (G4, G3)}

P(A) = n(A)n(S) = 1230 = 25(b) Let B1 and B2 be boys and G1 and G2 be girls.

1st student 2nd student Outcomes

B1B2 (B1, B2)G1 (B1, G1)G2 (B1, G2)

B2B1 (B2, B1)G1 (B2, G1)G2 (B2, G2)

G1B1 (G1, B1)B2 (G1, B2)G2 (G1, G2)

G2B1 (G2, B1)B2 (G2, B2)G1 (G2, G1)

Sample space, S = { (B1, B2), (B1, G1), (B1, G2), (B2, B1), (B2, G1), (B2, G2), (G1, B1), (G1, B2), (G1, G2), (G2, B1), (G2, B2), (G2, G1)}

n(S) = 12Let B = Event that two students are of

different gender = {(B1, G1), ( B1, G2), (B2, G1), (B2, G2),

(G1, B1), (G1, B2), (G2, B1), (G2, B2)}

P(B) = n(B)n(S) = 812 = 23

14



11. (a) Let M1 and M2 be Mathematics teachers, S1 and S2 be Science teachers and E be English teacher.

1st teacher 2nd teacher Outcomes

M1

M2 (M1, M2)S1 (M1, S1)S2 (M1, S2)E (M1, E)

M2

M1 (M2, M1)S1 (M2, S1)S2 (M2, S2)E (M2, E)

S1

M1 (S1, M1)M2 (S1, M2)S2 (S1, S2)E (S1, E)

S2

M1 (S2, M1)M2 (S2, M2)S1 (S2, S1)E (S2, E)

E

M1 (E, M1)M2 (E, M2)S1 (E, S1)S2 (E, S2)

Sample space, S = { M1, M2), (M1, S1), (M1, S2), (M1, E), (M2, M1), (M2, S1), (M2, S2), (M2, E), (S1, M1), (S1, M2), (S1, S2), (S1, E), (S2, M1), (S2, M2), (S2, S1), (S2, E), (E, M1), (E, M2), (E, S1), (E, S2)}

(b) (i) Let A = Event that a Mathematics teacher and an English teacher are chosen

= {(M1, E), (M2, E), (E, M1), (E, M2)} P(A) = n(A)n(S) = 420 = 15

(ii) Let B = Event that at least one Science teacher is chosen

= {(M1, S1), (M1, S2), (M2, S1), (M2, S2), (S1, M1), (S1, M2), (S1, S2), (S1, E), (S2, M1), (S2, M2), (S2, S1), (S2, E), (E, S1), (E, S2)}

P(B) = n(B)n(S) = 1420 = 710

12. (a) Let E1, E2 and E3 be excellent results, G1, G2, G3 and G4 be good results and P be poor result.

Group A Group B Outcomes

E1

E3 (E1, E3)G2 (E1, G2)G3 (E1, G3)G4 (E1, G4)

E2

E3 (E2, E3)G2 (E2, G2)G3 (E2, G3)G4 (E2, G4)

G1

E3 (G1, E3)G2 (G1, G2)G3 (G1, G3)G4 (G1, G4)

P

E3 (P, E3)G2 (P, G2)G3 (P, G3)G4 (P, G4)

Sample space, S = { (E1, E3), (E1, G2), (E1, G3), (E1, G4), (E2, E3), (E2, G2), (E2, G3), (E2, G4), (G1, E3), (G1, G2), (G1, G3), (G1, G4), (P, E3), (P, G2), (P, G3), (P, G4)}

15



(b) (i) Let A = Event that both students obtained excellent results

= {(E1, E3), (E2, E3)} P(A) = n(A)n(S) = 216 = 18(ii) Let B = Event that one student obtained

excellent result and another student obtained good result

= {(E1, G2), (E1, G3), (E1, G4), (E2, G2), (E2, G3), (E2, G4),

(G1, E3)} P(B) = n(B)n(S) = 716

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