New Trend Mathematics S3B — Junior From Supplement Exercises Solution Guide
Chapter 7 Introduction to Probability
15
12 300
1

820
2. P(Mr. Wong) 
Warm-up Exercise (page 7.1)
1. (a) Fraction required  180 
360 
1

2
(b) Fraction required  60 
360 
1

6
(c) Fraction required  210 
360 
7

12
60 
) cm2
360 
 25 .7 c m 2 (corr. to 3 sig. fig.)
2. (a) Area  (  7 2 
120 
) cm2
360 
 16 .8 c m 2 (corr. to 3 sig. fig.)
(b) Area  (  42 
210 
) cm2
360 
 66 .0 c m 2 (corr. to 3 sig. fig.)
(c) Area  (  62 
3. (a) Area  (18 2  12 2 ) c m 2
3. (a) P(letter ‘M’)  1
10
2
10
1

5
(b) P(letter ‘E’) 
4. P(February, April, June, August, October or
December)
2  2  2  2 1 3

32
3

8
5. P(black goldfish from fishpond A)
13

8  13
13

21
P(black goldfish from fishpond B)
26

18  26
13

22
13

21
 Fishpond A has a higher probability of
 180 c m 2
getting a black goldfish.
(b) Area  [  (4  4) 2    4 2 ] c m 2
 48  c m 2
(c) Area  (16  20 
6.
1
 16  9) c m 2
2
 248 c m 2
Build-up Exercise
Exercise 7A
Elementary Set (page 7.2)
1. P(Herbert) 
1
38
1
8
Number of tourists from Beijing 1

8
Total number of tourists
 Number of tourists from Beijing
 Total number of tourists  1
8
1
 328 
8
 41
 41 tourists from Beijing have entered
P(tourist from Beijing) 
Ocean Park over the past hour.
1
2
7.
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
8
13
Number of students with glasses
8

13
Total number of students
 Total number of students
 Number of students with glasses  13
8
13
 24 
8
 39
 There are 39 students in S3A.
P(student with glasses) 
8. (a) P(information technology department or
personnel department)
3 4

24
7

24
(b) Number
of
staff
members
from
warehouse department joining the
Christmas party
 24  4  3  12
5
P(neither personnel department nor
warehouse department)
24  4  5

24
5

8
9. (a) P(Chinese and English newspapers)
27  15  9

27
1

9
(b) P(subscribe one newspaper only)
15  9

27
8

9
10. According to the figure, 92 students score
80 marks below.
 P(grade A)  100  92
100
2

25
11. Let the number of toy trains made by
machine B be y.
5
P(machine A) 
8
x
5

x y 8
8x  5x  5 y
3x  5 y
3x
y
5
 The number of toy trains made by
machine B is 3x .
5
12. Let x be the number of tickets with prizes of
a doll each that should be added.
1
P(doll) 
7
10  x 1

100  x 7
70  7 x  100  x
6 x  30
x5
 5 tickets with prizes of a doll each
should be added.
13. Let the number of cartons of lemon tea on
the table be x, then there are (x  5) cartons
of apple juice on the table.
4
P(lemon tea) 
7
x
4

x  ( x  5) 7
7 x  8 x  20
x  20
 There are 20 cartons of lemon tea on the
table originally.
14. Let the number of red ball pens be x, then
the number of blue ball pens is x  3.
P(red ball pen)  0.4
x
 0 .4
x  ( x  3)
x  0 .8 x  1 .2
0.2 x  1.2
x6
 Total number of ball pens in the box
 x  ( x  3)
 6  (6  3)
 15
Chapter 7 Introduction to Probability
Advanced Set (page 7.4)
1. P(under 16)  3
24
1

8
600
1 764
50

147
2. P(Rachel) 
3. P(‘J’)  3
13
40  37
40
3

40
4. P(16 or above) 
5. P(flat occupied by less than four family
members)
280  64  40  28

280
37

70
6. P(girl from Mathematics club)  12
42  12
2

9
P(girl from Physics club)  8
30  8
4

19
2

9
 Mathematics club has a higher
probability of selecting a girl.
7.
2
17
Number of American
2

Total number of passengers 17
 Number of American
2
 Total number of passengers 
17
2
 204 
17
 24
 There are 24 American passengers in the
P(American) 
aeroplane.
8.
P(orange flavour) 
3
3
8
Number of packs of orange
flavour candies
3

Number of packs of all candies 8
 Number of packs of all candies
 Number of packs of orange
8
flavour candies 
3
8
 9
3
 24
 There are 24 packs of candies in the box.
9. (a) P(learning both piano and violin)
56  24  22

56
5

28
(b) P(learning only one kind of musical
instrument)
24  22

56
23

28
10. (a) P(Computer club or Mathematics club)
8  12

36
5

9
(b) P(Neither Computer club nor Sport club)
36  8  6

36
11

18
11. According to the figure, 36 students score
below 50 marks.
80  36
 P(pass) 
80
11

20
12. (a) Let the number of bulbs produced by
production line B be y.
5
P(production line A) 
12
x
5

Total number of bulbs 12
4
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
 Total number of bulbs  x  12
5
12 x

5
2
P(production line B) 
5
y
2

Total number of bulbs 5
 Total number of bulbs  y  5
2
5y

2
12 x 5 y

5
2
24 x
y
25
 The number of bulbs produced by
14. Let the number of VCDs be x, then the
number of DVDs is x  10 and the number of
CDs is x  4.
P(VCD)  10
27
x
10

x  ( x  10 )  ( x  4) 27
x
10

3 x  6 27
27 x  30 x  60
3 x  60
x  20

The number of VCDs on the shelf is 20.
4
4  5 1
2

5
15. (a) P(chocolate cake) 
production line B is 24 x .
25
4 1
(4  1)  5  1
1

3
(b) P(chocolate cake) 
5
12
x
5

x
12
x  24

440
25
(b) P(production line A) 
12 x  5 x 
24 x
 2 200
5
11 x
 2 200
5
x  1 000
 The number of bulbs produced by
production line A is 1 000.
13. Let the original number of lottery tickets be
x, then the present number of lottery tickets
is x + 240.
Number of lottery tickets with a prize 1

30
Original number of lottery tickets
Number
of
lottery
tickets
with
a
prize

1
 Original number of lottery tickets 
30
1
 x
30
x

30
Number of lottery tickets with a prize 1

90
Present number of lottery tickets
x
1
30

x  240 90
3 x  x  240
2 x  240
x  120
 The original number of lottery tickets
is 120.
11
(4  1)  5  (1  1)
0
(c) P(mango cake) 
Exercise 7B
Elementary Set (page 7.7)
1. Relative frequency of the number of days for
Philip being late for school
2

40
1

20
2. (a) Experimental probability of getting a
packet with 42 chocolates
12

100
3

25
(b) Experimental probability of getting a
standard packet
12  15  16  14  19

100
19

25
Chapter 7 Introduction to Probability
3. (a) Total frequency  8  10  6  4  2
 30
Relative frequency of not travelling by
bus in one day
8

30
4

15
(b) Relative frequency of travelling by bus
in one day
10  6  4  2

30
11

15
4. (a) Total number of students
 7  74  203  214  106  1
 605
Relative frequency with IQ lies between
96 and 105 inclusive
203  214

605
417

605
(b) Relative frequency with IQ higher than
105
106  1

605
107

605
Number of rotten eggs chosen
Number of eggs chosen
 Relative frequency of rotten eggs
4 1


x 8
x  32
5. (a) 
(b) Number of rotten eggs expected
1
 1 600 
8
 200
 The number of rotten eggs expected
is 200.
6. Yes, because when the number of
experiments
done
increases,
the
experimental probability for the coin to toss
a ‘head’ becomes nearer to the theoretical
probability for a fair coin to toss a ‘head’.
5
7. (a) (i) Experimental probability that the
weight of a moon cake lies between
221 g and 230 g inclusive
21

100
(ii) Experimental probability that the
weight of a moon cake lies between
241 g and 250 g inclusive
18

100
9

50
(iii) Experimental probability that the
weight of a moon cake is more than
230.5 g
16  18  21

100
55

100
11

20
(b) Estimated number of moon cakes which
weigh more than 230.5 g
11
 20 000 
20
 11 000
8. (a) (i) Experimental probability that the
person watches the news channel
most frequently
200

1 000
1

5
(ii) Experimental probability that the
person watches the drama channel
or entertainment channel most
frequently
220  250

1 000
47

100
(b) I will choose the entertainment channel
because it seems that people watch the
entertainment channel most frequently in
the evening.
(c) Estimated number of people who watch
the news channel most frequently
200
 300 000 
1 000
 60 000
6
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
Advanced Set (page 7.9)
1. (a) Total number of the group of Japanese
tourists
 644  576  325  195  107  153
 2 000
Relative frequency that it is the tourist’s
first time visiting Hong Kong
644

2 000
161

500
(b) Relative frequency that it is the tourist ’s
fourth time visiting Hong Kong
195

2 000
39

400
2. (a) Relative frequency for Derek to sleep
less than 8 hours at night
22

30
11

15
(b) Relative frequency for Derek to sleep at
least 7 hours but less than 9 hours
29  10

30
19

30
3. (a) Total number of students
 2  8  32  84  76  30  8
 240
Experimental probability that the height
of a S3 student lies between 140 cm and
169 cm inclusive
32  84  76

240
4

5
(b) Experimental probability that the height
of a S3 student is less than 140 cm
28

240
1

24
4. (a)  Relative frequency of defective
electronic components
1

75
Number of defective components 1


75
Number of components chosen
3
1

x 75
x  225
(b) Estimated number of defective electronic
components
1
 2 400 
75
 32
 The estimated number of defective
electronic components is 32.
5. (a)  Relative
copies
1

25
x
1


50 25
x2
frequency of misprinted
(b) Estimated number of misprinted copies
1
 1 200 
25
 48
 The estimated number of misprinted
copies is 48.
6. They are different because the numbers of
red balls and black balls drawn in the three
experiments
of
drawing
100
balls,
1 000 balls and 10 000 balls all differ from
each other a lot.
7. (a) Relative frequency of the crows in the
district with the rings
8

100
2

25
(b)  Relative frequency of the crows with
the rings
2

25
7
Chapter 7 Introduction to Probability
3. (a)
the district is 1 250.
8. (a) P(Orange Daily)
174  190

225  186  174  167  260
 288  94  190  294  122
91

500
Exercise 7C
Elementary Set (page 7.11)
1. (a) H, T
(b) HH, HT, TH, TT
2. The possible combinations are as follows.
Chinese History and Computer Studies,
Chinese History and Geography,
Chinese History and Economics,
History and Computer Studies,
History and Geography,
History and Economics
 There are 6 possible combinations.
Possible
outcomes
Y
Y
W
W
YY
YW
YW
W
Y
W
W
WY
WW
WW
W
Y
W
W
WY
WW
WW
M
F
2nd
child
3rd
child
Possible
outcomes
M
M
F
MMM
MMF
F
M
F
MFM
MFF
M
M
F
FMM
FMF
F
M
F
FFM
FFF
(b) 8 possible outcomes are obtained in (a).
(c) P(2 sons and 1 daughter)  3
8
(c) P(age  30) 
(d) From the result of the survey, Chiu Daily
is the most frequently read newspaper
with people of age 30 or below, and the
cost of posting advertisement on Chiu
Daily is reasonable. Thus the company
should choose Chiu Daily for posting the
advertisement.
2nd draw
1st
child
4. (a)
(b) P(Southern Daily, age  30)
288

288  94  190  294  122
72

247
186
186  94
93

140
1st draw
(b) P(WW)  4
9
5.
2nd dice
1
1st dice
Total number of crows
with the rings
2


Estimated total number 25
of crows
 Estimated total number of crows
 Total number of crows with the
25
rings 
2
25
 100 
2
 1 250
 The estimated number of crows in
2
3
4
5
6
1
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
2
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
3
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
6
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
(a) All possible outcomes of
‘the difference is 3’ are as follows.
(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)
6
 Probability required 
36
1

6
(b) All possible outcomes of
‘the product is less than 10’ are as follows.
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5),
(1, 6), (2, 1), (2, 2), (2, 3), (2, 4),
(3, 1), (3, 2), (3, 3), (4, 1), (4, 2),
(5, 1), (6, 1)
17
 Probability required 
36
8
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
9.
O
R
A
N
G
E
A
AO
AR
AA
AN
AG
AE
P
PO
PR
PA
PN
PG
PE
P
PO
PR
PA
PN
PG
PE
L
LO
LR
LA
LN
LG
LE
E
EO
ER
EA
EN
EG
EE
20
(b) (i) P(two letters are the same)  2
30
1

15
6
30
1

5
Box A
$10
Box B
$20
$50
$100
$500
20
20
50
$1 000
0
$0
$0
$0
$0
$0
$0
0.5
$5
$10
$25
$50
$250
$500
1
$10
$20
$50
$100
$500
$1 000
5
$50 $100 $250
10
$100 $200 $500 $1 000 $5 000 $10 000
$500 $2 500 $5 000
(20, 20)
(20, 20)
(20, 50)
(20, 50) (20, 100)
(20, 20)
(20, 50)
(20, 50) (20, 100)
(20, 50)
(20, 50) (20, 100)
20
(20, 20)
20
(20, 20)
(20, 20)
50
(50, 20)
(50, 20)
(50, 20)
50
(50, 20)
(50, 20)
(50, 20)
100
(100, 20) (100, 20) (100, 20) (100, 50) (100, 50)
10. (a)
32
3
(b) P(cash prize of $100) 
30
1

10
Buying
others
18  17
 28  5
 68
8. (a)
2nd card
1st card
F
F
U
R
FO
FU
FR
OU
OR
OF
U
UF
UO
R
RF
RO
(50, 50)
(50, 100)
Number of
customers
Buying
drink
32  1
 31
Buying
others
68
Buying
drink
32
Buying
others
68  1
 67
(b) (i) Number of outcomes of getting two
customers who buy drink most
frequently
 32  31
 992
O
O
(50, 50) (50, 100)
Number of
customers
Buying
drink
6
30
1

5
100
(b) P(two banknotes with different face
values)
22

30
11

15
(a) P(cash prize of $0)  6
30
1

5
(c) P(cash prize over $500) 
50
(a) P(two banknotes with equal face value)
8

30
4

15
(ii) P(two letters are vowels) 
7.
2nd banknote ($)
20
1st banknote ($)
6. (a)
UR
RU
(b) (i) P(two cards are the same)  0
(ii) P(two cards can form ‘OR’)  2
12
1

6
(ii) Total number of possible outcomes
of getting two customers
 100  99
 9 900
(c) P(two customers buying drink most
frequently)
992

9 900
248

2 475
9
Chapter 7 Introduction to Probability
Advanced Set (page 7.14)
1. Let R 1 and R 2 denote the 2 red scarves, and
W 1 , W 2 and W 3 denote the 3 white scarves.
2nd scarf
R1
R2
W1
W2
W3
R1
R1R1
R1R2
R 1 W1
R 1 W2
R 1 W3
R2
R2R1
R2R2
R 2 W1
R 2 W2
R 2 W3
W1
W1 R 1 W1 R 2 W1 W1 W1 W2 W1 W3
W2
W2 R 1 W2 R 2 W2 W1 W2 W2 W2 W3
W3
W3 R 1 W3 R 2 W3 W1 W3 W2 W3 W3
1st scarf
All possible outcomes of event B are as
follows.
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
6
P ( B) 
36
1

6
5

18
 Event A is more likely to occur.
4. (a)
(a) P(same scarf )  5
25
1

5
13
(b) P(scarves in the same colour) 
25
2. (a) 222, 223, 232, 233, 322, 323, 332, 333
3
(b) (i) P(only one digit is ‘2’) 
8
(ii) P(even number)  4
8
1

2
D
I
S
A
B
L
E
A
AD
AI
AS
AA
AB
AL
AE
B
BD
BI
BS
BA
BB
BL
BE
I
ID
II
IS
IA
IB
IL
IE
L
LD
LI
LS
LA
LB
LL
LE
I
ID
II
IS
IA
IB
IL
IE
T
TD
TI
TS
TA
TB
TL
TE
Y
YD
YI
YS
YA
YB
YL
YE
(b) (i) P(two letters are the same)  5
49
(ii) P(two letters are consonants) 
5. (a)
3.
2nd ball
2nd dice
2
3
4
R1
5
6
R1
R2
B1
B2
W1
W2
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
R2
R2R1
2
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
B1
B1R1 B1R2
3
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
4
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
5
6
R2B1 R2B2 R2W1 R2W2 R2W3
B1B2 B1W1 B1W2 B1W3
B2
B2R1 B2R2 B2B1
W1
W1R1 W1R2 W1B1 W1B2
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
W2
W2R1 W2R2 W2B1 W2B2 W2W1
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
W3
W3R1 W3R2 W3B1 W3B2 W3W1 W3W2
All possible outcomes of event A are as
follows.
(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3),
(4, 5), (5, 4), (5, 6), (6, 5)
10
P(A) 
36
5

18
W3
R1R2 R1B1 R1B2 R1W1 R1W2 R1W3
1
1st ball
1st dice
1
16
49
B2W1 B2W2 B2W3
W1W2 W1W3
2
(b) (i) P(two black balls) 
42
1

21
6
(ii) P(two white balls) 
42
1

7
(iii) P(two balls of the same colour)  10
42
5

21
W2W3
10
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
(iv) P(one black ball and one white ball)
12

42
2

7
6.
1
1
3
5
7
(1, 3)
(1, 5)
(1, 7)
(3, 5)
(3, 7)
3
(3, 1)
5
(5, 1)
(5, 3)
7
(7, 1)
(7, 3)
(5, 7)
(7, 5)
The possible outcomes of ‘a sum of 8’ are as
follows.
(1, 7), (3, 5), (5, 3), (7, 1)
 Probability required  4
12
1

3
P(not getting a storybook)  30
90
1

3
60
P(getting a storybook) 
90
2

3
P(getting a storybook)
 P(not getting a storybook)
2 1
 
3 3
1

3
 The probability of getting a storybook
is
1
higher than that of not getting a
3
storybook.
Number of
people
8. (a)
7. Let D denote doll, C denote chocolate,
P denote colour pencil and S denote
storybook.
2nd prize
D
1st prize
D
C
C
P
P
DC
DC
DP
DP
CC
CP
CP
CP
CP
C
CD
C
CD
CC
P
PD
PC
PC
P
PD
PC
PC
PP
P
PD
PC
PC
PP
PP
S
SD
SC
SC
SP
SP
S
SD
SC
SC
SP
SP
S
SD
SC
SC
SP
SP
S
SD
SC
SC
SP
SP
PP
1st prize
2nd prize
P
S
S
S
S
D
DP
DS
DS
DS
DS
C
CP
CS
CS
CS
CS
C
CP
CS
CS
CS
CS
P
PP
PS
PS
PS
PS
P
PP
PS
PS
PS
PS
PS
PS
PS
PS
SS
SS
SS
SS
SS
P
S
SP
S
SP
SS
S
SP
SS
SS
S
SP
SS
SS
SS
SS
Vote for
Jenny
33
100  33
Not vote
for Jenny  67
Number of
people
Vote for
Jenny
33  1
 32
Not vote
for Jenny
67
Vote for
Jenny
33
Not vote
for Jenny
67  1
 66
(b) (i) Number of possible outcomes of
getting two people who vote for
Jenny
 33  32
 1 056
(ii) Total number of possible outcomes
of getting two people
 100  99
 9 900
(c) P(getting two people vote for Jenny)
1 056

9 900
8

75
Chapter 7 Introduction to Probability
9. Let R denote red ball and W denote white
ball.
2nd ball
R
W
W
1st ball
Balls added
afterwards
R
R
RW
W
WR
W
WR
W
RW
RR
RW
WW
WR
WW
WR
WW
WW
(a) P(two balls in different colours) 
7
12
(b) P(getting white ball first, then getting
red ball)
4

12
1

3
Exercise 7D
Elementary Set (page 7.16)
1. Probability that the point locates in region I
Area of region I

Area of the square card
1

4
2.
A
3. (a) Area of the dartboard  (0.5  0.8) m 2
 0.4 m 2
 4 000 cm 2
Probability of hitting on circle A
Area of circle A

Area of the dartboard
40

4 000
1

100

(b) Probability of hitting on a circle
Area of four circles

Area of the dartboard
4  40

4 000
1

25
4. Probability of not stopping
circular region
Area of the white region

Area of the square
10  10    2 2

10  10
25  

25
within
the
5. (a) Expected value of the scores obtained by
the team in each match
 0.4  3  0.3  1  0.3  0
 1.5
B
O
20 cm
D
11
(b) Estimated total score of the team
 (20  1.5)
 30
C
Area of the square  2  Area of ABC
1
 2  (  AC  BO )
2
1
20
 (2   20  ) cm 2
2
2
 200 cm 2
20
Area of the circle  [  ( ) 2 ] cm 2
2
 100  cm 2
 Probability of hitting on the square region
Area of square

Area of circle
200

100 
2


6. (a) Area of the shaded region
1
5
 [    ( ) 2 ] cm2
2
2
25 
2

cm
8
Area of the dartboard  [ 1    (15 ) 2 ] cm 2
2
2
225 
2

cm
8
Probability of hitting the shaded region
Area of the shaded region

Area of the dartboard

25
8
225
8

1
9
12
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
(b) Area of the white region
225  25 
(

) cm 2
8
8
 25  cm 2
Probability of getting 2 marks
 Probability of hitting the white region
Area of the white region

Area of the dartboard
25 
 225
8

8
9
(c) Expected value of each throw
1
8
 (  5   2)
9
9
1
2
3
7. (a) Probability of winning 1st prize  1
500
(b) Probability of winning a prize
1  3  5  10

500
19

500
(c) Expected value of Stanley’s lucky draw
ticket
1
3
5
 $(
 5 000 
 1 000 
 500
500
500
500
10

 100 )
500
 $23
8. Expected value of the score of each question
1
3
  x   ( y )
4
4
x 3y
 
4 4
 Expected value of his score  0
x 3y
 40  (  )  0
4 4
x 3y

0
4 4
x 3y

4
4
x
3
y
 x : y  3 :1
9. (a) Expected value of the travelling time to
work of Mr. Cheung
7
10
3
 (  48 
 30 
 40 ) minutes
20
20
20
 37 .8 minutes
(b) Expected value of the fare to work of
Mr. Cheung
7
10
3
 $(  8.5 
 12 
 10 )
20
20
20
 $10 .475
10. All the possible outcomes each round are as
follows.
HH, HT, TH, TT
Expected value of the result obtained after
each round
1
1
2
  10   (20 )   5
4
4
4
0
Expected value of the result obtained after
ten rounds
 10  0
0
Advanced Set (page 7.18)
1. Probability that the point is marked on the
shaded region
Area of the shaded region

Area of the paper
3

6
1

2
2. Let the diameter of the circular dartboard
be d cm.
d cm
40 cm
80 cm
According to the Pythagoras’ theorem,
d 2  40 2  80 2
 8 000
d  8 000
8 000 2
) ] cm 2
2
 2 000  cm 2
Area of the dartboard  [  (
13
Chapter 7 Introduction to Probability
Area of the shaded region
 (2 000   40  80 ) cm 2
 (2 000   3 200 ) cm 2
Probability of hitting on the shaded region
Area of the shaded region

Area of the dartboard
2 000   3 200

2 000 
5  8

5
1
3. (a) Area of figure I  (  15  15) cm2
2
225

cm2
2
Area of figure II  2  Area of figure I
225
 (2 
) cm 2
2
 225 cm 2
Area of figure III  (10  10 ) cm
 100 cm 2
1
30
Area of figure IV  [    ( ) 2 ] cm 2
2
2
225 
2

cm
2
Area of the picture  (80  50 ) cm 2
2
 4 000 cm
Probability that the bug rests on either
figure II or IV
Area of figure II  Area of figure IV

Area of the picture
225
225  2

4 000
18  9

320
2
(b) Sum of the areas of the 4 figures
225
225 
(
 225  100 
) cm 2
2
2
875  225 

cm 2
2
Probability that the bug rests on neither
figure I, II, III nor IV
Area of the picture
 Sum of the areas of the 4 figures

Area of the picture
4 000  ( 8752225 )

4 000
8 000  875  225 

8 000
285  9

320
4. Number of banknotes in the wallet
 3 4 2
9
Expected value of the amount
banknote
3
4
2
 $(  20   50   100 )
9
9
9
1
 $51
9
of
the
5. (a) Probability of hitting region I
Area of region I

Area of the dartboard
1
   82
 4
  16 2
1

16
Probability of hitting region II
 Probability of hitting region I
1

16
Probability of hitting region III
Area of region III

Area of the dartboard


1
4
   16 2  14    8 2
  16 2
3
16
Probability of hitting region IV
 Probability of hitting region III
3

16
(b) Expected value of the points scored
1
1
3
3
  20   20   5   5
16
16
16
16
3
4
8
6. (a) Expected value of the amount spent by
Denise on breakfast each day
6
15
9
 $(  6 
 14 
 20 )
30
30
30
 $14 .2
(b) Expected value of the time spent by
Denise on breakfast each day
6
15
9
 ( 5
 12 
 15) minutes
30
30
30
 11.5 minutes
14
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
 Expected value of the points scored  0
x2
144  x 2

1

0

z2
z2
z 2  144

 ( 4)  0
z2
x 2  4( z 2  144 )  0
7. Let R denote red ball and W denote white
ball.
2nd ball
1st ball
R
W
W
W
R
RR
RW
RW
RW
W
WR
WW
WW
WW
W
WR
WW
WW
WW
x 2  2 2 ( z 2  144 )
W
WR
WW
WW
WW
x  2 z 2  144
1
(a) Probability of getting two red balls 
16
(b) Probability of getting
different colours
6

16
3

8
two
balls
(b) According to the result in part (a),
x  2 z 2  144
Since z is an integer greater than 12, the
least value of z is 13.
When z  13,
in
x  2 13 2  144
 2 169  144
(c) Probability of getting two white balls
9

16
(d) Expected value of the cash
obtained
1
3
9
 $(  100   20   0)
16
8
16
 $13 .75
 2 25
 10,
which is an integer smaller than 12
 x  10 and z  13 are one set of
possible integral values of x and z.
When z  14,
prize
x  2 14 2  144
 2 196  144
(e) No, because the fee for joining the lucky
draw is greater than the expected value
of the cash prize obtained.
8. (a) Probability of hitting region I
Area of region I

Area of the dartboard
  x2

  z2
x2
 2
z
Probability of hitting region II
Area of region II

Area of the dartboard
  12 2    x 2

  z2
144  x 2

z2
Probability of hitting region III
Area of region III

Area of the dartboard
  z 2    12 2

  z2
2
z  144

z2
 2 52 ,
which is a number greater than 12
 z  14 is not a possible integral value
of z.
For x  2 z 2  144 , where 0  x  12  z,
the value of x increases as the value of z
increases, i.e. when the value of z is
greater, the value of x is also greater.
Thus when z is an integer greater than 14,
the value of x is greater than 2 52 , i.e.
greater than 12.
 x  10 and z  13 is the only set of
possible integral values of x and z.
9. (a)
0
1
2
3
4
5
6
7
Time
8 (minute)
(b) Probability that a train stays in the
station when Raymond arrives
4 2

86
1

6
Chapter 7 Introduction to Probability
Chapter Test
8. (a) Total number of drink in the refrigerator
 12  14  30  24
 80
Probability of choosing a can of orange
juice
12

80
3

20
(page 7.20)
1. P(red)  3
12
1

4
2nd child
2. 1st child
Boy
Girl
Boy
Boy Boy
Boy Girl
Girl
Girl Boy
Girl Girl
Probability that both are boys 

(b) Expected value of the price of the drink
12
14
30
24
 $(  8.4   6.3 
 5.6 
 9.4)
80
80
80
80
 $7.282 5
1
4
3. There are 21 integers from 20 to 40 inclusive.
Within which all integers not divisible by 4
are as follows.
21, 22, 23, 25, 26, 27, 29, 30, 31, 33, 34,
35, 37, 38, 39
15
 P(not divisible by 4) 
21
5

7
3 x
3 x  x
3

4
4. Probability of hitting on region A 
5. Probability of getting a rotten orange
6  2 11 3  7

6  50
1

15
6. Probability that the student passed the test
8  5  4 1

12  10  8  5  4  1
9

20
7.
S
A
D
H
HS
HA
HD
A
AS
AA
AD
P
PS
PA
PD
P
PS
PA
PD
Y
YS
YA
YD
15
Probability of getting at least one ‘A’ 
9. (a) Probability that the vote is for singer C
360   45   126   54 

360 
3

8
(b) Estimated total number of votes for
singer C
3
 (640  80 ) 
8
 270
10. (a) Probability of hitting on the 3-point region
Area of the 3-point region

Area of the dartboard
  ( 12 ) 2

  ( 10
)2
2
1

100
(b) Probability of hitting on the 2-point region
Area of the 2-point region

Area of the dartboard
[  ( 10
) 2    ( 12 ) 2 ]  83
5

  52
297

800
11. (a) Number of students who are members of
both clubs
 30  18  40
8
 8 students are members of both clubs.
7
15
16
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
(b) Probability that the student is not a
member of Mathematics club
40  30

40
1

4
12. Obtaining two heads is a possible outcome
of tossing a coin twice.
 I is not an impossible event.
Obtaining a white sock and a blue sock is a
possible outcome of drawing two socks from
3 pairs of socks in blue, white and black
each.
 II is not an impossible event.
The sum of numbers obtained by tossing two
dice is at most 12.
 III is an impossible event.
 The answer is B.
P(E)
Number of outcomes favourable to event E

Total number of possible outcomes
 III is not correct.
 The answer is C.
14. Let M denote male and F denote female.
M
M
M
F
F
MM
MM
MM
MF
MF
MM
MM
MF
MF
MM
MF
MF
MF
MF
M
MM
M
MM
MM
M
MM
MM
MM
F
FM
FM
FM
FM
F
FM
FM
FM
FM
2nd card
)
2
FF
FF
3
2
4
5
8
10
12
15
6
3
6
4
8
12
5
10
15
20
20
P(product of the two numbers is an even
number)
10

12
5

6
P(product of the two numbers is a prime
number)
0

12
0
2nd card
2
1st card
Experimental probability of an event
Number of trials favourable to the event

Total number of trials
 II is correct.
M
15.
2
13. For any event E, 0  P( E )  1 .
 I is correct.
M

P (one male one female)  16
30
8

15
The answer is D.
1st card
(c) Probability that the student is a member
of either one of the clubs
40  8  5

40
27

40

3
4
5
(2, 3) (2, 4) (2, 5)
3
(3, 2)
(3, 4) (3, 5)
4
(4, 2) (4, 3)
5
(5, 2) (5, 3) (5, 4)
(4, 5)
P(two numbers are even numbers)  2
12
1

6
P(two numbers are prime numbers)  6
12
1

2
 ‘The product of two numbers is an even
number’ will happen with the highest
probability.
 The answer is A.
16. Expected value of buying an item by Adams
 $( 0.2  10  0.5  8  0.3  6)
 $7.8
 The answer is C.
Chapter 7 Introduction to Probability
17. Expected value of the return after 6 months
 $[ 0.4  2 400  0.6  (1 800 )]
 $120
 The answer is D.
18. 1st question


2nd question
Correct
Wrong
Correct
2
0
Wrong
0
2
Probability of obtaining 1 mark or above
1

4
The answer is C.
19. Let the number of white balls in the bag be n.
P(white ball)  0.4
n
 0.4
30
n  12
 Number of white balls in the bag  12
 The answer is C.
20. Probability that the
glasses
35  30

35  15  30  20
13

20
 The answer is D.
student is
wearing
22. There are 100 integers from 1 to 100
inclusive within which all the integers
divisible by 5 are as follows.
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55,
60, 65, 70, 75, 80, 85, 90, 95, 100
 P (divisible by 5)  20
100
1

5
 The answer is B.
23.
1 cm
1 cm
P(distance between the point and centre not
more than 1cm)
Area of the circle with radius 1 cm

Area of the circle with radius 2 cm
  12

  22
 0.25
 The answer is A.
n
33 n
3
P(white ball) 
33 n
24. P(black ball) 
21.
1st dice
2nd dice


)
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
3
36
1

12
17
1
4

P(black ball)  P(white ball) 

n
3
1


33 n 33 n 4
n3 1

6n 4
4n  12  6  n
3n  18
n6

Total number of balls in the bag
 336
 12
The answer is A.
P(sum of numbers  10) 
The answer is C.

18
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
25. Let R denote red ball and B denote blue ball.
2nd ball
1st ball
R
R
B
B
B
RB
RB
RB
BB
BB
B
BR
B
BR
BB
B
BR
BB
BB
BB
Probability that the 2nd ball drawn is red in
colour
3

12
1

4
 The answer is C.
26. Number of male ex-classmates of Kiki
3
 56 
4
 42
Number of female ex-classmates of Kiki
3
 (15  1) 
7
6
 Probability that she talks with her
ex-classmate
42  6

56  (15  1)
24

35
 The answer is A.
Chapter 7 Introduction to Probability

Open-ended questions have been broadly
advocated for education in Hong Kong secondary
schools following the current curriculum reform.
In view of this, new strategies are required not
only for teaching mathematics but also for
assessing student’s performance. At this
transitional stage, we have introduced in the
‘Open-ended Question Zone’ of New Trend
Mathematics – Junior Form Supplementary
Exercises a simple assessment scheme for
open-ended questions. At the same time, a more
detailed assessment scheme is also provided in
the article ‘Ideas for Mathematics Teaching’ in
our New Trend Mathematics S1 to S3 Teacher’s
CD. Teachers may refer to these two assessment
schemes, or others which they may come across,
and choose to adopt an appropriate one
according to their need.
There may be 2 black balls and 6 white
balls in the box.
Scoring: 2 marks for correct answer;
2 marks for clear explanation.
3. To satisfy condition (ii), the areas of all
regions should satisfy the following
conditions.
Area of region I  Area of region II
 Area of region III  Area of region IV
 Area of region V
Consider the following circular target
where O is the centre.
II
I
180
O
40
III
30
Exercise of Open-ended Questions 7
20
(page 7.28)
1. If the number of vowels is greater than the
number of consonants in a word, then the
probability of choosing a vowel from this
word is greater than that of choosing a
consonant.
 ‘You’ and ‘about’ are two required words.
Scoring: 2 marks for correct answer;
2 marks for clear explanation.
2. Suppose there are x black balls and y white
balls in the box, then the probability of
y
picking up a white ball 
.
x y
Since Helen works 6 days a week and puts
dresses on for work 4 or 5 days a week on
average, the probability of picking a white
4
5
ball randomly should lie between
and .
6
6
Suppose the probability of picking a white
4  5
3
ball is 6 6  ,
2
4
y
3
then

x y 4
4 y  3x  3 y
y  3x
Take x  2, y  3( 2)
6
19
IV
V



There are 5 regions in the target.
The target satisfies condition (i).
Angle at the centre of region I
 Angle at the centre of region II
 Angle at the centre of region III
 Angle at the centre of region IV
 Angle at the centre of region V
 Obviously,
area of region I  area of region II
 area of region III  area of region IV
 area of region V
 The target satisfies condition (ii).
 The above target satisfies the conditions
required.
Consider the following circular target
where O is the centre.
I
II III IV V
O
20
New Trend Mathematics S3B — Junior Form Supplementary Exercises Solution Guide
 There are 5 regions in the target.
 The target satisfies condition (i).
Let the radius of region V be r.
Area of region V    r 2
 r 2
Area of region IV    (2r ) 2    r 2
 3r 2
Area of region III    (3r ) 2    (2r ) 2
 5r 2
Area of region II    (4r ) 2    (3r ) 2
 7 r 2
Area of region I    (5r ) 2    (4r ) 2



 9r 2
Area of region I  Area of region II
 Area of region III  Area of region IV
 Area of region V
The target satisfies condition (ii).
The above target also satisfies the
conditions required.
Scoring: 2 marks for correct answer;
2 marks for clear explanation.
4. Under the situation of the question, all
possible outcomes are (Minibus, Bus),
(Minibus, MTR), (Tram, Bus), (Tram, MTR),
(Bus, Bus), (Bus, MTR), (MTR, Bus) and
(MTR, MTR).
In these 8 possible outcomes, 6 involve
taking two different means of transport.
 P(taking two different means of
transport)
6

8
3

4
 Wendy’s calculation is not correct.
Scoring: 2 marks for correct answer;
3 marks for clear explanation.
5. Suppose there are three $100 banknotes and
three $20 banknotes in the wallet, then all
the possible outcomes are listed as follows.
$100
$100
$100
$100
$20
$20
$20
$100,
$100
$100,
$100
$100,
$20
$100,
$20
$100,
$20
$100,
$100
$100,
$20
$100,
$20
$100,
$20
$100,
$20
$100,
$20
$100,
$20
$20,
$20
$20,
$20
$100
$100,
$100
$100
$100,
$100
$100,
$100
$20
$20,
$100
$20,
$100
$20,
$100
$20
$20,
$100
$20,
$100
$20,
$100
$20,
$20
$20
$20,
$100
$20,
$100
$20,
$100
$20,
$20
$20,
$20
$20,
$20
P(two banknotes with the same face value)
12

30
2

5
2

3
 There may be three $100 banknotes and
three $20 banknotes in the wallet.
Scoring: 3 marks for correct answer;
3 marks for clear explanation.