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More about Probability
20
p.119
More about Probability
1.
The event of getting a ‘1’ and the event of getting a
prime number are mutually exclusive events.

P(a ‘1’ or a prime number)
= P(a ‘1’  a prime number)
= P(a ‘1’) + P(a prime number)
1 3
= 
6 6
4
=
6
2
=
3
pp.112 – 141
p.112
(a) {a, b, e, f }
(b)
{a, b, e, f , g, h}
(c)
{b, e, f , g, h}
(d)
{b, e}
(e)

(f)
{f}
(g)
{a}
(h)
{a, b, e}
2.
The event of getting an even number and the event of
getting a prime number cannot occur at the same time.

Events A and C are mutually exclusive events.
The event of getting a multiple of 3 and the event of
getting a prime number cannot occur at the same time.

Events B and C are mutually exclusive events.
(a)
Mutually exclusive events
(b)
Complementary events; Mutually exclusive
p.123
(i)
1.
events
{g , h}
p.113
1.
Mutually exclusive events
(d)
Complementary events; Mutually exclusive
When tossing a coin, the possible outcomes are {H, T}.
The favourable outcome = {H}.
1

P ( H) 
2
events
2.
2.
(c)
When throwing a dice, the possible outcomes are {1, 2,
failing the test are complementary to each other.
5 3
P(fail the test)  1  
8 8
3, 4, 5, 6}.
(a)
(b)
The event of Peter passing the test and the event of
The favourable outcome = {1}.
1

P(a ‘1’) 
6
p.131
1.
The favourable outcome = {2, 4, 6}.
3 1

P(an even number)  
6 2
59
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20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
2.
1
P(all tails)  ;
8
(iii)
P(red pen)
2

24
1

3
(i)
P(black pen)
1

1 4 1
1

6
(ii)
P(black pen)
1

2  3 1
1

6
(iii)
P(black pen)
0

24
0
(i)
P(first blue and second red)
4 2
 
7 6
4

21
(ii)
P(first black and second blue)
1 4
 
7 6
2

21
3
P(1 head and 2 tails)  ;
8
1
P(no tails)  .
8
(b)
p.134
1.
(a)
P(getting an even number) 
(b)
P(getting a head) 
(c)
2.
1
2
1
2
P(getting an even number and a head)
1 1
 
2 2
1

4
Since the result of rolling a dice does not affect the
result of selecting a card, the two events are
independent events.
(c)
P(getting an odd number and a ‘heart’)
(a)
 P(odd number)  P(heart )
1 1

2 4
1

8

(b)
P(getting a ‘2’ and an ‘Ace’)
 P(a ‘2’)  P(an ‘Ace’)
1 1
 
6 13
1

78
pp.113 – 152
p.141
(a)
(i)
20.1 (a)
P(red pen)
1

1 4 1
1

6
P(a blue ball) 
(b)
(ii)
P(red pen)
2

2  3 1
1

3
© Hong Kong Educational Publishing Co.
Since all blue balls are the favourable
p.113
outcomes,
20.2 (a)
Since there is no black ball in the bag,
0
0
P(a black ball) 
15
x  1000  (120  200  120  180  130)
 1000  750
 250
60
3
1

3 48 5
p.115
More about Probability
(b)
Experimental probability of getting a multiple
(b)
of 2
200  250  130

1000
580

1000
29

50
20.3 Total number of books = 20
20.7 (a)
p.119
P(either a comic or a computer book)
 P (comic )  P (computer book )
5
9

20 20
14

20
7

10

20.4 (a)
(b)
Total number of students
 25  16  30  4  18  28  24  5
 150
(c)
P(from class B and fails the examination)
20  12

50
8

50
4

25
p.120
 P(a girl with blood typeO)
25
5


150 150
30

150
1

5
20.8 (a)
P(first spin is an odd number) 
P(third spin is an odd number) 

p.122
 P(female worker who lives in the New Territories)
400 350 170



950 950 950
580

950
58

95
(b)
P(first spin is ‘2’ or ‘5’) 
P(third spin is ‘2’ or ‘5’) 
p.124
61
3
5
3
5
2
5
P(second spin is ‘2’ or ‘5’) 

p.135
P(getting an odd number in all 3 times)
3 3 3
  
5 5 5
27

125
P(female worker or lives in the New Territories)
 P(female worker)  P(lives in the New Territories)
P(less than 3) = P(a ‘1’) + P(a ‘2’)
1 1
 
5 9
14

45
3
5
P(second spin is an odd number) 
New Territories
 350  180
 170
20.6 (a)
p.125
P(from class A and fails the examination)
30  18

50
12

50
6

25
P(either a boy with blood type A or a girl with
Number of female workers who live in the
P(from class B)
50  30

50
2

5
(b)
blood type O)
 P(a boy with blood typeA)
20.5
P(not less than 3) = 1 – P(less than 3)
14
 1
45
31

45
2
5
2
5
P(getting ‘2’ or ‘5’ in all 3 times)
2 2 2
  
5 5 5
8

125
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
20.9 (a)
Use ‘B’ to denote a boy and ‘G’ to
p.136
(b)
denote a girl.
P(a red marble is chosen from A)
3

5
P(a red marble is chosen from B after
a red marble is chosen from A)
3 3 3
  
5 6 10
P(a blue marble is chosen from A)
2

5
P(a red marble is chosen from B after
a blue marble is chosen from A)

(b)
20.10 (a)

P(different genders)
2 2 1 3
   
3 5 3 5
7

15
P(exactly one of them is rotten)
p.137
 P(an egg is rotten)  P(an egg is not rotten)
(b)
P(black | odd) 
P(at least one egg is rotten)
 1  P(all eggs are not rotten)
 1  (0.97) 3
(cor. to 3 sig. fig.)
20.11 P(first letter is a vowel) 
3 1

9 3
p.142
P(second letter is a vowel after the first one is a
vowel)
2

8
1

4
P(both letters are vowels)
20.14 (a)
P(a black marble)
2
 1
6
1

3
(b)
P(a female smoker is chosen)
 P(a female )  P(smoker a female )
P(a male smoker is chosen)
 P(a male )  P(smoker a male )
 0.52  (1  0.8)
 0.104
p.142
(c)
P(a smoker is chosen)
 P(a female smoker )  P(a male smoker )
 0.072  0.104
 0.176
© Hong Kong Educational Publishing Co.
62
p.144
P ( black  odd)
P (odd)
2
8

3
8
2

3
 (1  0.52)  0.15
 0.072
1 1
 
3 4
1

12
20.12 (a)
P (odd  black )
P ( black )
2
 8
5
8
2

5
P(odd | black) 
 0.03  0.97  0.97  3
 0.0847 (cor. to 3 sig. fig.)
 0.0873
P(a red marble)
3
2


10 15
13

30
20.13 (a)
 P(an egg is not rotten)  C13
(b)
2 2 2
 
5 6 15
p.145
More about Probability
20.15 By the multiplication law,
p.150
Example 20.2T
P(3 Chinese books)
 P(1st Chinese)  P(2nd Chinese)
 P(3rd Chinese)
4 3 2
 
9 8 7
1

21
(a)
Total number of students
 10  5  12  7  6
 40
(b)
Experimental probability of choosing badminton
7

40

20.16 Total number of permutations of 3 letters
out of 6 letters  P36  120
p.151
Example 20.3T
If the last letter is ‘D’, then the number of permutations
of the other two letters  P25  20
P(the last letter is ‘D’) 
(a)
P(even number) 
(b)
P(get a ‘5’) 
 P(even number) + P(get a ‘5’)
1 1
 
2 6
2

3
Example 20.4T
20.18 Total number of ways to select 5 students  C514 p.152
(a)
P(blood type B)
16  28

150
22

75
(b)
P(either blood type AB or O)
30  24  4  5

150
63

150
21

50
Number of ways to select 3 girls  C38
Number of ways to select 2 boys  C 26
P(3 girls and 2 boys)
C8 C6
 3 14 2
C5
60
143
pp.113 – 152
Example 20.1T
(b)
(c)
1
6
P(either an even number or a ‘5’)
(c)
P(same kind)  P ( A  B )
 P ( A)  P ( B )
1
5


21 42
1

6
(a)
3 1

6 2
p.152
B be the event of getting 3 lemons.
C4
C5
1
5
Then P ( A)  39 
and P ( B )  39 
C3 21
C3 42

p.119
20
1

120 6
20.17 Let A be the event of getting 3 oranges and

p.115
p.113
Example 20.5T
Since 3 and 6 are the favourable outcomes,
2 1
P(a multiple of 3)  
6 3
p.120
p.122
P(buy both brand A and brand N)
 (1  0.2)  0.6
 0.48
P(buy brand A or brand N)
 P(buy brand A)  P(buy brand N )
Since all the numbers on the dice are less than 10,
6
P(a number less than 10)   1
6
 P(buy both brand A and brand N )
 0.8  0.6  0.48
 0.92
 92%
Since there is no two-digit number on the dice,
0
P(a two-digit number)   0
6
63
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
Example 20.6T
(a)
(b)
p.124
Example 20.9T
P(Hong Kong or China will win)
2 3
 
7 5
31

35
(a)
p.136
Use ‘G’ to denote wearing glasses and ‘N’ to denote not
wearing glasses.
P(a draw)
 1  P(Hong Kong or China will win )
 1

31
35
4
35
Example 20.7T
p.125
(b)
(i)
14
35
2

5
(a)
P(has notebook computer) 
(b)
P(has desktop computer only) 
P(both students wear glasses)
 P(GG)
30 25

40 35
15

28

15  7
35
8

35
(ii)
P(one student wears glasses and one student
does not)
 P(GN)  P( NG)
30 10 10 25
 

40 35 40 35
11

28

(c)
P(has no computer)  1 

2 8

5 35
13
35
Example 20.10T
Example 20.8T
(a)
(b)
p.134
2 1

6 3
1
P(‘C’ from ‘CHAMPION’) 
8
1 1

P(both are ‘C’)  
3 8
1

24
(a)
P(one of them cannot function properly)
 0.02  (1  0.02)  2
 0.0392
(b)
P(both of them function properly)
 (1  0.02)  (1  0.02)
 0.9604
P(‘C’ from ‘SOCCER’) 
P(‘O’ from ‘SOCCER’) 

Example 20.11T
1
6
P(‘O’ from ‘CHAMPION’) 
P(first box has no present inside) 
1
8
p.141
9
10
P(second box has a present inside after the first box has no
1
present inside) 
9
1 1

6 8
1

48
P(both are ‘O’) 
© Hong Kong Educational Publishing Co.
p.137
P(gets the present at the second selection)
9 1
 
10 9
1

10
64
More about Probability
Example 20.12T
(a)
p.142
Example 20.14T
4 3

9 7
4

21
 4  5
P(boy failing the test)  1    1  
 9  6
5

54
4
5

P(student failing the test) 
21 54
107

378
P(1st card is a ‘spade’)
13

52
P(girl failing the test) 
P(2nd card is a ‘spade’ after the 1st card is a ‘spade’)
12

51
13 12 1
 

P(two ‘spades’) 
52 51 17
(b)
p.144
P(1st card is a ‘heart King’)
1

52
P(2nd card is a ‘heart’ after the 1st card is a ‘heart
Example 20.15T
King’)
12

51
Total number of children = 6 + 4 = 10
Number of ways to select 2 children out of 10  C210
P(1st card is a ‘King’ but not ‘heart King’)
3

52
(a)
Number of ways to get 2 boys out of 6 boys  C 26

P(2nd card is a ‘heart’ after the 1st card is a ‘King’
P(get 2 boys) 
but not ‘heart King’)
13

51

(a)
C 26
C 21 0
15
45
1

3

P(1st is a ‘King’ and 2nd is a ‘heart’)
1 12 3 13

  
52 51 52 51
1

52
Example 20.13T
p.150
(b)
Number of ways to get 2 girls out of 4 girls  C 24

P(get 2 girls) 
C 24
C 21 0
6
45
2

15

p.143
P(spade  picture)
P ( picture)
3
 52
12
52
1

4
P(spade  picture) 
Example 20.16T
p.151
Total number of permutations of 4 digits out of 6 digits
 P46
 360
If a number is divisible by 5, then the last digit is ‘5’.
(b)

P( picture  spade)
P(picture  spade) 
P(spade)
3
52

13
52
3

13
Number of permutations for the other three digits
 P35  60
60
360
1

6
P(the number is divisible by 5) 
65
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
Example 20.17T
4.
p.151
(a) Only 6 and 9 are contained in both A and B.

A  B = {6, 9}
Let R be the event of getting 3 red balls, G be the event of
getting 3 green balls and B be the event of getting 3 blue
(b) The elements contained in A but not in B are 2, 7
balls.
and 10.
C3
1
Then P ( R )  39  ,
C3 84

P (G ) 
C 34
4
1


and
C39 84 21
P( B) 
0
0
C39
5.
A \ B = {2, 7, 10}
X Y = {2, 4, 5, 6, 7, 8, 9, 10}
Only 6, 8 and 10 are contained in both X and Y.
 XY = {6, 8, 10}
The elements contained in X but not in Y are 2 and 4.
P(same colour)  P ( R  G  B )
 P ( R )  P (G )  P ( B )
1
1


0
84 21
5

84
 X \ Y = {2, 4}
6.
A  C = {1, 2, 5, 6, 8, 10, 12, 13}
The elements contained in B but not in C are 4, 5 and
10.
 B \ C = {4, 5, 10}
Example 20.18T
Only 5 and 10 are contained in both A and B.
p.152
 A  B = {5, 10}
Total number of ways to select 4 eggs  C 424
Number of eggs that are rotten = 4
7.
Since all blue balls are the favourable outcomes,
6
6 2


P(blue) 
7  6  8 21 7
8.
Since all reference books are the favourable outcomes,
12
12 2


P(a reference book) 
2  12  4 18 3
9.
(a)
Since all boys are the favourable outcomes,
25 5

P(a boy) 
40 8
(b)
Since all girls are the favourable outcomes,
40  25 15 3


P(a girl) 
40
40 8
(a)
Since 2, 4 and 6 are the favourable outcomes,
3 1
P(a multiple of 2)  
6 2
(b)
Since 1 and 2 are the favourable outcomes,
2 1
P(a number less than 3)  
6 3
Number of eggs that are not rotten = 24 – 4 = 20
C4
(a) P(all of them are rotten)  244
C4

(b)
P(2 of them are rotten) 

1
10 626
C 24 C 22 0
C 42 4
190
1771
20.1
pp.115  117
10.
p.115
1.
(a) spring, summer, autumn, winter
(b) 2, 3, 5, 7, 11, 13, 17, 19
2.
(a) male, female
11.
Since the red region is a quarter of a circle, P(red) 
12.
(a)
(b) 4, 8, 12, 16, 20, 24, 28
3.
S  T = {bus, lorry, car, bike, taxi, plane}
Since all 26 black cards are the favourable
S  T = {bus, lorry}
outcomes,
S \ T = {car, bike}
P(a black card) 
T \ S = {taxi, plane}
© Hong Kong Educational Publishing Co.
66
26 1

52 2
1
4
More about Probability
(b)
Since all 4 ‘King’ are the favourable outcomes,
4
1


P(a ‘King’) 
52 13
(c)
Since all 13 ‘heart’ are the favourable outcomes,
13 1


P(a ‘heart’) 
52 4
17. Let x be the number of pairs of white socks into the box.
1
(a) Since P(white)  ,
3
x
1

4 x 3
3x  4  x
2x  4
x2

13.
P(3 passengers or less)
12  16  10

12  16  10  7  5
38

50
19

25
18.
There are 2 pairs of white socks in the box.
5
5

4  2  5 11
(b)
P(black) 
(a)
Total number of red packets
 31  16  9  6  2
 64
(b)
Since ‘$50’ is the favourable outcome,
16 1


P($50) 
64 4
(c)
Since ‘$500’ and ‘$1000’ are the favourable
p.116
14.
(a)
n  12
6
n
n  12  6n
5n  12
n
12
5

12
is not a natural number,
5

6 is not an element of the set.
outcomes,

19.
(b)
Since
(a)
n  12
12 n  12
 1
,
is an integer if 12
n
n
n
P(more than $145) 
Since P(red) 
62 1

64
8
1
, we have
4
3
1

3 2 x 4
12  5  x
x7
is divisible by n. Therefore, n can only be 1, 2, 3,
4, 6 and 12. The integral elements are:
2, 3, 4, 5, 7, 13 (any three).
15.
(a)

(b)
(b)
Only ‘2’ is contained in both A and B.
P(black) 
7 1
8

3  2  7  4  1 17
A  B has only 1 element.
20.2
Only 2, 4, 6, …, 98 are contained in both A and
C.

16.
B  C has 25 elements.
(a)
Since ‘1’ and ‘3’ are the favourable outcomes,
12  11 23


P(a ‘1’ or a ‘3’) 
60
60
(b)
Since ‘3’ and ‘6’ are the favourable outcomes,

p.126
A  C has 49 elements.
(c)
pp.126  130
1.
Total number of discs  18  36  26
 80
P(either a music disc or a movie disc)
 P(music disc)  P(movie disc)
18 26


80 80
44

80
11

20
P(a number divisible by 3)
11  11

60
11

30
67
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
2.
Total number of letters of the word ‘MATHEMATICS’
(c)
= 11
(a)
(b)
3.
outcomes,
P(grade D or below)  0.18  0.10  0.08
Total number of ‘O’ = 0
0
P(‘O’)   0
11
 0.36
Alternative Solution
P(grade D or below)  1  P(grade C or above)
Total number of ‘A’ and ‘T’ = 4
4
P(either ‘A’ or ‘T’) 
11
 1  0.64
 0.36
8.
Total number of balls = 8 + 5 + 7 = 20
P(either red or black)
 P(red )  P(black )
8
7


20 20
15

20
3

4
4.
Since grades D, E and F are the favourable
P(joins the Music club or Science club)
 P(Music club )  P(Science club )  P(both clubs )
 0.2  0.15  0.02
 0.33
9.
P(buys both chicken and duck)
 P( buys chicken )  P (buys duck )
 P (buys either chicken or duck )
1 1 2
  
2 4 5
7

20
10.
(a)
P(either an ‘Ace’ or a ‘King’)
 P(Ace)  P(King)
4
4

52 52
8

52
2

13

Since 2, 4, 6, …, 24 are the favourable outcomes
of A, there are 12 elements in A.
12
P ( A) 

25
Since 2, 3, 5, 7, 11, 13, 17, 21, 23 are the
5.
(a)
P(a ‘4’) 
favourable outcomes of B, there are 9 elements
1
6
in B.

(b)
6.
Since 2, 3, 4 and 6 are the favourable outcomes,
4 2
P(a multiple of 2 or 3)  
6 3
(a)
(b)
P( A  B)  P( A)  P( B)  P( A  B)
12 9
1



25 25 25
20

25
4

5
(a)
Since A contains 4 elements, P ( A) 
P(either an ‘L’ or ‘I’) = P(‘L’) + P(‘I’)
2 2
 
9 9
4

9
11.
7.
9
25
Only 2 is contained in both A and B.
1
P( A  B) 

25
Total number of letters of the word ‘BRILLIANT’ = 9
3 1
(a) P(a vowel)  
9 3
(b)
P( B) 
x  1  (0.12  0.3  0.18  0.10  0.08)
 0.22
4
1

52 13
Since B contains 13 elements, P( B ) 
13 1

52 4
Only ‘Ace of hearts’ is contained in both A and
(b)
Since grades A, B and C are the favourable
B.
outcomes,
P(grade C or above)  0.12  0.3  0.22

 0.64
© Hong Kong Educational Publishing Co.
68
P( A  B) 
1
52
More about Probability
(b)
P( A  B)  P( A)  P( B)  P( A  B)
(c)
1 1 1
  
13 4 52
4

13
12.
P(sum less than 7)
 1  P(sum of 7)  P(sum greater than 7)
 1
5
12

Total number of fruit = 6 + 2 + 1 = 9
P(not a pear)  1  P(a pear)
17.
1
 1
9
8

9
Total number of two-digit numbers = 99 – 9 = 90
(a)
Total number of two-digit numbers less than 40
= 39 – 9 = 30

P(less than 40)
30

90
1

3
Alternative Solution
P(not a pear)  P(an apple)  P(an orange )
6 2

9 9
8

9

13.
(b)
Total number of multiples of 5 = 18

P(a multiple of 5)
18

90
1

5
P(not delayed)  1  P(delayed)
 1  0.13
 0.87
14.
P(fail the test)  1  P(pass the test)
(c)
 1  0.77
 0.23
15.
P(not a multiple of 5)
 1  P(a multiple of 5)
 1

P(on time) = 0.38, P(early) = 0.48.
(a)
P(not late)  P(on time)  P(early )
 0.38  0.48
 0.86
(b)
1 5

6 12
18.
1
5
4
5
(a)
P(either one of them will win)
 P (James win )  P (Patrick win )
 P (Andy win)
1 2 3

 
10 9 8
251

360
(b)
P(none of them will win)
 1  P(either one of them will win )
P(late)  1  P(not late)
 1  0.86
 0.14
p.128
16.
(a)
Since (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1)
251
360
109

360
are the favourable outcomes,
6
P(sum of 7) 
36
1

6
 1
19.
(b)
15
36
5

12
P (sum greater th an 7) 
(a)
Let x be the number of cards marked with a ‘8’,
1
Since P(‘8’)  , we have
4
x 1

8 4
x2

69
The value of y is 8.
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
(b)
Total number of cards = 8 + 2 = 10
(i)
23.
(a)
P(male or Mathematics teacher)
 P(male )  P)(M athematics teacher )
 P (M athematics male teacher )
38 2 8

 
70 7 70
5

7
(b)
Number of female Mathematics teachers
2
 70   8
7
 12
Since 8, 8, 8, 9, 10 are the favourable
outcomes.
P(greater than 6) 
(ii)
20.
(a)
5 1

10 2
Since 3 and 9 are the favourable outcomes,
2 1
P(an odd number ) 

10 5
P(coke) 
6
 0.3
20
P(female or Mathematics)
 P(female )  P(M athematics)
 P(female M athematics teacher )
32 2 12

 
70 7 70
4

7
P(coke or orange juice)
 P(coke)  P(orange juice )
 0.3  0.2
 0.5
(b)
P(not coke)
 1  P(coke)
 1  0.3
 0.7
21.
(a)
(b)
(c)
32  2  32  12
 1   
70  7 
70
32 5 20

 
70 7 70
31

35
x  1  (0.3  0.45  0.1)
 0.15
(i)
P(female or not teaching Mathematics)
 P(female )  P(not teaching M athematics)
 P(female teacher not teaching M athematics)

P(either a ‘$2’ or a ‘$5’ coin)
 P(‘$2’) +P(‘$5’)
 0.45  0.15
 0 .6
24.
(ii)
(a)
16
400
1

25
(i)
P ( rotten) 
(ii)
P (normal )  1  P(rotten)
Since ‘$5’ and ‘$10’ are the favourable
outcomes,
P(greater than $2)
 0.15  0.1
 0.25
22.
 1

Number of female who are left-handed
= 120 – 48
1
25
24
25
= 72
(b)
P(a female or left-handed)
(i)
Expected number of normal eggs
24

 75
25
 72
(ii)
Expected profit he can make
 $(72  0.9)  $45
 P(a female) + P(left-handed)
 P(a left-hand female)
475 120
72


1000 1000 1000
523

1000
 0.523

 $19.8
Expected profit percentage
19.8
100%
45
 44%

© Hong Kong Educational Publishing Co.
70
More about Probability
25.
(a)
Area of the cake  (6) 2 cm 2
 36  cm
(b)
(i)
(5  2r ) 2  (3  2r ) 2
4

81
(12  2r ) 2
(b)
2
8r 2  32r  34
4

2
4r  48r  144 81
81(8r 2  32r  34)  4(4r 2  48r  144)
Area of a circle with radius 2 cm
316r 2  1200r  1089  0
= 4 cm2
P(within 2 cm of the centre)
4

36 
1

9
(ii)
r
 (1200)  (1200) 2  4(316)(1089)
2(316)
729
r  1.5 or
(rejected)
316
(
r cannot be greater than 1.5)
Area of a circle with radius 4 cm
pp.137  140
20.3
= 16 cm2
P(at least 4 cm away from the centre)
 1  P( within 4 cm from the centre)
p.137
16
36
20

36
5

9
 1
26.
1.
Since 2, 3 and 5 are the favourable outcomes,
3 1
P(first one is a prime number)  
6 2
1
P(second one is a prime number) 
2
1
P(third one is a prime number) 
2
1 1 1 1

P(three prime numbers)    
2 2 2 8
2.
P(a boy from A) 
The centre of the coin must lie within the squares
ABCD or PQRS such that the coin lies entirely on the
coloured region.
1
2
2
P(a boy from B) 
3

3.
P(two boys) 
1 2 1
 
2 3 3
The first school can choose any month, second school
only can choose the month the first school selected.
1
1

P(the same month)  1 
12 12
4.
(a)
Case 1:
P(an orange from A) 
P(lie entirely on the coloured region)
[5  2(1)]2  [3  2(1)]2
[12  2(1)]2
9 1

100
1

10
 0.1
P(a mango from B) 

3
4
2
3
Case 2:
1
4
1
P(an orange from B) 
3
P(a mango from A) 

71
P(two different types of fruit)
3 2 1 1
   
4 3 4 3
7

12
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
5.
(a)
26 1

52 2
26 1

P(second card is black) 
52 2

(b)
Use ‘B’ to denote bad and ‘G’ to denote not bad.
P(both cards are black)
1 1
 
2 2
1

4
4
1

52 13
36 9

P(second card is a ‘number’) 
52 13
P(first card is an ‘Ace’) 

6.
7.
P(first card is black) 
P(1st is ‘Ace’, 2nd is ‘number’)
1 9
 
13 13
9

169
Use ‘T’ to denote pass and ‘F’ to denote fail.
8.
(a)
P(both are bad)
 0.04  0.03
 0.0012
(b)
P(apple is bad but orange is not bad)
 0.04  0.97
 0.0388
P(red) = 0.4
P(green) = 1 – 0.4 = 0.6
(a)
P(only the first light is red)
 P(red )  P(green )  P(green )
 0.4  0.6  0.6
 0.144
(b)
(a)
(b)
P(all of them are green)
 P(green )  P(green )  P(green )
 0.6  0.6  0.6
 0.216
P(pass both papers)
2 5
 
5 8
1

4
p.138
9.
P(pass only one paper)
2 3 3 5
   
5 8 5 8
3 3


20 8
21

40
(a)
P(none of them is ‘6’)
 1  1
  1    1  
 6  6
5 5
 
6 6
25

36
(b)
Since (3, 6), (4, 5), (5, 4) and (6, 3) are the
favourable outcomes,
4
P(the sum is 9) 
36
1

9
© Hong Kong Educational Publishing Co.
72
More about Probability
(c)
Since (6, 5), (6, 4), (6, 3), …, (2, 1) are the
(b)
favourable outcomes,
9 
 2 
  1    1  
 5   20 
3 11
 
5 20
33

100
P(1st number is greater than 2nd number)
15

36
5

12
10.
(a)
P(1st is blue) 
2 1

6 3
(c)
2 1
P(2nd is blue)  
6 3
P(3rd is biue) 

(b)
P(red) 

P(at least one of them hits the target)
 1  P(both of them miss the target )
33
100
67

100
 1
2 1

6 3
1 1 1
P(all are blue)   
3 3 3
1

27
P(blue) 
P(both of them miss the target)
 P(Leo misses)  P(Jack misses)
12.
(a)
2 1

6 3
1
6
P(two are blue and one is red)
1 1 1
   3
3 3 6
1

18
3
(c)
1
1
P(all are blue)    
27
 3
3
1
1
P(all are red)    
216
6
3
1 1
P(all are white)    
2 8

11.
(a)
(b)
P(the same colour)
1
1
1



27 216 8
1

6
(i)
Since (6, 9), (8, 8), (8, 9), (9, 6), (9, 8) and
(9, 9) are the favourable outcomes,
P(the sum is greater than 14)
6

16
3

8
P(Leo hits while Jack misses)
 P(Leo hits)  P(Jack misses)
2 
9 
 1  
5  20 
2 11
 
5 20
11

50

73
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
(ii)
15.
Since (3, 6), (3, 8), (6, 3), …, (9, 8) are the
(a)
favourable outcomes, there are 12
elements.
P(the product is even)
12

16
3

4
P(only two of them have mobile phones)
 P(has a mobile phone)  P(has a mobile
phone)  P(does not have a mobile phone)  3
 0.68  0.68  (1  0.68)  3
 0.444 (cor. to 3 d. p.)
(b)
P(none of them has a mobile phone)
 P(does not have mobile phone)
 P(does not have a mobile phone)
(iii)
 P(does not have a mobile phone)
Since (3, 3), (6, 6), (6, 8), (8, 6), (8, 8)
(8, 9), (9, 8) and (9, 9) are the favourable
 (1  0.68)  (1  0.68)  (1  0.68)
outcomes,
 0.033 (cor. to 3 d. p.)
P(the difference is less than 3)
8

16
1

2
13.
(a)
(c)
P(at least one of them has a mobile phone)
 1  P(none of them has a mobile phone)
 1  0.032768
 0.967 (cor. to 3 d. p.)
P(Bobby can but Calvin cannot)
 0.8  (1  0.65)
16.
(a)
P(MTR or mini-bus)
 P(MTR) + P(mini-bus)
 0.18  0.06
 0.8  0.35
 0.28
 0.24
(b)
P(only one of them can)
 0.28  (1  0.8)  0.65
(b)
(i)
 0.28  0.13
 0.41
(c)
14.
 0.3  0.3
 0.09
P(at least one of them can)
 1  P(both of them cannot )
 1  (1  0.8)(1  0.65)
 1  0.07
 0.93
(a)
P(all of them wear seat belts)
 P ( wear seat belt )  P ( wear seat belt )
 P ( wear seat belt )
 0.9  0.9  0.9
 0.729
(b)
P(only one of them wears a seat belt)
 P( wear seat belt )  P(not wear seat belt )
 P(not wear seat belt )  3
 0.9  (1  0.9)  (1  0.9)  3
 0.027
© Hong Kong Educational Publishing Co.
P(both by bus)
 P(bus)  P(bus)
(ii)
P(one by bus and the other by car)
 P(bus)  P(car )  2
 0.3  0.08  2
 0.048
(iii)
74
P(neither of them walk to school)
 [1  P(on foot )]  [1  P(on foot )]
 (1  0.25)  (1  0.25)
 0.75  0.75
 0.5625
More about Probability
4.
pp.145  149
20.4
4
45
P(first one is a boy) 
(a)
4
9

p.145
P(second one is a boy after the first one is a
1.
Since there are 2 ‘O’,
2 1
P(1st is ‘O’)  
6 3
boy)
3
35
3

8

P(2nd is ‘O’ after 1st is ‘O’) 
1
5
P(both of them are boys)
4 3
 
9 8
1

6
P(two letters are the same)
1 1
 
3 5
1

15
2.
(a)
P(first one is a football) 
6
3

64 5
(b)
Case 1:
P(first one is a girl) 
P(second one is a football after the first one is
a football)

5
54
5

9

5
45
5
9
P(second one is a boy after the first one is a girl)
4

44
1

2
P(both of them are footballs)
3 5
 
5 9
1

3
Case 2:
P(first one is a boy) 
4
9
P(second one is a girl after the first one is a boy)
(b)
6
3

P(first one is a football) 
64 5
5
35
5

8

P(first is football and second is volleyball)
3 4
 
5 9
4

15
3.
P(different genders)
5 1 4 5
   
9 2 9 8
5

9
Case 1:
P(rainy tomorrow) 
3
4
5.
1
P(go to library on a rainy day) 
5
P(at least one card is even)
 1  P(both cards are odd)
 1  P(first card is odd)  P(second card is odd after
Case 2:
the first card is odd)
3 1
P(sunny tomorrow)  1  
4 4
P(go to library on a sunny day) 
 3  3  1 
 1   

 5  5  1 
7

10
3
5
3 1 1 3
  
4 5 4 5
3

10
P(go to library) 
75
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
6.
(a)
P(first one is rotten) 
2
1

20 10
(b)
P(boy  left )
P(left )
4

43
4

7
P(boy  left) 
P(second one is rotten after the first one is
rotten)

1
19
1 1

10 19
1

190
P(both are rotten) 
(b)
8.
P(prime  even)
P(prime  even )
P(even )
1
 6
3
6
1

3

Case 1:
P(second one is not rotten after the first one is
rotten)
18

19
Case 2:
9.
P(multiple of 5  yellow)
P( yellow)
2
9

5
9
2

5

rotten)
2

19
P(one is not rotten and the other is rotten)
1 18 
1 2
  1   
10 19  10  19
18 18


190 190
18

95

(c)
P(multiple of 5  yellow)
(a)
P(second one is rotten after the first one is not
P(green  even)
(b)
P(green  even )
P(even )
2
9

5
9
2

5

P(second one is not rotten after the first one is
not rotten)
17

19
P(both are not rotten)
1  17

 1   
10

 19
153

190
p.147
10.
7.
(a)
P(left  boy)
P(boy)
4

4  16
1

5
(a)
P(1st is ‘King’ and 2nd is ‘Queen’)
4 4


52 51
4

663
(b)
P(both of them are face cards)
12 11


52 51
11

221
P(left  boy) 
© Hong Kong Educational Publishing Co.
76
More about Probability
11.
(c)
P(none of them is an ‘Ace’)
48 47


52 51
188

221
(b)
P(none of $20 is drawn)
7 6
 
9 8
7

12
(a)
Case 1:
(c)
P(second is greater than the first)
 P ($10, $20)  P ($10, $50)  P($20, $50)
3
P(a red ball is drawn from A) 
5
4 2 4 3 2 3
    
9 8 9 8 9 8
8  12  6

72
13

36

P(a red ball is drawn from B after a red ball is
drawn from A)
2
5

Case 2:
P(a white ball is drawn from A) 
2
5
13.
(a)
(2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1),
P(a red ball is drawn from B after a white ball
(4, 2) and (5, 1) are the favourable outcomes,
is drawn from A)
1

5

(b)

12.
(a)
there are 15 elements.
15
36
5

12
P(less than or equal to 6) 
P(the last ball drawn is red)
3 2 2 1
   
5 5 5 5
6
2


25 25
8

25
P(any ball is drawn from A) 
Since (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1),
P(same  less than or equal to 6)
(b)
P(same  less than or equal to 6)
P(less than or equal to 6)
3
36

15
36
1

5

5
1
5
P(the last ball drawn is black)
 P(a black ball is drawn from B after
any ball is drawn from A)
2
 1
5
2

5
14.
(a)
Since ($10, $10), ($10, $20) and ($20, $10) are
(b)
not the favourable outcomes,
P(greater than $30)
4 3 4 2 2 4
 1       
9 8 9 8 9 8
1 1 1
 1  
6 9 9
11

18
(c)
77
1 1
P(point B)  
3 2
1

6
1 1 1 1
P(point C)    
3 2 3 2
1

3
1 1 1
 
3 2 2
1

12
P(the aquarium) 
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
15.
Number of users who have installed software = 50 000
(b)
(i)
P(both balls are blue)
1 1
 
5 5
1

25
Number of users who have not install software
= 800 000 – 50 000 = 750 000
P(the user will get the junk mail)

50 000  0.1  750 000  0.8
800 000
(ii)
 0.756 (cor. to 3 sig. fig.)
P(both are red and come from the same
bag)
 P (both red balls from A)
16.
(a)
 P (both red balls from C )
1 2 1 2 1  1 
           1    1 
3 5 3 5 3  3 
29

225
18.
(a)
P(Peter loses on the first pick) 
(b)
P(ends before the third pick)
 P(ends in the 1st pick)
1
6
P(late for work on a particular day)
5 4  5 1
  1   
8 5  8 9
13

24

 P(ends in the 2nd pick)
1 5 1
  
6 6 5
1

3

(b)
17.
(a)
P(late for work on two particular days)
13 13


24 24
169

576
(i)
(c)
5 4 3 2
  
6 5 4 3
1

3

P(a blue ball)
 P(get a blue ball from Bag B)
1
3
 
3 23
1

5
(ii)
(iii)
P(a white ball)
 P( white ball from A)
 P ( white ball from B )
1
3
1
2
 
 
3 23 3 23
1

3
19.
(d)
P(John loses on the sixth pick)
5 4 3 2 1 1
     
6 5 4 3 2 1
1

6
(a)
(i)
P(a teacher)
 P(a teacher from Room B)
1 1
 
3 2
1

6
P(a red ball)
 P(red ball from A)  P(red ball from C )
1
2
1
 
 1
3 23 3
7

15
© Hong Kong Educational Publishing Co.
P(the game continues until the fifth pick)
 P( the game does not end on the fourth pick)
78
More about Probability
(ii)
(c)
P(a student)
 P (a student from Room A)
 P (a student from Room B )
1 2 1 1
   
3 3 3 2
7

18
 1

(d)
(b)
(iii)
P(a social worker)
 P(a social worker from Room A)
 P(a social worker from Room C )
1 1 1
   1
3 3 3
4

9
(i)
P(both of them talk to a teacher)
1 1
 
6 6
1

36
(ii)
1.
pp.152  155
Total number of ways to select 2 boys and 2 girls
 C 24
Number of ways to select the gender first  C12
social worker in Room C )

P(they are sitting alternately)

C12
C 24
2
6
1

3

Let x be the number of mint chocolates in box B.
1
x
1 8 5
  
P(mint)  
3 6 x 3 8 9
2.
1 x
 5
 1 

3 6 x  9
x6 x 5

6 x
3
6 x  18  30  5 x
x  12
(b)
P ( box A  white)
P ( white)
1 8

 3 12
1
3
2
 9
1
3
2

3
P(box A  white) 
p.152
1 1 1 1 1  1 
          1   1
 3 3 3 3  3  3 
10

81

1
9
P(both of them talk to a social worker in
in Room A)  P(both of them talk to a
(a)
5 1

9 3
20.5
the same room)
 P(both of them talk to a social worker
20.
P(milk)  1  P(mint )  P( white)
(a)
Total number of ways to select 2 letters  C 226
Number of ways to select ‘A’ and ‘T’  C 22
P(‘A’ and ‘T’) 
C 22
C 22 6

1
325
There are 12 mint chocolates in box B.
(b)
1
8
1
6
 
P(white)  
3 8  4 3 6  12
2 1
 
9 9
1

3
Number of ways to select 2 vowels  C 25
P(two vowels) 
C 25
C 22 6
10
325
2

65

79
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
3.
Total number of ways to choose 3 fruit  C39
7.
Number of ways to choose an apple  C13
Number of ways to select ‘T’, ‘O’ and ‘M’  C33
Number of ways to choose a pear  C11

Number of ways to choose a mango  C15

P(get the word ‘TOM’) 

P(get an apple, a pear and a mango)

C13  C11  C15
C39
8.
15
84
5

28

4.
Number of ways to arrange 3 letters  P36
C33
P36
1
120
Number of ways to arrange 3 boys  P38
Number of ways to select Jason, Bobby and Eric  C33

Total number of ways to choose 5 cards  C552
P(in the order of Jason, Bobby and Eric)

C 33
P38

1
336
Since only 4 kinds of royal flush can be formed,
number of ways to choose royal flush  C14

P(get a royal flush) 
C14
C552
9.
digits
4
2 598 960
1

649 740
 P35

5.
Total number of permutations of 3 digits out of 5
 60
If the tens digit is ‘1’, then the number of permutations
for the other two digits  P24  12
P(the tens digit is ‘1’) 
Total number of children = 6 + 4 = 10
12 1

60 5
Number of ways to select 2 children  C210
Number of ways to select 2 boys  C 26

10.
C6
P(select 2 boys)  120
C2
If the first letter is ‘M’, then the number of
permutations for the other 3 letters  P36  120
15
45
1

3

6.
Total number of permutations of 4 letters out of 7
letters  P47  840
P(first letter is ‘M’) 
120 1

840 7
Number of even numbers from 1 to 11 = 5
Number of ways to select 2 numbers  C211
11.
Number of ways to select 2 even numbers  C 25

P(get 2 even numbers) 
Number of ways to select 2 people to have intensive
training  C211
C 25
C 21 1

10

55
2

11
© Hong Kong Educational Publishing Co.
Total number of ways to select 2 people  C 220
P(both Andy and Billy are chosen)

C 21 1
C 22 0
55
190
11

38

80
More about Probability
15.
p.153
12.
(a)
Total number of ways to select 4 TV sets  C 420
Number of ways to select 3 defective sets
 C33  C117
Let A be the event of getting 3 cartons of orange juice
and B be the event of getting 3 cartons of lemon juice.

C34
4
1


Then P( A)  10
and
C3
120 30
P(gets 3 defective sets) 

C36
20 1
P ( B )  10


C3
120 6
P(the same kind of juice)  P ( A  B )
 P ( A)  P ( B )
1 1


30 6
1

5
13.
(b)
1
285
Number of ways to select 2 defective sets
 C23  C217

P(gets 2 defective sets) 

C 23  C 21 7
C 42 0
8
95
Let A be the event of getting 3 storybooks, B be the
16.
event of getting 3 comic books and C be the event of
(a)
Then P( A) 
C35 10 5


,
C39 84 42
P( B) 
C33
1

and
9
C3 84
P(C ) 
0
0
C39

P(All given to boys) 

(b)
C38
C312
14
55
Number of ways to select 2 boys and a girl
 C28  C14
P(3 books of the same kind)
 P( A  B  C )
 P ( A)  P ( B )  P (C )
5
1


0
42 84
11

84
(a)
Total number of ways to select 3 students  C312
Number of ways to select 3 boys  C38
getting 3 atlas.
14.
C33  C117
C420

17.
Total number of combinations
(a)
 C 552
 2 598 960
P(2 given to boys and 1 given to girl)

C 28  C14
C312

28
55
P(at least 2 of them get different points)
 1  P(all of them get the same point)
1
 1  12
C12  C112  C112
1
144
143

144
 1
(b)
Number of ways to select a suit  C
4
1
Number of ways to select 5 cards in a suit  C513

P(5 cards are of the same suit)

C14 C513
2 598 960
(b)
P(Ann gets more points than Tom)
1 11 1 10
1 1
    
12 12 12 12
12 12
1 1
  (11  10    1)
12 12
1  (11  1)11 


144 
2

11

24

33

16 660
81
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
18.
(a)
Let x be the number of lorries, then the number
(b)
must be ‘4’ or ‘6’.
of vans is 3x.
Number of ways to choose the last card  C12
x
1

12  4 x 10
10 x  12  4 x
6 x  12
x2

P(divisible by 2)

(i)
120
360
1

3
Total number of ways to choose two
vehicles  C 220
20.
(a)
Number of ways to choose two private
cars  C212

C12  P35
P46

The number of lorries is 2 and the number
of vans is 6.
(b)
To form a number divisible by 2, the last card
Total number of ways to arrange 3 out of 14
numbers  P314
P(first prize) 
C 12
P(both are private cars)  22 0
C2

33

95
(ii)
(b)
Number of ways to choose a lorry and a
van  C12  C16

1
2184
P(second prize) 
3!  1
P314

5
2184
C2 C6
P(a lorry and a van)  1 2 0 1
C2

6
95
(c)
Expected award
 $6000 
(iii)
1
P314
Number of ways to choose the same
1
5
 $1000 
2184
2184
 $5.04 (cor. to 2 d. p.)
vehicle
 C212  C22  C26

21.
 1
C 212  C 22  C 26
C 220
 1
41
95

19.
(a)
(a)
P(not the same kind)
 1  P(same kind )
Total number of ways to select 5 cards  C552
P(Four of a Kind) 

54
95
(b)
P(Full House) 

Total number of ways to arrange 4 cards  P
6
4
C113  C148
C552
1
4165
C113  C34  C112  C 24
C552
6
4165
To form a number less than 5000, the first card
must be ‘1’, ‘3’ or ‘4’.
(c)
Number of ways to choose the first card  C13
Kind’ is less than a hand of ‘Full House’. So
‘Four of a Kind’ beats ‘Full House’ in the game
P(less than 5000)

of poker.
C13  P35
P46
180
360
1

2

© Hong Kong Educational Publishing Co.
The probability of getting a hand of ‘Four of a
82
More about Probability
4.
(a)
pp.160  171
Since 13 and 19 are the favourable outcomes,
P(prime) 
2
9
p.160
1.
(a)
(b)
Only ‘9’ is contained in both A and B.
A  B  {9}

Since 10, 18, 20, 22 and 24 are the favourable
outcomes,
P(even) 
(c)
5
9
Since 15, 18 and 24 are the favourable outcomes,
P(multiple of 3) 
(b)
The elements contained in A but not in B are 1, 3
5.
and 7.

A \ B  {1, 3, 7}
(b)
(c)
(a)
12
40
3

10
P(a black card) 
(b)
P(a ‘spade’ or a ‘heart’)
13  13

52
1

2
(c)
P(a ‘King of spades’)
1

52
and 6.

B \ A  {2, 4, 6}
2.
The elements contained in N but not in A are 2, 4, 6, …

The elements in N \ A are all positive even
numbers.
3.
(a)
P(‘E’) 
3
8
(b)
P(‘A’) 
0
0
8
83
26
52
1

2
(a)
The elements contained in B but not in A are 2, 4
7.
Total number of students in S6A
 15  5  12  8
 40
P(Chinese) 
A  B  {1, 2, 3, 4, 6, 7, 9}
6.
(d)
3 1

9 3
P(varies less than 300 points)
 1  P(rising by more than 300 points)
 P(falling by more than 300 points)
5 1
 1 
8 4
1

8
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
8.
(a)
(b)
9.
(a)
(b)
10.
11.
P (orange juice or apple juice)
3 4
 
12 12
7

12
12.
P (lemon juice)
7
1
12
5

12
(b)
Favourable outcomes: (1, 5), (2, 4), (3, 3), (4, 2),
(5, 1), (2, 6), (3, 5), (4, 4), (5, 3) and (6, 2).
Number of favourable outcomes = 10

P (the sum is 6 or 8)
10

36
5

18
13.
Sum is 4: (1, 3), (2, 2), (3, 1)
Sum is 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
Sum is 12: (6, 6)

P (the sum is multiple of 4)
9

36
1

4
P(‘heart’) 
14.
(c)
P(‘Queen of heart’) 
15.
 

 P(‘Queen of hearts’)
1 1
1
  
4 13 52
4

13
© Hong Kong Educational Publishing Co.
1
4
3 1
P(two boys)  
5 4
3

20
P(mobile phone) 
1
3
84
3
4
P(gets the mobile phone but not the flash disk)
1 3
 
3 4
1

4
2
1
2
1
P(2nd call) 
2
1
P(3rd call) 
2
P(1st call) 

1
52
 P(‘heart’) + P(‘Queen’)

P(a boy from B) 

P(either a ‘heart’ or a ‘Queen’)
(d)
3
5
P(not the flash disk) 
4
1

52 13
P(‘Queen’) 
P(greater than 5 or odd number)
 P(greater than 5)  P (odd number )
 P(odd number greater than 5)
5
5 1

 
10 10 5
4

5
P(a boy from A) 

13 1

52 4
(b)
Since 7 and 9 are the favourable outcomes,
P(odd number greater than 5)
2

10
1

5
P(only one of them is late)
 P(Sally late but Edwin not late)
 P(Sally not late but Edwin late)
 0.65  (1  0.35)  (1  0.65)  0.35
 0.545
(a)
(a)
P(3 calls to the same doctor)
1 1
 1 
2 2
1

4
More about Probability
16.
17.
4
1
4
1
P(2nd letter) 
4
1
P(3rd letter) 
4
Case 2:

= 1 – 0.6 = 0.4
P(1st letter) 
(a)
(b)
P(James cannot play in the coming match)
= 1 – 0.6 = 0.4
P(win the match  James is absent)
P(3 letters put in the same box)
1 1
 1 
4 4
1

16

21.
P (product is even)
1 P (product is odd)
3 2
1 
5 4
7

10
P(volleyball  basketball)
P( volleball  basketball)
P(basketball)
0.16

0.28
4

7

Favourable outcomes: (1, 5), (5, 1)
Number of favourable outcomes = 2
Total number of possible outcomes  5  4  20

P (difference is greater than 3)
2

20
1

10
22.
The 11th trial is independent of the previous trials.

23.
1
P(1st trial correct) 
8
P(2nd trial correct) 

19.
P(head in the 11th trial) 
7 1 1
 
8 7 8
P(Billy, Cherry, David, Andy and Emily)
1
 5
P5

1 1

8 8
1

4
1
2
Total number of ways to arrange 5 people  P55

18.
P(the team will win the coming match)
 0.6  0.85  0.4  0.4
 0.67
1
120
P(win) 
24.
(a)
Total number of ways to select 4 numbers out of
14 numbers  C414
P(win the lottery if not considering the order)
1
 14
C4
Since ‘C’ and ‘E’ are the favourable outcomes,
P(two letters are the same)

 P(2 ‘C’) + P(2 ‘E’)
2 1 2 1
  
7 6 7 6
2

21
1
1001

20.
(b)
Total number of ways to arrange 4 out of 14
numbers  P414
P(win the lottery if considering the order)
1
 14
P4
Case 1:
P(James can play in the coming match) = 0.6

P(win the match  James is present) = 0.85
85
1
24 024
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
25.
A  B  {a, b, e, i, o, u}
Total number of balls = 6 + 3 = 9
Number of ways to select 3 balls out of 9 balls  C39
A  C  {a, b, e, i, o, u}
Number of ways to select 3 red balls out of 6 red balls
 C36
Number of ways to select 3 green balls out of 3 green
balls  C33

29.
(a)
P(have the same colour)
C3  C6
 3 9 3
C3
( A  B)  ( A  C)  {a, b, e, i, o, u}

A  ( B  C )  ( A  B)  ( A  C )
Total number of balls = 4 + 6 = 10
Number of ways to select 2 balls out of 10 balls
 C210
Number of ways to select 1 white ball and 1
black ball  C14  C16
1
4



26.
Total number of children = 8 + 6 = 14
P(1 white ball and 1 black ball)

Number of ways to select 4 children out of 14 children
24
45
8

15

 C414
Number of ways to select 2 boys out of 6 boys  C 26
Number of ways to select 2 girls out of 8 girls  C 28

P(2 boys and 2 girls)

C 26  C 28
C 41 4
(b)
15  28
1001
60

143
Number of ways to select 2 white balls out of 4
white balls  C 24


P(at least 1 white ball)
 P(1 white ball and 1 black ball)
 P(2 white balls )

(a)
8 C24

15 C210
8
6

15 45
2

3

p.163
27.
C14  C16
C 21 0
Since 10  n  15,
A  {10 2  1, 112  1, 12 2  1, 132  1, 14 2
 1, 15 2  1}
 {101, 122, 145, 170, 197, 226}
(b)
30.
(a)
P(both are not chosen) 
C 23 8
C 24 0

703
780
Only 122, 170 and 226 are contained in both A
and N.

28.
A  N have 3 elements.
(c)
122 or 170 or 226
(a)
Only e and u are contained in both A and B.

A  B  {e, u}
(b)

31.
(a)
C12  C13 8
C 24 0
19
195
P(cancelled) = 0.8
P(not cancelled) = 1 – 0.8 = 0.2
Only a and u are contained in both A and C.

A  C  {a, u}
(b)
P(only one is chosen) 
P(training on all the three days)
 0.2  0.2  0.2
 0.008
Yes, they are equal.
Only b and u are contained in both B and C.

B  C  {b, u}

(b)
A  (B  C)  {a, b, e, i, o, u}
© Hong Kong Educational Publishing Co.
86
P(cancelled one of the three days)
 0.2  0.2  0.8  3
 0.096
More about Probability
(c)
P(training at least once in the three days)
 1  P(no training on all the three days)
34.
(a)
P(female or type O)
 P(female )  P( typeO)
 P( typeO female )
400 310 200



650 650 650
51

65
(b)
P(female but not of type O)
400  200

650
4

13
(c)
P(male but not of type O)
 1  P(female or typeO)
 1  0.8  0.8  0.8
 0.488
32.
Total number of people sitting at the chief table
= 1 + 6 + 3 + 2 = 12
Number of ways to select 2 people out of 12 people
 C212
(a)
Number of ways to select 2 teachers out of 3
teachers  C 23
P(two teachers get the prizes)
C3
 122
C2
3
66
1

22

 1

(b)
Number of ways to select the principal and 1
guest  C11  C16
35.
P(the principal and a guest get the prizes)
C1  C 6
 1 12 1
C2
(a)
51
65
14
65
Number of ways to select 4 out of 10
transistors  C410
P(all are defective) 
C 44
C 41 0

1
210
6
66
1

11

(b)
(c)
P(two are defective) 
Number of ways to select 2 people out of 12
people not including the students  C210
90
210
3

7

P(at least one student will get a prize)
 1  P(no student will get the prize)
 1
C 210
C 212
36.
45
 1
66
7

22
33.
C 24  C 26
C 41 0
(a)
Number of ways to arrange 4 digits  P48
To form an even number, the last digit must be
‘2’, ‘6’ or ‘8’.
Number of ways to select the last digit  C13
7
3
P(even)  P3  C1
8
P4
P(class A or foreign child)
 P(class A)  P(foreign child )
 P(foreign child in class A)
225 250 75



500 500 500
400

500
4

5
 0.8
630
1680
3

8

87
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
(b)
To form a number greater than 8000, the first
(b)
P (use the same entrance)
= P (both use entrance A) + P (both use entrance B)
2 1 3 2
   
5 3 5 3
8

15
(c)
P (use different entrances)
1 P (use the same entrance)
8
1
15
7

15
P (use the same exit)
= P (Tom will use any exit) P (Jenny will use the
same exit)
1
 1
3
1

3
P (different entrances, same exit)
7 1
 
15 3
7

45
(a)
P (patient is still in room A)
= P (chooses a doctor or a nurse in 1st trial)
1 2

1 2 1
3

4
(b)
P (patient in room B)
= P (chooses a patient in 1st trial and chooses
others in 2nd trial)
1 4
 
4 5
1

5
(c)
P (patient in room C)
= P (chooses the patient in both trials)
1 1
 
4 5
1

20
(d)
P (all are nurses in room C)
2 3 2 2
   
4 5 4 5
1

2
digit must be either ‘8’ or ‘9’.
Number of ways to select the first digit  C12
2
7
P(greater than 8000)  C1  P3
8
P4
420
1680
1

4

37.
(a)
P (miss the next throw | miss a particular
throw)
1
 1
4
3

4
(b)
P (three successful throws | hits the target in a
particular throw)
3 3 3
  
5 5 5
27

125
(c)
(i)
P (three successful throws)
1 3 3
  
2 5 5
9

50
(ii)
P (he misses once)
= P (misses 1st throw only) + P (misses
2nd throw only) + P (misses 3rd throw
only)
1 1 3 1  3 1
     1   
2 4 5 2  5 4
39.
1 3  3
  1  
2 5  5
49

200

(iii)
38.
(a)
P (misses all three throws)
1  1  1
  1    1  
2  4  4
9

32
P (uses any one exit) 
1
3
2 3

5 5
P (Tom will use entrance B and exit Q)
3 1
 
5 3
1

5
P (Tom will use entrance B)  1 

© Hong Kong Educational Publishing Co.
88
More about Probability
40.
(a)
(b)
41.
Let x be the number of boys, then
the number of girls = (1 + 20%)x = 1.2x
Total number of students = x + 1.2x = 2.2x
30  2.2 x  40
13.6  x  18.2
Since x and 1.2x must be integers, x = 15
Number of girls  1.2  15
 18
(ii)
Number of ways to select 1 English
channel and 1 Chinese channel  C13  C14
 P(one English channel and one Chinese
channel)
18
18  15
6

11

C13  C14
C 27

4
7
P (a girl) 
43.
18 15 15 18
  
33 32 33 32
45

88
(c)
P (one boy and one girl) 
(a)
P(correct) = 0.7
(a)
P (3  0)  0.5  0.9  0.75
 0.3375
(b)
P(3 – 1)
 P (loses in Game 1 only)
 P (loses in Game 2 only)
 P (loses in Game 3 only)
 (1  0.5)  0.9  0.75  0.35
 0.5 1  0.9  0.75  0.35
 0.5  0.9 1  0.75  0.35
P(3 marks)
 P(3 correct and 1 wrong)
 0.1706 (cor. to 4 d. p.)
 0.7 3  (1  0.7)  C14
 0.4116
(c)
(b)
P(no marks)  P (all wrong)
 P(loses in Game 1, 3 only)
 P(loses in Game 1, 4 only)
 P(loses in Game 2, 3 only)
 P(loses in Game 2, 4 only)
 P(loses in Game 3, 4 only)
 (1  0.7) 4
 0.0081
(c)
P(1 mark)
 P(1 correct and 3 wrong)
 (1  0.5)  (1  0.9)  0.75  0.35  0.7
 (1  0.5)  0.9  (1  0.75)  0.35  0.7
 0.7  (1  0.7)  C
3
P(3 – 2)
 P(loses in Game 1, 2 only)
4
1
 (1  0.5)  0.9  0.75  (1  0.35)  0.7
 0.0756
 0.5  (1  0.9)  (1  0.75)  0.35  0.7
 0.5  (1  0.9)  0.75  (1  0.35)  0.7
(d)
 0.5  0.9  (1  0.75)  (1  0.35)  0.7
P(more than 2 marks)
 P(3 marks )  P(4 marks )
 0.261625
 0.4116  0.7 4
 0.6517
 P(3  0)  P(3  1)  P(3  2)
P(China wins the match)
 0.7697 (cor. to 4 d. p.)
42.
(a)
(b)
3
P(English channel) 
7
44  46
(HKCEE Questions)
Number of ways to select 2 buttons out of 7
buttons  C 27
(i)
Number of ways to select 2 buttons out of
3 buttons  C 23
P(both are English channels) 

C 23
C 27
1
7
89
© Hong Kong Educational Publishing Co.
20
New Progress in Senior Mathematics 6A (Compulsory Part) Solution Guide
Extended Question
47.
(a)
p.168
49.
It is assumed that the probability of having a
birthday is the same on each day in a year for
everyone.
P (all of them have different birthdays)
365 364 363
343



 
365 365 365
365
 0.493
P (at least two of them have the same birthday)
 1  0.493
 0.507
The required probability is 50.7%.
Faces
1
2
3
4
Probability
0.2
0.3
0.35
0.15
Mark
30
25
20
40
(or other reasonable answers)
Expected mark
 30  0.2  25  0.3  20  0.35  40  0.15
 26.5 ( 25)
Multiple-choice Questions
1.
pp.169  171
B
2 2
 1.14  1
3

2.
A
3.
B
For I,
There is no red ball in a pack of black balls.
(b)
For II,
Since 7 is not an element of a fair dice, P(7) 
Since level 1, 2, 3, 4 and 5 are the elements in the
HKDSE, P(Level 5) > 0

1
2
C 2n  C 24 1

C 2n  4
2
n!
6
(n  2)!2!

(n  4)!
(n  2)!2!
[n!12(n  2)!](n  2)!

(n  2)!(n  4)!
[n(n  1)  12](n  2)!

(n  4)!
1
2
1
2
1
2
2n 2  2n  24  (n 2  7 n  12)  0
n 2  9n  12  0
1.63  n  7.37
The possible values of n are any integral values
from 2 to 7.
© Hong Kong Educational Publishing Co.
C
5.
B
6.
D
7.
D
P (at least one head)
1 P (all tail)
1  1  1 
 1    
2  2  2 
1
 1
8
7

8
n 2  n  12
1

(n  4)(n  3) 2

I and II only.
By the definition of probability, 0  P(E)  1.
p.169
P(two cans of same juice) 
48.
0
0.
6
For III,
4.
Open-ended Questions
P (red ball )  0

P (all of them have different birthdays)
365 364 363
327



 
365 365 365
365
 0.1218
P (at least two of them have the

same birthday)
 1  0.1218
 0.8782
 0.878 (correct to 3 sig. fig.)
The required probability is 87.8%.
90
More about Probability
8.
9.
B
Let the three numbers be A, B and C where A < B < C.
All possible arrangements:
ABC, ACB, BAC, BCA, CAB, CBA
1

P (in ascending order) 
6
15.
P (the problem will be solved)
1 P (both of them cannot solve the problem)
 1  (1  p)(1  q)
 1  (1  p  q  pq)
 p  q  pq
D
16.
P(different colours)
A
P (does not wear glasses)
= P (he is a boy and does not wear glasses)
+ P (she is a girl and she does not wear glasses)
2
40
3
20




2  3 100 2  3 100
7

25
2 1 4 3
   
6 4 6 4
7

12
10.
D
A
P(Eric sits opposite to Betty)
17.
2!
( 4  1)!
2

6
1

3

11.
12.
A
There are 3 face cards of ‘spade’.

P (face card or ‘spade’)
4  3  13  3

52
22

52
11

26
p 2  p  12  0
( p  4)( p  3)  0
p  4 or 3 (rejected)
18.
B
19.
C
P(only one key can open the safe)


C 22
C 26

C
P (at least one head)
1 P (all are tails)
1
 1  
2
1
 1 n
2
C12  C14
C 26
P(both keys can open the safe)
D
P (only one of them will pass the test)
2  3   2  3 
 1    1   
3  4   3  4 
5

12
14.
D
P (win at least one of three games)
1 P (lose in all three games)
 1  (0.4)(0.4)(0.4)
 0.936
P (‘B’ is chosen but ‘S’ is not chosen)
1 2 2 1
   
4 3 4 3
1

3
13.
B
Total number of balls  p  9  p  9
p p 1
P (two white balls)  
9
8
p( p  1)

72
p( p  1) 1

72
6
p( p  1)  12
P(at least one key can open the safe)

C12  C14 C 22
 6
C 26
C2

C14  C12  C 22
C 26
n
91
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