11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
(c)
11
Binomial Distribution
P( X  3)
 1  P ( X  3)
 1  [ P ( X  0)  P ( X  1)  P ( X  2)]
1
11
  3  0  5 1 2
 3 5
 1  C01 2      C11 2    
8 8
  8   8 
2
10
3 5 
 C21 2     
 8   8  
 0.8865
pp.91 – 93
p.91
(a)
Yes. Define X as the number of odd numbers.
Then f ( x)  0.5 x 0.51  x for x  0, 1.
(d)
(b)
Yes. Define X as the number of days that Peter forgets
to take his lunch box. Then f ( x)  0.2 x 0.81  x for
x  0, 1.
(c)
Yes. Define X as the number of patients who are
diagnosed with cancer. Then f ( x)  0.25x 0.751  x for
x  0, 1.
p.93
(a)
x
10  x
for x  0, 1, …, 10
f ( x)  C 10
x (0.1) (0.9)
(b)
f ( x)  C x6 (0.5) x (0.5) 6  x for x  0, 1, …, 6
(c)
f ( x)  C x30 (0.24) x (0.76) 30  x for x  0, 1, …, 30
(d)
f ( x)  C x36 (0.45) x (0.55) 36  x for x  0, 1, …, 36
(e)
x
14  x
for x  0, 1, …, 14
f ( x)  C14
x (0.8) (0.2)
P( X  n)  0.001 838 3
11.2 (a)
p.95
C (0.35) (0.65)  0.001 838 3
n
n
n
0
0.35n  0.001 838 3
log 0.35n  log 0.001 838 3
n log 0.35  log 0.001 838 3
log 0.001 838 3
log 0.35
6
n
(b)
P( X  4)
 C46 (0.35) 4 (0.65) 2
 0.0951 (cor. to 4 d. p.)
pp.94 – 110
11.1 (a)
P ( X  10)
12  3 
 C10
 
8
 0.0014
(b)
10
11.3 (a)
p.94
2
5
 
8
(cor. to 4 d. p.)
10
p.95
The probability function of Y is
f ( y )  P(Y  y )
12  10
y
6 y
3 
 3  
 C y6   1  
20
20
  

 C y6 (0.15) y (0.85) 6  y ,
P ( 2  X  3)
 P( X  2)  P( X  3)
3
 C212  
8
 0.2532
3
)
20
Y ~ B(6,
where y = 0, 1, 2, …, 6
3
5
12  3   5 
   C3    
8
8 8
(cor. to 4 d. p.)
9
(b)
P(2 defective light bulbs)
 C26 (0.15) 2 (0.85) 4
 0.1762 (cor. to 4 d. p.)
11.4
E (Y )  150(0.2)
 30
Var ( X )  150(0.2)(0.8)
 24
© Hong Kong Educational Publishing Co.
54
p.100
Binomial Distribution
Var ( X )  9.6
40 p (1  p )  9.6
11.5 (a)
p.100
11.8 (a)
X ~ B ( 20, 0.8)
E ( X )  20(0.8)
p.102
 16
40 p  40 p 2  9.6
25 p  25 p  6  0
Var ( X )  20(0.8)(0.2)
(5 p  2)(5 p  3)  0
 3.2
2
p  0.4 or 0.6
(b)
(b)
For p  0.4,
E ( X )  40(0.4)
 16


2 16

32
3.2
3.2
2 

P X 

 


32 

 P X 
3.2 

 P( X  17.888 544)
For p  0.6,
E ( X )  40(0.6)
 24
 P( X  18)  P( X  19)  P( X  20)
E (Z )  9
11.6
2
 C1820 (0.8)18 (0.2) 2  C1920 (0.8)19 (0.2)  (0.8) 20
p.101
 0.2061 (cor. to 4 d. p.)
np  9................(1)
Var ( Z )  0.21n
np(1  p)  0.21n
11.9 (a)
p(1  p)  0.21
p.106
diagnosed correctly.
p 2  p  0.21  0
Then X ~ B(20, 0.914).
 (1)  (1)  4(1)(0.21)
P(more than 90% of the patients are
2
p
Let X be the number of patients are
diagnosed correctly)
 P ( X  18)
 P ( X  19)  P ( X  20)
2(1)
 0.3 or 0.7
For p  0.3,
 C1290 (0.914)1 9 (0.086)  (0.914) 2 0
Substituting p  0.3 into (1),
n(0.3)  9
n  30
 0.477 082 503
 0.4771 (cor. to 4 d. p.)
For p  0.7,
Substituting p  0.7 into (1),
(b)
n(0.7)  9
incorrectly)
 P( X  18)
90
n
, which is not an integer.
7


11.7 (a)
 P( X  18)  P( X  19)
The case of p  0.7 is rejected.
n  30 and p  0.3
Let X be the number of staff
P(less than 3 patients are diagnosed
 C1820 (0.914)18 (0.086)2  0.477 082 503
 0.7556 (cor. to 4 d. p.)
p.101
11.10 (a)
wearing glasses in the sample.
39 

Then X ~ B 30,

50 

Let X be the number of missiles hit the
p.107
target.
Then X ~ B(9, 0.75).
P(at least 2 of missiles hit the target)
 P ( X  2)
E( X )
 39 
 30 
 50 
 23.4
 1  P ( X  2)
 1  P ( X  0)  P( X  1)
 1  (0.25) 9  C19 (0.75)(0.25) 8
 0.9999 (cor. to 4 d. p.)
(b)
Var(X)
 39  39 
 30 1  
 50  50 
 5.148
55
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
(b)
Let Y be the number of missiles hit the
(c)
target when n of missiles are fired.
in a sample of 10 products.
Then Y ~ B(n, 0.75).
Then Y ~ B(10, 0.0625)
P(hitting the target)  0.95
P(Y  1)  0.95
P(the production line is considered effective
at stage II)
 P( X  1)  P(Y  0)
1  P(Y  0)  0.95
1  (0.25) n  0.95
 0.366 745 053  (0.9375)10
 0.192 343 284
0.25  0.05
n log 0.25  log 0.05
n
 0.1923 (cor. to 4 d. p.)
log 0.05
n
log 0.25
 2.160 964 047

11.11 (a)
Let Y be the number of defective products
(d)
P(the production line is considered effective)
 P(the production line is considered effective at
The least number of n is 3.
stage I) + P(the production line is
Let X be the number of times that the
considered effective at stage II)
 0.275 058 789  0.192 343 284
p.108
 0.4674 (cor. to 4 d. p.)
number chosen by the player appears
on the dice.
Then X ~ B(4,
1
).
6
Expected gain
 2  P( X  0)  1  P( X  1)  2  P( X  2)
pp.93 – 110
 3  P( X  3)  4  P( X  4)
4
3
2
5
 1  5 
1 5
 2     1  C14     2  C24    
6
6
6
 
  
6 6
1
 3  C34  
6
193

coin
648
3
5
1
   4 
6
6
Example 11.1T
2
(a)
4
11.12 (a)

The expected gain  0.

It is not fair to the player.
Let X be the number of defective
P (Y  8)
1
 C815  
4
 0.0131
(b)
(b)
p.93
15  8
3
 
4
(cor. to 4 d. p.)
P (Y  3)
 P(Y  0)  P(Y  1)  P(Y  2)
1
 C015  
4
 0.2361
p.110
8
0
15  0
1
15  1
3
1 3
 C115    
 
4
4 4
(cor. to 4 d. p.)
2
15  2
1 3
 C215    
4 4
products in a sample of 20 products.
(c)
Then X ~ B(20, 0.0625).
P(the production line is considered
effective at stage I)
 P( X  0)
 (0.9375) 20
 0.275 058 789
 0.2751 (cor. to 4 d. p.)
(b)
P(the second sample for inspection is needed)
 P( X  1)
Example 11.2T
 C120 (0.0625)(0.9375)19
(a)
p.94
P(Y  n)  0.100 112 9
C (0.75) n (0.25)0  0.100 112 9
n
n
 0.366 745 053
0.75n  0.100 112 9
n log 0.75  log 0.100 112 9
 0.3667 (cor. to 4 d. p.)
log 0.100 112 9
log 0.75
8
n
© Hong Kong Educational Publishing Co.
56
Binomial Distribution
(b)
P (Y  7)
Example 11.7T
p.101
 C78 (0.75) 7 (0.25)
 0.2670 (cor. to 4 d. p.)
Let X be the number of days that Leo is late.
Then X ~ B(22, 0.12).
Example 11.3T
(a)
Expected number of days that he is late
p.95
 22(0.12)
 2.64
Let X be the number of boys selected.
Variance  22(0.12)(0.88)
 2.3232
Then X ~ B(7, 0.53).
Hence the probability function of X is
f ( x)  P( X  x)
 C x7 (0.53) x (0.47) 7  x ,
Example 11.8T
p.102
where x  0, 1, 2, 3, 4, 5, 6, 7
(a)
(b)
ends.
P(exactly 3 boys)
 P ( X  3)
Then X ~ B(10, 0.2).
E( X )
 10(0.2)
2
 C37 (0.53) 3 (0.47) 4
 0.2543 (cor. to 4 d. p.)
Example 11.4T
Let X be the number of rats die before the experiment
p.100
(b)
E (Y )  40(0.37)
  10(0.2)(1  0.2)
 1.6
 14.8
P( X     )
Var (Y )  40(0.37)(0.63)
 P ( X  2  1.6 )
 P ( X  3.264911)
 9.324
 P ( X  0)  P ( X  1)  P( X  2)  P( X  3)
Example 11.5T
(a)
 (0.8)10  C110 (0.2)(0.8) 9  C 210 (0.2) 2 (0.8) 8
p.100
 C310 (0.2) 3 (0.8) 7
E ( Z )  21
 0.8791 (cor. to 4 d. p.)
30 p  21
p  0.7
Example 11.9T
(b)
Var(Z)
(a)
 30(0.7) (1  0.7)
 6.3
Example 11.6T
p.105
Let X be the number of games that Andy wins.
Then X ~ B(10, 0.323).
P(Andy wins at least 3 games)
 P( X  3)
 1  P( X  3)
p.101
 1  P( X  0)  P( X  1)  P( X  2)
E ( Z )  14.4
 1  (0.677)10  C110 (0.323)(0.677) 9
np  14.4.............(1)
 C 210 (0.323) 2 (0.677) 8
Var ( Z )  25.6 p
 0.676111
 0.6761 (cor. to 4 d. p.)
np (1  p )  25.6 p
n(1  p )  25.6
n  np  25.6.............(2)
(b)
Substituting (1) into (2),
n  14.4  25.6
n  40
P(Chris wins at least 3 games)
 P ( X  7)
 1  P ( X  7)
 1  P( X  8)  P( X  9)  P( X  10)
Substituting n  40 into (1),
40 p  14.4
p  0.36
 1  C810 (0.323) 8 (0.677) 2  C910 (0.323) 9 (0.677)
 (0.323)10
 1  0.002715
 0.9973 (cor. to 4 d. p.)
57
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
(c)
P(Andy wins more than Chris)
 P( X  6)
 P( X  6)  P( X  7)  P( X  7)
Example 11.12T
(a)
 C (0.323) (0.677)  C (0.323) (0.677)
10
6
6
4
10
7
7
p.110
Let X be the number of defective items in a sample of
10 items.
3
Then X ~ B(10, 0.1).
 0.002715
 0.0665 (cor. to 4 d. p.)
P(a batch is rejected)
 P(X  2 in stage I)
Example 11.10T
(a)
+ P(X  1 in stage I and X  2 in stage II)
p.106
 1  (0.9)1 0  C11 0 (0.1)(0.9)9  C11 0 (0.1)(0.9)9
 [1  (0.9)1 0  C11 0 (0.1)(0.9)9 ]
Let X be the number of people who are overweight.
 0.263 901 07  0.387 420 489  0.263 901 07
Then X ~ B(n, 0.6).
P( X  n  1)  0.9
 0.3661 (cor. to 4 d. p.)
1  P( X  n)  0.9
1  ( 0 .6 ) n  0 . 9
(b)
0 .6 n  0 .1
10
 10  P( X  0)  20  P( X  1)   10  P( X  x)
n log 0.6  log 0.1
x2
10
log 0.1
n
log 0.6
 4.507576

Mean number
  10  P( X  x)  10  P( X  1)
x 0
10
 10  P( X  x)  10  C110 (0.1)(0.9) 9
x 0
The minimum value of n is 5.
 10  1  10  C110 (0.1)(0.9) 9
 13.8742 (cor. to 4 d. p.)
(b)
X ~ B(5, 0.6)
Expected number of people who are overweight
(c)
 5(0.6)
3
Let Y be the number of months that the manufacturer
rejects the batch.
Then Y ~ B(6, 0.3661).
Example 11.11T
(a)
P(the manufacturer stops buying within
p.107
the first half year)
 P(Y  2 except the manufacturer rejects in the
Let X be the number of games that Amy loses.
first month and the sixth month)
Then X ~ B(4, 0.4).
 1  P(Y  2)  P(the manufacturer rejects in the
Expected number of games she loses
 4(0.4)
 1 .6
(b)
first month and the sixth month)
 1  (1  0.3661) 6  C16 (0.3661)(1  0.3661) 5
(1  0.3661) 4 (0.3661) 2
 0.6886 (cor. to 4 d. p.)
Her expected gain
 5  P ( X  0)  4  P ( X  1)  3  P ( X  2)
 2  P ( X  3)  ( 2)  P ( X  4)
 5  (0.6) 4  4  C14 (0.4)(0.6) 3  3  C24 (0.4) 2 (0.6) 2
11.1
 2  C34 (0.4) 3 (0.6)  ( 2)  (0.4) 4
pp.96 – 98
 $3.3232
pp.96  97
(c)
Let $y be the money she should pay for losing all 4
1.
games such that the game is fair.
Her expected gain  0
5  P( X  0)  4  P( X  1)  3  P( X  2)
 2  P( X  3)  ( y )  P( X  4)  0
5  (0.6) 4  4  C14 (0.4)(0.6) 3  3  C24 (0.4) 2 (0.6) 2
 2  C34 (0.4) 3 (0.6)  ( y )  (0.4) 4  0
3.3744  0.0256 y
y  131.8125
 She should pay $132. (cor. to the nearest dollar)
© Hong Kong Educational Publishing Co.
58
(a)
Binomial Distribution
(d)
(b)
P( X  3)
 1  P( X  3)
 1  0.794 569
 0.2054 (cor. to 4 d. p.)
4.
(a)
P (Y  4)
 C49 (0.8)4 (0.2)5
 0.016 515 1
 0.0165 (cor. to 4 d. p.)
(c)
(b)
P (Y  2)
 P(Y  0)  P(Y  1)  P(Y  2)
 C09 (0.8) 0 (0.2) 9  C19 (0.8)1 (0.2) 8
 C 29 (0.8) 2 (0.2) 7
 0.0003 (cor. to 4 d. p.)
2.
(a)
(c)
P( X  8)
 C816 (0.25)8 (0.75)16  8
 0.016 515 1  C59 (0.8)5 (0.2) 4
 0.0197 (cor. to 4 d. p.)
(b)
 0.0826 (cor. to 4 d. p.)
P( X  3)
 1  P( X  3)
(d)
 1  [ P( X  0)  P( X  1)  P( X  2)  P( X  3)]
16  0
 1  [C (0.25) (0.75)
16
0
0
P(3  Y  5)
 P (Y  4)  P (Y  5)
 C79 (0.8) 7 (0.2) 2  C89 (0.8) 8 (0.2)1
16  1
 C (0.25) (0.75)
16
1
1
P (Y  7)
 P(Y  7)  P(Y  8)  P(Y  9)
 C99 (0.8) 9 (0.2) 0
 0.7382 (cor. to 4 d. p.)
 C216 (0.25) 2 (0.75)16  2  C316 (0.25)3 (0.75)16  3 ]
 0.5950 (cor. to 4 d. p.)
5.
3.
(a)
(a)
P( X  2)
 C1020 (0.4)10 (0.6)10
 C (0.2) (0.8)
12
2
2
 0.1171 (cor. to 4 d. p.)
10
 0.283468
 0.2835 (cor. to 4 d. p.)
(b)
(b)
P( X  3)
 P( X  0)  P( X  1)  P( X  2)  P( X  3)
0
12
12
1
1
P( X  3)
 P( X  0)  P( X  1)  P( X  2)  P( X  3)
 C020 (0.4) 0 (0.6) 20  C120 (0.4)1 (0.6)19
 C220 (0.4) 2 (0.6)18  C320 (0.4) 3 (0.6)17
 0.0160 (cor. to 4 d. p.)
 C (0.2) (0.8)  C (0.2) (0.8)  0.283468
12
0
P ( X  10)
11
 C312 (0.2) 3 (0.8) 9
 0.794569
(c)
 0.7946 (cor. to 4 d. p.)
P(14  X  16)
 P( X  14)  P( X  15)  P( X  16)
 C1420 (0.4)14 (0.6) 6  C1520 (0.4)15 (0.6) 5
(c)
P (6  X  8)
 C1620 (0.4)16 (0.6) 4
 0.0064 (cor. to 4 d. p.)
 P( X  6)  P( X  7)  P( X  8)
 C612 (0.2) 6 (0.8) 6  C712 (0.2) 7 (0.8) 5
 C812 (0.2)8 (0.8) 4
 0.0193 (cor. to 4 d. p.)
(d)
P( X  2)
 1  P( X  2)
 1  P( X  0)  P( X  1)  P( X  2)
 1  C020 (0.4)0 (0.6) 20  C120 (0.4)1 (0.6)19
 C220 (0.4) 2 (0.6)18
 0.9964 (cor. to 4 d. p.)
59
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
6.
X ~ B(8, 0.5)

12.
The probability function of X is
f ( x)  C x8 (0.5) x (0.5)8  x
Let X be the number of light bulbs that are still working
after 800 hours of use.
Then X ~ B(7, 0.4).
 C x8 (0.5) 8 , where x  0, 1, 2, …, 8
7.
P(exactly 3 of light bulbs are working after
X ~ B(20, 0.25)
800 hours of use)
 P ( X  3)

 C37 (0.4) 3 (0.6) 4
The probability function of X is
f ( x)  C x20 (0.25) x (0.75) 20  x ,
 0.2903 (cor. to 4 d. p.)
where x  0, 1, 2, …, 20
pp.97  98
8.
(a)
1
X ~ B (13, )
6

13.
The probability function of X is
x
x
13  x
1 5
f ( x)  C    
6 6
13
x
(a)
P( X  x)
,
where x  0, 1, 2, …, 13
(b)
(b)
P(the number of ‘3’s obtained is 6)
 P( X  6)
1
 C613  
6
 0.0103
6
0
1
2
3
4
5
6
0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0689
P( X  2)
 P ( X  0)  P ( X  1)  P ( X  2)
 0.0135  0.0725  0.1757
 0.2617
7
5
 
6
(cor. to 4 d. p.)
 0.5
P( X  3)
 P( X  2)  P( X  3)
 0.2617  0.2522
9.
(a)
 0.5139
X ~ B(6, 0.95)

 0.5
The probability function of X is
f ( x)  C x6 (0.95) x (0.05) 6  x ,

The least value of x is 3.
where x  0, 1, 2, …, 6
14.
(b)
(a)
Z ~ B(n, 0.4)
The probability function of Z is

f ( z )  Czn (0.4) z (0.6) n  z ,where z  0, 1, 2, …, n
P(5 cups of yogurt do not go bad)
 P ( X  5)
 C56 (0.95) 5 (0.05)1
 0.2321 (cor. to 4 d. p.)
(b)
(i)
P ( Z  4)
 C45 (0.4) 4 (0.6) 5  4
10.
 0.0768
Let X be the number of blank pages which are printed
out.
(ii)
Then X ~ B(4, 0.4).
P ( Z  4)
 C46 (0.4) 4 (0.6) 6  4
P(More than 2 blank pages are printed out)
 P( X  2)
 0.1382 (cor. to 4 d. p.)
 P( X  3)  P( X  4)
 C34 (0.4) 3 (0.6)1  C 44 (0.4) 4 (0.6) 0
(iii)
 0.1792
P ( Z  4)
 C47 (0.4) 4 (0.6) 7  4
 0.1935 (cor. to 4 d. p.)
11.
Let X be the number of females being selected.
Then X ~ B(12, 0.55).
15.
W ~ B(15, p)
The probability function of W is
f (w)  Cw15 p w (1  p)15  w ,
 C 41 2 (0.55) 4 (0.45) 8
where w  0, 1, 2, …, 15
 0.0762 (cor. to 4 d. p.)
© Hong Kong Educational Publishing Co.
(a)

P(exactly 4 females)
 P ( X  4)
60
Binomial Distribution
(b)
P (W  4)
(i)
19.
(a)
(i)
 C415 (0.3) 4 (1  0.3)15 4
Then X ~ B(5, 0.25).
 0.2186 (cor. to 4 d. p.)
P(no hearts)
 P( X  0)
 C05 (0.25)0 (0.75)5
P (W  4)
(ii)
 C415 (0.5) 4 (1  0.5)15 4
 0.237 305
 0.0417 (cor. to 4 d. p.)
 0.2373 (cor. to 4 d. p.)
P (W  4)
(iii)
Let X be the number of hearts obtained.
 C415 (0.7) 4 (1  0.7)15 4
(ii)
P(exactly one heart)
 P( X  1)
 0.0006 (cor. to 4 d. p.)
 C15 (0.25)1 (0.75) 4
 0.395 508
16.
P(Y  5)  0.370 74
(a)
 0.3955 (cor. to 4 d. p.)
C p (1  p)  0.370 74
5
5
5
0
p 5  0.370 74
(iii)
1
5
p  (0.370 74)
 0.82 (cor. to 2 d. p.)
(b)
P(exactly two hearts)
 P( X  2)
 C25 (0.25) 2 (0.75)3
 0.263 672
 0.2637 (cor. to 4 d. p.)
P (Y  3)
 C35 (0.82) 3 (0.18) 2
 0.1786 (cor. to 4 d. p.)
17.
(b)
 P ( X  0)  P ( X  1)  P ( X  2)
P( X  0)  0.175 656
(a)
 0.237 305  0.395 508  0.263 672
C (0.22) (0.78)  0.175 656
n
0
0
P(at most two hearts)
 P ( X  2)
n
 0.8965 (cor. to 4 d. p.)
0.78n  0.175 656
n log 0.78  log 0.175 656
20.
log 0.175 656
log 0.78
 7 (cor. to the nearest integer)
n
(b)
(a)
target.
 1
Then X ~ B 6,  .
 4
P( X  4)
P(hit exactly 3 times)
 P( X  3)
 C47 (0.22) 4 (0.78) 3
 0.0389 (cor. to 4 d. p.)
18.
1
 C36  
4
 0.1318
Let X be the number of seeds germinate.
p 

Then X ~ B 6,
.
 100 
(b)
P( X  0)  0.000 1911
0
Let X be the number of times that Patrick hits the
6
p 
 p  
C 06 
 1 
  0.000 1911
 100   100 
6
(c)
p  76 (cor. to the nearest integer)
2
4
3
 
4
(cor. to 4 d. p.)
P(hit exactly once)
 P( X  1)
1
1
 C16  
4
 0.3560
61
3
3
 
4
(cor. to 4 d. p.)
P(hit exactly twice)
 P ( X  2)
1
 C 26  
4
 0.2966
 100  p 

  0.000 1911
 100 
100  p  23.999 937 71
3
5
3
 
4
(cor. to 4 d. p.)
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
(d)
P(hit at least 4 times)
 P( X  4)
3.
(a)
Mean  n(0.65)
 12(0.65)
 7.8
 P( X  4)  P( X  5)  P( X  6)
1
 C46  
4
 0.0376
4
2
5
1
6
3
1 3
1 3
   C56      C66    
4
4 4
4 4
(cor. to 4 d. p.)
0
Standard deviation
 n(0.65)(0.35)
 12(0.65)(0.35)
 1.6523 (cor. to 4 d. p.)
11.2
pp.102 – 104
(b)
Mean  n(0.65)
 20(0.65)
 13
pp.102  104
1.
(a)
E ( X )  16(0.4)
Standard deviation
 6.4
 n(0.65)(0.35)
Var ( X )  16(0.4)(0.6)
 20(0.65)(0.35)
 3.84
(b)
 2.1331 (cor. to 4 d. p.)
E ( X )  60(0.7)
 42
(c)
Mean  n(0.65)
 600(0.65)
 390
Var ( X )  60(0.7)(0.3)
 12.6
(c)
Standard deviation
 n(0.65)(0.35)
E ( X )  120(0.32)
 600(0.65)(0.35)
 38.4
 11.6833 (cor. to 4 d. p.)
Var ( X )  120(0.32)(0.68)
 26.112
Var ( X )  10.5
4.
2.
(a)
n(0.7)(0.3)  10.5
E (Y )  40 p
 40(0.3)
n  50
 12
5.
Var (Y )  40 p(1  p)
E (Y )  7.68
12 p  7.68
 40(0.3)(0.7)
p  0.64
 8.4
(b)
6.
E (Y )  40 p
 40(0.5)
(a)
Z ~ B(n, 0.3)
E ( Z )  15
n(0.3)  15
 20
n  50
Var (Y )  40 p(1  p)
 40(0.5)(0.5)
(b)
 10
(c)
Var ( Z )
 50(0.3)(0.7)
 10.5
E (Y )  40 p
 40(0.7)
 28
7.
(a)
E (Y )  26.4
80 p  26.4
Var (Y )  40 p(1  p)
p  0.33
 40(0.7)(0.3)
 8.4
(b)
Standard deviation of Y
 80(0.33)(0.67)
 4.2057 (cor. to 4 d. p.)
© Hong Kong Educational Publishing Co.
62
Binomial Distribution
8.
(a)
Var ( Z )  6
p.104
 3  2 
n    6
 5  5 
n  25
(b)
Var ( X )  3.75
np (1  p )  3.75...........(2)
3
E (Z )  25 
5
(2)  (1):
 15
9.
E( X )  5
np  5................(1)
14.
1  p  0.75
p  0.25
Substituting p  0.25 into (1),
n(0.25)  5
n  20
Let X be the number of girls in the Lee family.
Then X ~ B(3, 0.5).
Mean  3(0.5)
 1.5
15.
Mean  8
np  8…………..(1)
Variance  3(0.5)(0.5)
 0.75
Variance  5.44
np(1 – p)  5.44……(2)
(2)  (1):
10.
Let X be the number of students who catch influenza
after the vaccination.
1  p  0.68
p  0.32
Substituting p  0.32 into (1),
Then X ~ B(40, 0.05).
n(0.32)  8
Mean  40(0.05)
2
n  25
Variance  40(0.05)(0.95)
 1.9
16.
E ( X )  12.6
np  12.6..............(1)
Var ( X )  E ( X 2 )  [ E ( X )]2
11.
Let X be the number of cockroaches that are killed.
 168.714  12.6 2
Then X ~ B(8, 0.9).
 9.954
Mean  8(0.9)
 7.2

Variance  8(0.9)(0.1)
 0.72
12.
np(1  p)  9.954……….(2)
(2)  (1):
1  p  0.79
p  0.21
Substituting p  0.21 into (1),
n(0.21)  12.6
n  60
Let X be the number of smokers being selected.
Then X ~ B(16, 0.4).
Mean  16(0.4)
17.
 6.4
Var (Y )  11.2
np(1  p)  11.2...............(1)
Var (Y )  E (Y 2 )  [ E (Y )]2
Variance  16(0.4)(0.6)
 3.84
11.2  3147.2  [ E (Y )]2
E (Y )  56
np  56.................(2)
13.
X ~ B(15, 0.125)
E ( X )  15(0.125)
 1.875
(1)  (2):
Var(X)  15(0.125)(1  0.125)
 1.640 625
Substituting p  0.8 into (2),
n(0.8)  56
n  70
63
1  p  0.2
p  0.8
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
E ( Z )  36
18.
21.
(a)
np  36................(1)
John gets.
Var ( Z )  0.2496n
1

Then X ~ B 20, .
4

  E( X )
np(1  p)  0.2496n
p(1  p)  0.2496
p  p  0.2496  0
1
 20 
4
5
2
 (1)  (1)  4(1)(0.2496)
2
p
2(1)
 0.48 or 0.52
  Var ( X )
For p  0.48,
 1  3 
 20  
 4  4 
 1.936 492
Substituting p  0.48 into (1),
n(0.48)  36
n  75
 1.9365 (cor. to 4 d. p.)
For p  0.52,
Substituting p  0.52 into (1),
(b)
n(0.52)  36
n  69.230 769 , which is not an integer.


Let X be the number of correct answers
P( X     )
 P ( X  5  1.936 492)
 P ( X  3.063 508)
p  0.52 is rejected.
n  75 and p  0.48
 1  P ( X  3)
 1  P ( X  0)  P ( X  1)  P ( X  2)  P ( X  3)
20
19.
19
2
18
3
 1  3 
1 3
 1     C12 0     C22 0    
4
 4  4 
4 4
E ( X )  67.5
np  67.5...............(1)
3
17
1 3
 C32 0    
4 4
 0.7748 (cor. to 4 d. p.)
Var ( X )  82.5 p
np (1  p )  82.5 p
n(1  p )  82.5
n  np  82.5..............( 2)
22.
Substituting (1) into (2),
n  67.5  82.5
n  150
(a)
  E (Y )
 130(0.014)
 1.82
Substituting n  150 into (1),
150 p  67.5
  Var (Y )
2
 130(0.014)(1  0.014)
p  0.45
20.
 1.794 52
Let X be the number of heads.
(b)
Then X ~ B(n, p).
 P(Y  3.159 597)
 1  P(Y  3)
np  52.5...................(1)
Variance  24.9375
 1  P(Y  0)  P(Y  1)  P(Y  2)  P(Y  3)
np(1  p)  24.9375…....(2)
 1  (0.986)130  C1130(0.014)(0.986)129
1  p  0.475
p  0.525
 C2130(0.014) 2 (0.986)128  C3130(0.014)3 (0.986)127
 0.1106 (cor. to 4 d. p.)
Substituting p  0.525 into (1),
n(0.525)  52.5
n  100
© Hong Kong Educational Publishing Co.
P(Y     )
 P(Y  1.82  1.794 52
Mean  52.5
(2)  (1):
Y ~ B(130, 0.014)
64
Binomial Distribution
4.
11.3
(a)
Let X be the number of respondents agree.
pp.110 – 115
Then X ~ B(5, 0.35).
pp.110  113
P(exactly 3 respondents agree)
 P( X  3)
 C35 (0.35)3 (0.65) 2
1.
Let X be the number of boys they have.
 0.181 147
Then X ~ B(4, 0.49).
 0.1811 (cor. to 4 d. p.)
P(they have 2 boys)
 P ( X  2)
(b)
 C 24 (0.49) 2 (0.51) 2
 0.3747 (cor. to 4 d. p.)
P(at least 3 respondents agree)
 P( X  3)
 P( X  3)  P ( X  4)  P( X  5)
 0.181 147  C45 (0.35) 4 (0.65)  (0.35)5
2.
(a)
(b)
Y ~ B(3, 0.4)
 0.231 695
The probability function of Y is
f ( y )  C y3 (0.4) y (0.6) 3  y , where y  0, 1, 2, 3.
 0.2352 (cor. to 4 d. p.)
(c)
Mean  3(0.4)
 1.2
 1  P ( X  3)
Variance  3(0.4)(0.6)
 1  0.235 169 5
 0.72
3.
(a)
P(fewer than 3 respondents agree)
 P ( X  3)
 0.7648 (cor. to 4 d. p.)
Let X be the number of patients recover from the
5.
(a)
disease.
X ~ B(2, p)
P(a system of large engines fails)
 P( X  2)
Then X ~ B(10, 0.6).
P(exactly 2 patients recover)
 p2
 P( X  2)
 C210 (0.6) 2 (0.4)8
 0.010 617
(b)
Y ~ B(4, 2p)
P(a system of small engines fails)
 P (Y  2)
 0.0106 (cor. to 4 d. p.)
 P (Y  3)  P (Y  4)
(b)
P(at most 2 patients recover)
 C34 ( 2 p ) 3 (1  2 p )  ( 2 p ) 4
 P ( X  2)
 32 p 3  64 p 4  16 p 4
 P ( X  0)  P ( X  1)  P ( X  2)
 32 p 3  48 p 4
 (0.4)1 0  C11 0 (0.6)(0.4)9  0.010 617
 0.001 678  0.010 617
(c)
 0.0123 (cor. to 4 d. p.)
(c)
P(at least 2 patients recover)

Both systems have an equal chance of

failure.
P( X  2)  P(Y  2)
p 2  32 p 3  48 p 4
 P( X  2)
48 p 2  32 p  1  0
 1  P( X  2)
 1  [ P( X  0)  P( X  1)]
p
 (32)  (32) 2  4(48)(1)
2(48)
 0.0329 or 0.6338 (rejected)
 1  0.001 678
 0.9983 (cor. to 4 d. p.)
(cor. to 4 d. p.)
65
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
6.
(a)
Let X be the number of defective batteries in a
9.
(a)
package of 10.
1 P (all 3 attempts fail)
Then X ~ B(10, 0.02).
 1  (1  0.6) 3
Proportion
 P ( X  1)
 0.936
 1  P ( X  1)
(b)
 1  P ( X  0)  P ( X  1)
Let X be the number of members who are
qualified in a team.
 1  (0.98)1 0  C11 0 (0.02)(0.98) 9
Then X ~ B(5, 0.936).
 0.0162 (cor. to 4 d. p.)
(b)
P(an athlete will be qualified for the final)
P(a team is qualified for the final)
 P( X  4)
Let Y be the number of packages need to be
 P( X  4)  P( X  5)
replaced.
 C45 (0.936) 4 (1  0.936)  (0.936) 5
Then Y ~ B(10, 0.0162) approximately.
 0.9640 (cor. to 4 d. p.)
P(3 of the packages need to be replaced)
 P (Y  3)
10.
 C31 0 (0.0162) 3 (0.9838) 7
(a)
(i)
 0.0005 (cor. to 4 d. p.)
X ~ B(8, 0.3)
The probability function of X is

f ( x)  C x8 (0.3) x (0.7)8  x ,
where x  0, 1, 2, …, 8.
7.
Let X be the number of games she wins.
Then X ~ B(5, p).
P( X  4)  3 p4
(ii)
of relevant experience)
C45 p 4 (1  p )  p 5  3 p 4
 P ( X  3)
5 p 4  4 p5  3 p 4
 C38 (0.3) 3 (0.7) 5
4p  2p  0
2 p 1  0
5
4
p
8.
(a)
 0.2541 (cor. to 4 d. p.)
1
2
(b)
(i)
where x  0, 1, 2, …, 8.
Then X ~ B(12, 0.05).
(ii)
Proportion
 P( X  2)
 1  P ( X  2)
 1  P ( X  0)  P ( X  1)  P( X  2)
 P(Y  5)
 1  P(Y  5)
 1  P(Y  6)  P(Y  7)  P(Y  8)
 C 212 (0.05) 2 (0.95)10
 1  C68 (0.7) 6 (0.3) 2  C78 (0.7) 7 (0.3)  (0.7) 8
 0.0196 (cor. to 4 d. p.)
 0.4482 (cor. to 4 d. p.)
Let Y be the number of defective discs in a
sample of 12 by using the new machines.
Then Y ~ B(12, 0.02).
Proportion  P(Y  1)
 1  P(Y  1)
 1  P(Y  0)  P(Y  1)
 1  (0.98)12  C112 (0.02)(0.98)11
 0.0231
 P( X  2)
The proportion increases.
© Hong Kong Educational Publishing Co.
P(at least 3 applicants have less than
10 years of relevant experience)
 P( X  3)
 1  (0.95)12  C112 (0.05)(0.95)11

Y ~ B(8, 0.7)
The probability function of Y is

f ( y )  C y8 (0.7) y (0.3) 8  y ,
Let X be the number of defective discs in a
sample of 12.
(b)
P(3 applicants have less than 10 years
66
Binomial Distribution
11.
(a)
pp.113  115
Let A be the event that a bird is diagnosed to
have H5N1 virus and B be the event that a bird
13.
has the virus.
(a)
P(exactly 3 students can type more than 30
Chinese characters per minute)
The required probability
 P( B | A)
P( B  A)

P( A)
P( A | B) P( B)

P( A | B) P( B)  P( A | B) P( B)
0.02  0.95

0.98  0.05  0.02  0.95
19

68
 P(2 boys and 1 girl can do this)
+ P(1 boy and 2 girls can do this)
 (0.4  0.4)  (0.5  0.5  2)
(0.4  0.6  2)  (0.5  0.5)
 0.2
(b)
Let A be the event that a student can type more
than 30 Chinese characters per minute.
P(A)  P(boy and A) + P(girl and A)
 0.6  0.4 + 0.4  0.5
(b)
Let X be the number of the birds have the virus if
 0.44
they are diagnosed to have the virus.
 49 
X ~ B 3,

 68 

30 Chinese characters per minute)
 0.44 n
P(at least two birds have the virus)
 P ( X  2)
 P ( X  2)  P ( X  3)
2
 49   19   49 
C     
 68   68   68 
(c)
sample of m students.
Then X ~ B(m, 0.44).
P ( X  m  1)  0.95
1  P ( X  m)  0.95
 0.8094 (cor. to 4 d. p.)
(a)
1  0.44m  0.95
P(all 4 forgotten digits are different)
0.44m  0.05
m log 0.44  log 0.05
P410
10 4
 0.504

(b)
log 0.05
log 0.44
m  3.648 969
m
P(forgotten digits only consist of 1, 2, 5 or

9 and can appear more than once)
44
 4
10
 0.0256
(c)
Let X be the number of students can type more
than 30 Chinese characters per minute in a
3
3
2
12.
P(All n student can type more than
14.
(a)
Let X be the number of forgotten digits have
The minimum number is 4.
Let X be the number of games Aaron wins.
 2
Then X ~ B 5,  .
 3
already appeared among the first 3 digits.
P(Aaron wins)
 P( X  3)
Then X ~ B(4, 0.3).
 P( X  3)  P( X  4)  P( X  5)
P(exactly 2 of them)
 P ( X  2)
 2 1
 2 1  2
 C35      C45       
 3  3
 3  3  3
3
 C 24 (0.3) 2 (0.7) 2
2
4
5
 0.7901 (cor. to 4 d. p.)
 0.2646
(b)
67
Aaron’s expected gain
 2

 $ 5   p 
 3

10
$ p
3
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
(c)
(d)
Calvin’s expected gain
 1

 $ 5   q 
3


5
$ q
3
16.
(a)
Let X be the number of imperfect fans.
Then X ~ B(10, 0.01).
P(Victor rejects)
 P( X  1)
 1  P( X  0)  P( X  1)
 1  (0.99)10  C110 (0.01)(0.99)9
 0.0043 (cor. to 4 d. p.)
In order to be a fair game, their expected gains
should be equal.
10
5
p q

3
3
q  2p
(b)
Let Y be the number of imperfect fans.
Then Y ~ B(10, 0.1).
P(Victor accepts)
 P(Y  1)
15.
(a)
Let X be the number of successful throws.
 P(Y  0)  P(Y  1)
Then X ~ B(10, 0.8).
The mean  E ( X )
 10(0.8)
8
 (0.9)10  C110 (0.1)(0.9)9
 0.7361 (cor. to 4 d. p.)
(c)
(b)
Yes, since even though 10% of the fans are
imperfect, it seems to have a good chance
P( X  7)  C710 (0.8) 7 (0.2) 3
(0.7361) that Victor accept the claim ‘only 1% of
 0.201 326 592
the fans are imperfect’.
P( X  8)  C810 (0.8)8 (0.2) 2
 0.301 989 888
17.
P( X  9)  C910 (0.8) 9 (0.2)
Let X be the number of dishes fail the test in a sample
of 5 dishes.
 0.268 435 456
Let Y be the number of dishes fail the test in a sample
P( X  10)  (0.8)
 0.107 374 182
10
of 3 dishes.
Then X ~ B(5, 0.05) and Y ~ B(3, 0.05).
P( X  6)
 1  P( X  7)  P( X  8)  P( X  9)  P( X  10)
P(accepting a batch)
 P ( X  0)  P ( X  1) P (Y  0)
 0.120 873 882
 (0.95) 5  C15 (0.05)(0.95) 4  (0.95) 3
Michael’s expected gain
 $[50  P( X  6)  14  P( X  7)  16  P( X  8)
 0.9484 (cor. to 4 d. p.)
 18  P( X  9)  20  P( X  10)]
18.
 $9 (cor. to the nearest dollar)
(a)
Let X be the number of times a person correctly
answers.
Then X ~ B(8, 0.5).
(c)
It is not a fair game.
P(a person fails the test)
 P ( X  7)
Let $y be the amount that Michael should pay.
Michael’s expected gain  $0
  y  P( X  6)  14  P( X  7)  16  P( X  8)
 1  P( X  7)  P( X  8)
 1  C78 (0.5) 7 (0.5)  (0.5)8
18  P( X  9)  20  P( X  10)  0
 0.9648 (cor. to 4 d. p.)
0.120 873 882 y  14.629 732 34  0
y  121.033 03
(b)
 Michael should pay $121.
Let Y be the number of times he gets the
outcomes of a rolled dice.
1
Then Y ~ B(4,
).
6
P(He passes)  P(Y  3)
 P(Y  3)  P(Y  4)
(cor. to the nearest dollar)
3
1 5 1
 C 34       
6 6 6
 0.016 203 7
4
 0.0162 (cor. to 4 d. p.)
© Hong Kong Educational Publishing Co.
68
Binomial Distribution
(c)
P(William passes the test)
 P( X  7)  P( X  6) P(Y  3)
(b)
Let Y be the number of batches being rejected.
Then Y ~ B(20, 0.058 153 8) approximately.
 C78 (0.5)7 (0.5)  (0.5)8  C68 (0.5)6 (0.5)2
P(no more than 2 out of 20 batches
 0.016 203 7
tested will be rejected)
 P(Y  2)
 0.0369 (cor. to 4 d. p.)
 P(Y  0)  P(Y  1)  P(Y  2)
19.
(a)
(i)
 (0.941 846 2) 20  C120 (0.058 153 8)(0.941 846 2)19
Let X be the number of patients agree
 C220 (0.058 153 8) 2 (0.941 846 2)18
the trial.
 0.8929 (cor. to 4 d. p.)
Then X ~ B(10, 0.4).
P(exactly 2 patients agree)
 P( X  2)
 C (0.4) (0.6)
10
2
2
(c)
8
Let Z be the number of defective modules
found in a sample of n modules.
 0.120 932
Then Z ~ B(n, 0.04).
 0.1209 (cor. to 4 d. p.)
P(a batch is accepted)
 P ( Z  1)
 P ( Z  0)  P ( Z  1)
(ii)
P(at most 2 patients agree)
 (0.96) n  C1n (0.04)(0.96) n  1
 P ( X  2)
 (0.96) n  (0.04n)(0.96) n  1
 P ( X  0)  P ( X  1)  P ( X  2)
 (0.96  0.04n)  0.96 n  1
 (0.6)1 0  C11 0 (0.4)(0.6)9  0.120 932
When n  13,
P( Z  1)
 0.167 289
 0.1673 (cor. to 4 d. p.)
 (0.96  0.04  13)  0.9612
 0.906 810 4
(iii)
 0.9
P(at least 2 patients agree)
 P( X  2)
When n  14,
P( Z  1)
 P( X  2)  P( X  2)
 P( X  2)  [1  P( X  2)]
 (0.96  0.04  14)  0.961 3
 0.120 932  (1  0.167 289)
 0.894 066 1
 0.9536 (cor. to 4 d. p.)
 0 .9

(b)
(i)
The maximum value of n is 13.
No, since the patients are receiving two
kind of trials with the different probability
(ii)
of withdrawing.
pp.118 – 126
No, since the number of patients to be
pp.118  120
invited is not a constant.
20.
(a)
1.
Let X be the number of defective modules found
(a)
in a sample.
Then X ~ B(10, 0.04).
P(a batch is rejected)
 P ( X  1)
 1  P ( X  0)  P ( X  1)
 1  (0.96)1 0  C11 0 (0.04)(0.96)9
 0.058 153 8
 0.0582 (cor. to 4 d. p.)
69
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
(b)
(c)
P( X  4)
 1  P( X  4)
 1  P( X  3)  P( X  3)  P( X  4)
 1  0.678 543  0.207 642  C48 (0.25)4 (0.75)4
 0.0273 (cor. to 4 d. p.)
4.
(a)
P (Y  5)
 C515 (0.35)5 (0.65)10
 0.212 339
(c)
 0.2123
(b)
P (Y  5)
 P(Y  0)  P(Y  1)  P (Y  2)  P(Y  3)
 P(Y  4)
 (0.65)15  C115 (0.35)(0.65)14  C215 (0.35) 2 (0.65)13
 C315 (0.35)3 (0.65)12  C415 (0.35) 4 (0.65)11
 0.351 943
 0.3519 (cor. to 4 d. p.)
2.
(a)
P (Y  0)
 C018 (0.24) 0 (0.76)18  0
(c)
 0.0072 (cor. to 4 d. p.)
(b)
P (Y  6)
 1  P ( X  6)
 1  P ( X  5)  P ( X  5)
 1  0.212 339  0.351 943
P(Y  6)
 0.4357 (cor. to 4 d. p.)
 C618 (0.24) 6 (0.76)18  6
 0.1317 (cor. to 4 d. p.)
5.
(c)
(a)
P (Y  11)
 C28 (0.73) 2 (0.27)6  C38 (0.73)3 (0.27)5
 P(Y  12)  P(Y  13)  P(Y  14)  P(Y  15)
 C48 (0.73) 4 (0.27) 4  C58 (0.73)5 (0.27)3
 P(Y  16)  P(Y  17)  P(Y  18)
 0.005 780 8  0.031 259 0  0.105 643 9
18
18
 C12
(0.24)12 (0.76) 6  C13
(0.24)13 (0.76) 5
 0.228 503 9
18
18
C14
(0.24)14 (0.76) 4  C15
(0.24)15 (0.76) 3
 0.3712 (cor. to 4 d. p.)
18
18
 C16
(0.24)16 (0.76) 2  C17
(0.24)17 (0.76)1
18
 C18
(0.24)18 (0.76) 0
 0.0002 (cor. to 4 d. p.)
3.
(a)
P ( 2  X  5)
 P ( X  2)  P ( X  3)  P ( X  4)  P ( X  5)
(b)
P (3  X  7)
 P( X  4)  P( X  5)  P( X  6)
 0.105 643 9  0.228 503 9  C68 (0.73)6 (0.27) 2
P( X  3)
 0.6431 (cor. to 4 d. p.)
 C38 (0.25)3 (0.75)5
 0.207 642
 0.2076 (cor. to 4 d. p.)
(b)
6.
(a)
P(1  X  5)
 P( X  2)  P( X  3)  P( X  4)  P( X  5)
 C218 (0.31) 2 (0.69)16  C318 (0.31) 3 (0.69)15
P( X  3)
 C418 (0.31) 4 (0.69)14  C518 (0.31) 5 (0.69)13
 0.4856 (cor. to 4 d. p.)
 P( X  0)  P( X  1)  P( X  2)
 (0.75)8  C18 (0.25)(0.75)7  C28 (0.25)2 (0.75)6
 0.678 543
(b)
 0.6785 (cor. to 4 d. p.)
P(11  X  15)
 P( X  11)  P( X  12)  P( X  13)  P( X  14)
18
18
 C11
(0.31)11 (0.69) 7  C12
(0.31)12 (0.69) 6
18
18
 C13
(0.31)13 (0.69) 5  C14
(0.31)14 (0.69) 4
 0.0080 (cor. to 4 d. p.)
© Hong Kong Educational Publishing Co.
70
Binomial Distribution
7.
(a)
Let X be the number of sunny days in April.
12.
X ~ B(200, 0.67)
Mean  E(X)
Then X ~ B(30, 0.4).
 200(0.67)
 134
P(exactly 4 sunny days)
 P ( X  4)
 C 430 (0.4) 4 (0.6) 26
Standard deviation
 0.0012 (cor. to 4 d. p.)
 Var ( X )
 200(0.67)(0.33)
(b)
P(exactly 6 days that are not sunny)
 P ( X  24)
 6.6498 (cor. to 4 d. p.)
 C 2340 (0.4) 2 4 (0.6) 6
13.
 7.80 10 6 (cor. to 3 sig. fig.)
E (Y )  27
36 p  27
p  0.75
8.
Let X be the number of guests drive to the venue.
Then X ~ B(15, 0.46).
14.
P(all parking spaces are occupied)
 P ( X  10)
(a)
Var(Z)  2.88
12 p(1  p)  2.88
12 p 2  12 p  2.88  0
 P ( X  10)  P( X  11)  P ( X  12)  P ( X  13)
 (12)  (12) 2  4(12)(2.88)
2(12)
 0.4 or 0.6
 P ( X  14)  P ( X  15)
p
15
15
 C10
(0.46)10 (0.54) 5  C11
(0.46)11(0.54) 4
15
15
 C12
(0.46)12 (0.54) 3  C13
(0.46)13 (0.54) 2
15
 C14
(0.46)14 (0.54)  (0.46)15
(b)
 0.0890 (cor. to 4 d. p.)
9.
(a)
P(Z  4)
 C412 (0.4) 4 (0.6)8
X ~ B(4, 0.52)
 0.2128 (cor. to 4 d. p.)
The probability function of X is
f ( x)  C x4 (0.52) x (0.48) 4  x ,
For p  0.6,
P(Z  4)
 C412 (0.6) 4 (0.4)8
where x  0, 1, 2, 3, 4
(b)
 0.0420 (cor. to 4 d. p.)
P(the family has exactly 1 girl)
 P ( X  1)
15.
 C14 (0.52)(0.48) 3
(a)
 0.2300 (cor. to 4 d. p.)
10.
E ( X )  48
n(0.32)  48
n  150
E ( X )  50(0.85)
(b)
 42.5
P( X  24)
150
 C24
(0.32) 24 (0.68)126
Var ( X )  50(0.85)(0.15)
 4.0622 10 6 (cor. to 5 sig. fig.)
 6.375
11.
For p  0.4,
16.
E ( X )  120(0.15)
 18
Let X be the number of members who are university
graduates.
Then X ~ B(12, 0.28).
E ( X 2 )  Var ( X )  [ E ( X )]2
Mean  E(X)
 120(0.15)(0.85)  18 2
 12(0.28)
 3.36
 339.3
Standard deviation
 Var ( X )
 12(0.28)(0.72)
 1.5554 (cor. to 4 d. p.)
71
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
17.
Let X be the number of students who are late.
21.
Then X ~ B(45, 0.1).
Then X ~ B(8, 0.9).
Mean  E(X)
P(all the hostages are rescued)
 P( X  6)
 45(0.1)
 4.5
 P( X  6)  P( X  7)  P( X  8)
 C68 (0.9) 6 (0.1) 2  C78 (0.9) 7 (0.1)  (0.9)8
Variance  Var(X)
 0.9619 (cor. to 4 d. p.)
 45(0.1)(0.9)
 4.05
22.
18.
Let X be the number of helicopters arrive successfully.
(a)
Let X be the number of customers using an
X ~ B(15, 0.25)
Octopus card.
Mean  E(X)
Then X ~ B(25, 0.7).
 15(0.25)
 3.75
P(exactly 12 customers pay the bill
using Octopus card)
 P( X  12)
Standard deviation
 C1225 (0.7)12 (0.3)13
 Var ( X )
 0.0115 (cor. to 4 d. p.)
 15(0.25)(0.75)
 1.6771 (cor. to 4 d. p.)
(b)
P(at most 20 customers pay the bill
using an Octopus card)
19.
(a)
Let X be the number of defective food
 P ( X  20)
containers.
 1  P ( X  20)
Then X ~ B(6, 0.1).
 1  P ( X  21)  P ( X  22)  P ( X  23)
P(a box contains 2 defective food containers)
 P ( X  2)
 1  C 2215 (0.7) 2 1(0.3) 4  C 2225 (0.7) 2 2 (0.3) 3
 P ( X  24)  P ( X  25)
 C 2235 (0.7) 2 3 (0.3) 2  C 2245 (0.7) 2 4 (0.3)  (0.7) 2 5
 C 26 (0.1) 2 (0.9) 4
 0.9095 (cor. to 4 d. p.)
 0.0984 (cor. to 4 d. p.)
(b)
pp.120  125
P(a box contains more than 2
defective food containers)
 P ( X  2)
23.
P( X  0)  0.008 393
(1  p)10  0.008 393
 1  P ( X  2)
1
 1  P ( X  0)  P ( X  1)  P ( X  2)
1  p  (0.008 393)10
 1  (0.9) 6  C16 (0.1)(0.9) 5  C 26 (0.1) 2 (0.9) 4
p  0.38 (cor. to 2 d. p.)
 0.0159 (cor. to 4 d. p.)
24.
20.
(a)
P( X  n)  0.005 489
(0.42) n  0.005 489
Let X be the number of defective items in the batch.
n log 0.42  log 0.005 489
log 0.005 489
n
log 0.42
Then X ~ B(12, 0.15).
P(a batch will be returned)
 P ( X  2)
 1  P ( X  2)
 1  P ( X  0)  P( X  1)
 6 (cor. to the nearest integer)
 1  (0.85)12  C112 (0.15)(0.85)11
(b)
 0.5565 (cor. to 4 d. p.)
P( X  3)
 C36 (0.42)3 (0.58)3
 0.2891 (cor. to 4 d. p.)
© Hong Kong Educational Publishing Co.
72
Binomial Distribution
25.
E (Z ) 
np 
1
1 p 
p
6
Var ( Z )
5
6
np (1  p )
5
6
(1  p )
5
5
6
1
6
(b)
 P(repairing 8 air-conditioners)
+ P(replacing 1 air-conditioner)
 C880 (0.2)8 (1  0.2  0.1) 72
 C180 (0.1)(1  0.1  0.2) 79
 5.218  10 7 (cor. to 4 sig. fig.)
28.
(a)
Then X ~ B(5, 0.7).
P(get more than 10 points)
 P( X  3)
 P( X  3)  P( X  4)  P( X  5)
 C35 (0.7) 3 (0.3) 2  C45 (0.7) 4 (0.3)  (0.7) 5
 0.8369 (cor. to 4 d. p.)
P( X  k )
P( X  k  1)

(b)
nk
C ( p ) (1  p )
C kn  1 ( p ) k  1 (1  p ) n  k  1
n
k
k
Let X be the number of shots that she hits the
orange region.
25
9
25
np (1  p ) 
9
 1  1  25
n 1   
 6  6  9
n  20
Var ( Z ) 
26.
P(total cost is $4000)
P(get exactly 11 points)
 P ( X  3)
 C35 (0.7) 3 (0.3) 2
 0.3087
n!
(1  p )
( n  k )!k!

n!
( p)
( n  k  1)!( k  1)!
n!(1  p ) ( n  k  1)!( k  1)!


( n  k )!k!
n! p
( k  1)(1  p )

(n  k ) p
29.
(a)
1
)
6
P( X  1)  P( X  0)
X ~ B(5,
5
5
 
6
 0.401 878
 0.4019 (cor. to 4 d. p.)
27.
(a)
Let X be the number of air-conditioners needed
P( X  1)
to be repaired and Y be the number of
 P ( X  0)  P ( X  1)
air-conditioners need to be replaced every year.
5
5
 1  5 
    C15   
6
 6  6 
 0.803 755
Then X ~ B(80, 0.2) and Y ~ B(80, 0.1).
Expected costs of repairing
 $ E (500 X )
4
 0.8038 (cor. to 4 d. p.)
 $[500 E ( X )]
 $[500  80(0.2)]
 $8000
(b)
Expected costs of replacing
 $ E (4000Y )
P( B)  P( X  1)
 1  P ( X  1)
 1  0.401 878
 0.598 122
 $[4000 E (Y )]
P( A  B)  P( X  1)
 1  P( X  1)
 $4000  80  0.1
 $32 000
 1  0.803 755
 0.196 245
P( A  B)
P ( B)
0.196 245

0.598 122
P( A | B) 
 0.3281 (cor. to 4 d. p.)
73
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
30.
(a)
Let X be the number of defective toys in a box.
(c)
Then X ~ B(5, 0.1).
 1  P( X  8)
 1  P( X  0)  P( X  1)  P( X  2)
P(a box delivered contains 2 defective toys)
 P(2 toys chosen are not defective | X  2) 
 P( X  3)  P( X  4)  P( X  5)
 P( X  6)  P( X  7)
P(X  2)
C3
 25  C25 (0.1) 2 (0.9)3
C2
6
14
7
13
1 3
1 3
 0.382 827 3  C620      C720    
4 4
4 4
 0.1018 (cor. to 4 d. p.)
 0.021 87
(b)
P(Peter passes the test)
 P( X  8)
P(a box delivered contains a defective toy)
 P(2 toys chosen are not defective | X  1) 
(d)
P(X  1)
C4
 25  C15 (0.1)(0.9) 4
C2
If the total number of questions is 10,
let Y be the correct answers he gets.
1

Then Y ~ 10,  .
4

 0.196 83
P(Peter passes the test)
 P(Y  4)
(c)
 1  P(Y  4)
P(a box delivered contains defective toy(s))
 1  P(Y  0)  P(Y  1)  P(Y  2)  P(Y  3)
 P(2 toys chosen are not defective and X  1)
10
3
1 3
 C310    
4 4
C2
 25  C35 (0.1)3 (0.9) 2
C2

(a)
Let X be the number of correct answers Peter
32.
gets.
(a)
8
7
Let A be the event that a battery passes the test
Korea.
P ( B | A)
P( A  B)

P ( A)
P(he gets 5 correct answers)
 P( X  5)
5
The answer in (c) will be increased.
and B be the event that a battery is produced in
1

Then X ~  20,  .
4

15
1 3
 C520    
4 4
 0.2023 (cor. to 4 d. p.)
(b)
2
 0.2241 (cor. to 4 d. p.)
 0.219 51
31.
9
3
 1  3 
1 3
 1     C110     C210    
4
 4  4 
4 4
C4
C3
 25  C15 (0.1)(0.9) 4  25  C25 (0.1) 2 (0.9)3
C2
C2
0 .5  0 .4
0 . 5  0 . 4  0 . 2  0 . 8  0 . 3  0 .6
10

27

P(he gets more than 5 correct answers)
 P ( X  5)
(b)
 1  P ( X  5)
produced in Korea in the selected batteries.
 10 
Then X ~ B 4,  .
 27 
 1  P ( X  0)  P ( X  1)  P ( X  2)
 P ( X  3)  P ( X  4)  P ( X  5)
20
19
2
18
3
 1  3 
1 3
 1     C12 0     C22 0    
4
 4  4 
4 4
3
17
4
P(at least 2 of them are from USA or Japan)
 P ( X  2)
 P ( X  0)  P ( X  1)  P ( X  2)
16
1 3
1 3
 C32 0      C42 0    
4 4
4 4
5
4
3
2
 17 
 10  17 
 10   17 
    C14     C 24    
 27 
 27  27 
 27   27 
 0.8532 (cor. to 4 d. p.)
15
1 3
 C52 0    
4 4
 0.382 827 3
 0.3828 (cor. to 4 d. p.)
© Hong Kong Educational Publishing Co.
Let X be the number of batteries that are
74
2
Binomial Distribution
33.
(a)
Let X and Y be the number of components fail
35.
(a)
Let X be the number of routers function in a
in devices A and B respectively.
3-router network.
Then X ~ B(2, q) and Y ~ B(4, q).
Then X ~ B(3, p).
P(failure of A)
P(a 3-router network operates)
 P ( X  2)
 P ( X  2)
 P ( X  2)  P ( X  3)
 q2
 C 23 p 2 (1  p )  p 3
P(failure of B)
 P(Y  2)
 3 p2  3 p3  p3
 3 p2  2 p3
 P(Y  3)  P(Y  4)
 C34 q 3 (1  q)  q 4
 q 3 ( 4  4q  q )
(b)
 q 3 (4  3q)
Let Y be the number of routers function in
a 5-router network.
Then Y ~ B(5, p).
(b)
P(a 5-router network operates)
P(failure of A)  P(failure of B)
q 2  q 3 (4  3q )
1  q ( 4  3q )
 P (Y  3)
 P (Y  3)  P (Y  4)  P (Y  5)
 C35 p 3 (1  p ) 2  C 45 p 4 (1  p)  p 5
3q 2  4q  1  0
 10 p 3 (1  2 p  p 2 )  5 p 4  5 p 5  p 5
(3q  1)(q  1)  0
 10 p 3  20 p 4  10 p 5  5 p 4  5 p 5  p 5
1
 q 1
3
 10 p 3  15 p 4  6 p 5
For
34.
(a)
P(a 5-router network operates) >
P(a 3-router network operates)
10 p3 15 p 4  6 p5  3 p 2  2 p3
Let X be the number of defective components
in the sample.
6 p 3  15 p 2  12 p  3  0
Then X ~ B(30, p).
P ( X  1)  0.04
2 p3  5 p 2  4 p 1  0
( p  1)(2 p 2  3 p  1)  0
C130 p(1  p) 29  0.04.................(1)
( p  1) 2 (2 p  1)  0
P( X  2)  0.1
C p (1  p)  0.1.................(2)
30
2
2
(2)  (1):
1
 p 1
2
28
30
2
C p
 2 .5
C13 0 (1  p )
14.5 p  2.5(1  p)
36.
(a)
17 p  2.5
p  0.15 (cor. to 2 d. p.)
(b)
P(cannot win)
 P(three ‘1’) + P(two ‘1’ and one ‘3’)
3
2
5
5 3
    C23    
9
9 9
350

729
Let Y and Z be the number of defective
components in instruments A and B respectively.
Then Y ~ B(3, 0.15) and
Z ~ B(4, 0.15) approximately.
(b)
P(type A functions properly)
 P(Y  1)
 P(Y  0)  P(Y  1)
P(Y  500)
 P( X  10)
 P(two ‘3’ and one ‘5’) + P(two ‘5’ and one ‘1’)
+ P(two ‘5’ and one ‘3’) + P(three ‘5’)
 (0.85)3  C13 (0.15)(0.85) 2
2
2
2
3 1
1 5
1 3
 C23      C23      C23    
9 9
9 9
9 9
 0.939
P(type B functions properly)
1
 
9
52

729
 P( Z  2)
 P( Z  0)  P( Z  1)  P( Z  2)
 (0.85) 4  C14 (0.15)(0.85)3  C24 (0.15) 2 (0.85) 2
3
 0.988

Type B is more reliable.
75
© Hong Kong Educational Publishing Co.
11
New Progress in Senior Mathematics Module 1 Book 2 (Extended Part) Solution Guide
P(Y  100)
 1  P(Y  0)  P(Y  500)
Extended Questions
350 52

729 729
109

243
42.
1

(c)
(i)
(a)
p.126
Let X be the number of guinea pigs which are
infected.
Then X ~ B(7, 0.7).
P(4 out of 7 guinea pigs are infected)
The probability function of Y is
 350
 729 for y  0

 109 for y  100
f ( y )   243
 52
for y  500

 729
0
otherwise
 P ( X  4)
 C 47 (0.7) 4 (0.3) 3
 0.2269 (cor. to 4 d. p.)
(b)
P(the 7th guinea pig is the 4th one infected)
 P(X  4 and the 7th guinea pig is infected)
4
 C47 (0.7) 4 (0.3) 3 
7
 0.1297 (cor. to 4 d. p.)
Let Z be the number of customers get no
prize.
350 

Then Z ~ B 20,
.
729 

43.
(a)
(i)
Total number of matches have been
selected  4
(ii)
Among the 4 matches selected, one is from
P(less than 5 customers get no price)
 P( Z  5)
 P( Z  0)  P( Z  1)  P( Z  2)
 P( Z  3)  P( Z  4)
20
any box and the other three are from
another box.
19
 379 
 350  379 

  C120 


729


 729  729 
2
Number of combinations
 C 34
18
 350   379 
 C220 
 

 729   729 
3
17
4
16
4
 350   379 
 C320 


 729   729 
(iii)
find one of the box is empty)
 350   379 
 C420 
 

 729   729 
 0.0096 (cor. to 4 d. p.)
(ii)
 P(picks 4 matches and one box is empty)
1
 C34  
2
1

4
Let A be the number of customers getting a
$500 coupon.
52 

Then A ~ B 20,
.
729 

(b)
1
 C N2 N  k  
2
19
 52  677 
C 


 729  729 
 0.3497 (cor. to 4 d. p.)
39  41
(HKALE Questions)
© Hong Kong Educational Publishing Co.
P(exactly k matches in the other box |
 P(pick 2N  k matches and one box is empty)
20
1
(HKASLE Questions)
4
find one of the box is empty)
P(only 1 customer getting a $500 coupon)
 P ( A  1)
37  38
P(exactly 2 matches in the other box |
76
2N  k
Binomial Distribution
Open-ended Questions
44.
p.126
For n  10, p  0.4, m  8, q  0.5:
E ( X )  10(0.4)
4
E (Y )  8(0.5)
4

E(X)  E(Y)
For n = 20, p  0.3, m  30, q  0.2:
E ( X )  20(0.3)
6
E (Y )  30(0.2)
6

E(X)  E(Y)
The two possible sets of values:
n  10, p  0.4, m  8, q  0.5;
n  20, p  0.3, m  30, q  0.2.
(or other reasonable answers)
45.
For p  0.1 or 0.9,
Var(Z)  50(0.1)(0.9)
 4.5
For p  0.25 or 0.75,
Var(Z)  50(0.25)(0.75)
 9.375

Two possible pairs of values of p:
p  0.1 or 0.9; p  0.25 or 0.75 and their
variances are 4.5 and 9.375 respectively.
(or other reasonable answers)
77
© Hong Kong Educational Publishing Co.