Chapter 1
1.1
(a)
1
2
1
n
i
i
n

(b)
0 1 2 3
0 1 2 3
0 1 2
0 1
 13
1.2 | 1 1
2 2 1
n i n n n
  
1
i
  | 1
i

n  | n2
1.3 (a)
n300  4
n301  4
n302  3
n303  2
n310  4
n311  3
n312  2
n313  1 | n320  3
n321  2
n322  1
n330  2
n331  1
(b)  |  4  4  3...  2 1  321.4
1 2
2 1 2 3
1 1
n n
i j
n n n n
 
 
2 Mathematical Statistics, 8E
1.5 (b) 6, 20, and 70
“2 out of 3” m = 2
1 2
2 2(1 2) 6
1 1
   
             
“3 out of 5” m = 3
2 3 4
2 2(1 3 6) 20
2 2 2
     
                   
“4 out of 7” m = 4
3 4 5 6
2 2(1 4 10 20) 70
3 3 3 3
       
                         
1.6 (a)
10
10 10
10! 0 (7.92665)(3.678797) (7.92665)(454,002.49) 3,598,719
e
  π    
   
% error 3.6288 3.5987 100 0.83%
3.6288

  
12
12 12
(8.683215)(4.41455) 475,683,224 | 12! 24
e
   
  | 
4.7800 4.7568
% error  | 100 0.69%
  |  π   
4.7900
(b)
52
13 39
    | 13 13! 39!
     
13  | 3952 52 52
13 39 39 39
52
104
52 52!
26 78
13 4 4
639 billion
19.5 13 13 3 19.5 3
e
e e
π
π π
π π
 
   
       

  
 
1.7 Using Stirling’s formula in
2 2 !
! !
n n
n n n
 
     yields
2
2
n
n
  | 2
n
 
2
2
n
  | 2 2
2
n
n
 
   
4
n n
n e n
π π
π
       1
2
n
e
π
π
   
 
nr and 123 = 1,728 |    1.8 1.9
1 5 3 1 7
and 21
5 5
r n
     |      
  |   |   |     r
      1.10 Substitute r  n for r into result of 1.9
1 1 5 1 4
and 6
5 3 2
r n n r
r n r n
            
          
          
Chapter 1 3
1.11 (b) Seventh row is 1, 6, 15, 20, 15, 6, 1
Eighth row is 1, 7, 21, 35, 35, 21, 7, 1
(x  y)6  x6  6x5y 15x4 y2  20x2 y3 15x2 y4  6xy5  y6
(x  y)7  x7  7x6 y  21x5 y2  35x4 y3  35x3y 4  21x2 y5  7xy6  y7
1.14 (a) Set x = 1 and y = 1
(b) Set x = 1 and y = 1
(c) Set x = 1 and y = a  1
1.19 (a)
1 1 3 5
2 2 2 2
         | 15
   | and 24 384
     
   ( 3)( 4)( 5) 10
6
  
 
(b)
1 2
1
5 2 1
4
 
     
2 3
1 1 1 1 1 1 1 3 1
2 1
2 4 2 2 4 2 2 2 4
1 1 3 512 64 8 3
2 1 2
8 64 512 512
571
2 2.23
512
             
                            
    
         
  

1142
2.230
512

1.20 (a) ( 1)( 2)...( ) ( 1)
!
r r
r
  
 
(b) ( )( 1)...( 1) ( 1) ( 1)...( 1)
!
r  | !
r | r
 
( 1)...( 1)
r
n r n n
    | 1
r
n r
   
( 1) ( 1)
      | r
 
!
r  |  r
n n n n r n n n r
          
    
1.21 8! 8 7 6 5 4 560
2! 3! 3! 2 6
   
 

1.22 9! 23 32 ( 4)3 9 8 7 6 5 4 8 9 64 23,224,320
3! 2! 3! 12
    
         
4 Mathematical Statistics, 8E
1.24 Note: If there are 0 turn-ons the first night, 6 turn-ons in four nights can only occur if there are 2
turn-ons on each of the subsequent three nights. Thus, we need to show only that part of the tree
following this event.
Chapter 1 5
1.25
1.26 (a) (b)
1.27 (a) 5
(b) 4
1.28
1.29
1.30 (a) 65  30; (b) 66  36;
1.31 (a) 6; (b) 65  30; (c) 5 4  20 first one fixed; (d) 6  30  20  56
6 Mathematical Statistics, 8E
1.32 (a) 45 2  40; (b) 5 63  90
1.33 (a) 5 4  20; (b) 5 4 3  60
1.34 315  14,348,907
1.35 15 14 105
2 1



1.36 (a) 1098 7  5040; (b) 5040 210
24

1.37 (a) 14 13 91;
2 1



(b) 14 13 12 364
3 2 1
 

 
1.38 6!  720
1.39 6! 720 90
2! 2! 2! 8
 
1.40 5!  120 and 120  2 4!  72
1.41 7!  5040
1.42 (a) 5!  120; (b) 5! 60
2!

1.43 10! 3628800 50,400
3! 3! 2! 72
  and 8! 40320 3360
3! 2! 12
 
1.44 10! 3628800 1,260
5! 4! 120 24
 

1.45 8! 40320 280
3! 4! 6 24
 

1.46 (a)
20
77,520
7
 
     ; (b)
20
184,755
10
 
    
(c)
20 20 20 20
1140 190 20 1 1351
17 18 19 20
       
                       
1.47 (a)
7
21
2
 
     ; (b)
4
6
2
 
     ; (c) 3 4  12
Chapter 1 7
1.48
3 7 3 7
3 21 1 7 63 7 70
2 2 3 1
       
                       
1.49
4 7 3
6 35 3 630
2 3 1
     
               
1.50
13 13 13 13
1287 286 286 78 8,211,173,256
5 3 3 2
       
                    
1.51 7! 5040 420
3! 2! 12
 
1.52 310  59,049
1.53 55  15,625
1.54
12 6 1 17 17
6,188
12 12 5
       
              
1.55
12 1 11
462
6 6
    
         
1.56
14 3 1 16
120
14 14
     
         
1.57
2 1 1
1 1
r n n r n
n n
        
    
     
1 10
45
1 2
r n
 n
    
     
    
8 ScholarStock
Chapter 2
2.1 (a) P[A]  P[( A B)  ( A B)]  P( A B)  P( A B)  P( A B)
(b) A B  ( A B)  ( A B)  ( A  B)  A (A  B)
2.6 P(A)  P( A B)  (a  b)  a  b  P( A B)
P(A B)  P( A)  P(A  B)  P( A)
2.7 1 ( ) ( ) ( ) ( ) ( ) ( )
( )
P A P B P A B a b c d a b a c a d
P A B
             
   
2.8 [( ) ( )] ( ) ) 2
( ) ( ) 2 ( ) Refer to Figure 2.6
P A B A B b c a b a c a
P A P B P A B
           
   
2.9 (a) P(A)  P(B)  P(A B)  0  P(A B)  P(A)  P(B)
(b) P(A)  P(B)  P(A B)  1 ( P A B)  P(A)  P(B) 1
2.10 Refer to Figure 2.7 P(A)  1  e  c  f  0
P(B)  1  d  f  g  0
P(C)  1  b  e  g  0
Therefore P(A)  a  b  d  g  a  1 QED
2.11 ( ) ( ) [ )
( ) ( ) ( ) ( )
( ) ( ) ( ) QED
P A B P A P A B
P A P A B P A B P A B
P A P B P A B
    
       
   
2.12 ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ( ) ( ) ( )
( ) ( )
( )
P A P B P C P A B P A C P B C P A B C
a b d g a b c e a c d f a b
a d a c a a b c d e f
P A B C D
          
             
          
   
Chapter 2 9
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2.13 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
P A P B P C P D P A B P A C P A D P B C
P B D P C D P A B C P A B D
P A C D P B C D P A B C D
          
         
         
( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
a b d g i j l o a b c e i j k m
a c d f i k l n a b c d e f g h
a b i j a d i l a b d g
a c i k a b c e a c d f
a i a b a d a c a
a b c d e f g h i j k l m n o
P
               
               
           
           
        
              
 ( A B  C  D)
2.14 For n  2, P(E1  E2)  P(E1)  P(E2)  P(E1  E2)  P(E1)  P(E2)
Assume that for some n :
1 2
1
( ) ( )
n
n j
j
P E E E P E

     , then
 1 2 1  1 2 1
1
1 2 1 ( )
1
( ) ( )
( ) ( )
j
n n n n
n
n n P E
j
P E E E E P E E E E
P E E E P E
 



        
      
 

where the first inequality follows from the first step of the induction, and the second inequality
comes from the second step of the induction.
2.15
1
p A
p B


, pb  A Ap , PA pB  A, p( A B)  A , p A
A B


2.16 (a) P:ostulate 1 P(A) a 0
a b
 

(B) Postulate 2 P(A) a
a b


, P(A ) b
a b
 

P(A) P(A ) a b 1 P(S)
a b a b
     
 
2.17 (a) ( ) ( ) 0;
( )
P A B
P A B
P B

  (b) ( ) ( ) ( ) 1
( ) ( )
P B B P B
P B B
P B P B

  
(c) 1 2 1 2
1 2
1 2
[ ) ]
( )
( )
( ) ( )
( ) ( )
( ) ( )
P A A B
P A A B
P B
P A B P A B
P B P B
P A B P A B
  
  
 
  
  




10 Mathematical Statistics, 8E
ScholarStock
2.18 For example
(a)  | If P(A B)  P( A B)  P( A  B)
1
( )
 P A  B   | so that4
1
( ) ,
2
P B A  ( ) 1 ,
2
P B A  and
P(B A)  P(B A)  1
(b) If ( ) ( ) ( ) 1
5
P A B  P A B  P A  B 
and ( ) 2
5
P A  B 
1
( ) ,
2
P B A  ( ) 1 ,
3
P B A  and
5
( ) ( )
6
P B A  P B A 
2.19 ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
P A B C D P A B C P D A B C
P A B P C A B P D A B C
P A P B A P C A B P D A B C
       
    
   
2.20 ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
P A B C P B C P A B C P A B
P C A B P C B
P A B P B P B C P B
P A B C P A B
     
     
 
  
2.21 ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
P A B P A B
P B A P B P A P A B P A
P A P B
 
     
2.22 (a) P(B)  P(A B)  P(A  B)  P(A)P(B)  P( A  B)
P(A  B)  P(B)  P( A)P(B)  P(B)[(1 P( A)]  P(B)P( A) QED
(b) P(B)  P(A B)  P( A  B)  P(A  B)  P(B)P( A)
P(A  B)  P(B)  P(B)P( A B)  P(B)[(1 P(A)]  P(B)P(A) QED
2.23 Assume that A and B are independent and show that this leads to contradiction.
P(A)  P(A B)  P(A B)  P(A B)  P(A)P(B)
P(A B)  P( A)  P( A)P(B)  P( A)[1 P(B)]  P( A)P(B) and A and B are independent
2.24 P(A)  0.60, ( P B)  0.80, ( P C)  0.50, ( P A B)  0.48, ( P AC)  0.30
P(B C)  0.38, ( P A B C)  0.24
P(A B C)  0.24, ( P A)(B)(C)  (0.6)(0.8)(0.5)  0.24
P(B C)  0.38, ( P B)P(C)  (0.8)(0.5)  0.40 B and C not independent
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2.25 Refer to 2.21
( P A B)  0.48, ( P A)P(B)  (0.6)(0.8)  0.48 A and B independent
( P AC)  0.30, ( P A)P(C)  (0.6)(0.5)  0.30 A and C independent
( P B C)  0.38, ( P B)P(C)  (0.8)(0.5)  0.40 B and C not independent
2.26 (Refer to 2.24 and 2.25) Already showed that A and B independent,.
A and C independent
P[( A(B C)]  0.54, ( P A)  0.60, ( P B C)  0.92, (0.6)(0.92)  0.552  0.54
2.27 (a)
(b)  | P[( A(B C)]  P( A B C)  P( A)P(B)P(C)  P(A)P(B C) QED
[( ( )] [( ) ( )]
P A B C P A B A C
     ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( )[ ( ) ( ) ( )]
( ) ( ) QED
P A B P A C P A B C
P A P B P A P C P A P B P C
P A P B P C P B C
P A P B C
      
  
   
 
2.28 ( )
P A B
  | ( )
P A
 | P B P B A P B
  
( )  | (  | )  | ( )( ) P A B
2.29 Proof by induction: If n = 2, then P(A1  A2)  P( A1)  P( A2)  P( A1) P( A2)
and
1[1 P( A1)][1 P( A2)]  11 P(A1)  P( A2)  P( A1)P( A2).
Assuming P(A1  A2   An )  1[1 P( A1)][1 P( A2)] [1 P( An )].
we can write
1 2 1 1 2 1
1 2 1
1 2 1 1
( ) ( ) ( )
( ) ( )
( ) [1 ( )] ( )
n n n n
n n
n n n
P A A A A P A A A P A
P A A A P A
P A A A P A P A
 

 
        
   
      
 


1 2 1 1
1 2 1
{1 [ ( )] [1 ( )] [1 ( )]} [1 ( )] ( )
1 [1 ( )] [1 ( )] [1 ( )] [1 ( )]
n n n
n n
P A P A P A P A P A
P A P A P A P A
 

         
         


2.30 Two at time
k 2
 
   
Three at time
k 3
 
   
k at time
k k
 
   
2 2 1
2 3 0 1
k k k k k k k
k
k
         
                            
2.31 P(A)  P( A) P( A)  P(A) P(), since ( P  A)  P()  0.
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2.32 Since
B1  B2   Bk  S, A (B1  B2   Bk )  A. Thus, by the distributive property,
( A B1) ( A B2)   ( A Bk )  A , and
P(A)  P(B1)P(A B1)  P(B2)P(A B2)   P(Bk )P( A Bk ) QED
2.33 The probability of no matches on any given trial is n 1
 n
; since the trials are independent, the
probability of no match in n trials is n 1 n 1 1 n
n n
    
          .
2.34 ( ) ( ) ( ) ( ) [1 ( )] [1 ( )] ( )
1 ( ) ( ) [1 ( )].
P A B P A P B P A B P A P B P A B
P A P B P A B
            
       
Since 1 P( A B)  0, ( P A B)  1 P( A)  P(B) QED
2.35 (a) {6, 8, 9}; (b) {8}; (c) {1, 2, 3, 4, 5, 8}; (d) {1, 5};
(e)  | (2, 4, 8};  | (f) 
2.36 (a)
(b)
(c)
(f)
(h)  | Los Angeles, Long Beach, Pasadena, Anaheim, Santa Maria, Westwood;
San Diego, Long Beach, Pasadena, Anaheim, Santa Maria, Westwood;
Santa Barbara; (d) ; (e) San Diego, Long Beach, Santa Barbara, Anaheim;
San Diego, Santa Barbara, Long Beach; (g) Los Angeles, Santa Barbara, Anaheim;
Los Angeles, Pasadena, Santa Maria, Westwood; (i) Los Angeles, Pasadena,
Santa Maria, Westwood.2.37 (a) {5, 6, 7, 8}; (b) {2, 4, 5, 7}; (c) {1, 8} (d) (3, 4, 7, 8}
2.38 (a) He chooses a car with air conditioning.
(b)
(c)
(d)
2.39 (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)  | He chooses a car with bucket seats or no power steering.
He chooses a car with bucket seats that is 2 or 3 years old.
He chooses a car with bucket seats that is 2 or 3 years old.
House has fewer than three baths;
does not have fire place;
does not cost more than $200,000
is not new;
has three or more baths and fire place;
has three more baths and costs more than $200,000
costs more than $200,000 but has no fire place;
is new or costs more than $200,000
is new or costs $200,000 or less
has 3 or more baths and/or fire place;
has 3 or more baths and/or costs more than $200,000;
is new and costs more than $200,000
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2.41 (a) (H,1), (H,2), (H,3), (H,4), (H,5), (H,6)
(T,H,H), (T,H,T), (T,T,H), (T,T,T)
(b)
(c)
2.42 (a)  | (H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,H,T), (T,T,H)
(H,5), (H,6), (T,H,T), (T,T,H), (T,T,T)
S = {(0,0,0)(1,1,1)}
A = {(1,0,1),(0,1,1),(1,1,1)}
B = {(0,1,1)}
C = {(1,0,1)}
A & B not mutually exclusive, A & C not mutually exclusive, B & C are mutually exclusive.
(b) 2.43
3, 3, 3, 3, x1 x1x2 x1x2x3  where x1  1,2,4,5,6, for all i
(a) 5k1; (b) 1 5 5 5 1
4
k
k 
  
2.44 S  {(x, y) (x  2)2  ( y  3)2  9}
2.45 (a) (x 3  x  10) ; (b) (x 5  x  8) ; (c) (x 3  x  5);
(d) (x 0  x  3 or 5  x  10)
2.46 1 A driver has liability insurance and collision insurance.
2 A driver has liability insurance but not collision insurance.
3 A driver has collision insurance but not liability insurance.
4 A driver has neither liability insurance nor collision insurance.
2.47 (a)
(b)
(c)
(d)  | A driver has liability insurance.
A driver does not have collision insurance.
A driver has either liability or collision insurance, but not both.
A driver does not have both kinds of insurance.2.48 (a) A car brought to the garage needs engine overhaul, transmission
repairs, and new tires.
(b) A car brought to the garage needs transmission repairs, new
tires, but no engine overhaul.
(c) A car brought to the garage needs engine overhaul, but neither
transmission repairs nor new tires.
(d) A car brought to the garage needs engine overhaul and new
tires.
(e) A car brought to the garage needs transmission repairs, but no
new tires.
(f) A car brought to the garage does not need engine overhaul.
2.49 (a) 5; (b) 1 and 2 together (c) 3, 5, and 6 together; (d) 1, 3, 4, and 6 together
14 Mathematical Statistics, 8E
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2.50 500  (308 266) 103  29  59 results are inconsistent
2.51 200  (1.38 115)  91  38
2.52 (a) 12; (b) 6; (c) 20
2.53 (a) permissible;
(b)
(c)
(d)
(e)  | not permissible because the sum of the probabilities exceeds 1;
permissible;
not permissible because P(E) is negative
not permissible because the sum of the probabilities is less than 1.
2.54 (a)
(d)
(f)  | 1 0.37  0.63;  | (b) 1 0.44  0.56 ;  | (c)  | 0.37  0.44  0.81;
0;  | (e) 0.37, P(A B)  P(A) for mutually exclusive events;
1 0.81  0.192.55 (a) Probability cannot be negative.
(b)
(c)
(d)  | 0.77  0.08  0.85  0.95
0.12  0.25 0.36  0.14  0.09  0.07  1.03  1
0.08 0.21 0.29  0.40  0.98  1
2.56 (a)
(c)  | 0.12  0.17  0.29;  | (b)  | 0.17  0.34  0.29  0.80
(d) 0.34  0.29  0.08  0.71
0.34  0.17  0.12  0.63;
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2.57 (0,0), (1,0), (2,0), (3,0), (4,0), (5,0), (0,1), (1,1), (2,1), (3,1),
(4,1), (5,1), (0,2), (1,2), (2,2), (3,2), (4,2), (5,2), (0,3), (1,3),
(2,3), (3,3), (4,3), (5,3), (0,4), (1,4), (2,4), (3,4), (4,4), (5,4)
(a) 10 1 ;
30 3
 (b) 5 1 ;
30 6
 (c) 15 1 ;
30 2
 (d) 10 1
30 3

2.58 (a) 20 10 3; 4 5 1
80 8 80 4
 
  ; (c) 2 4 1 ;
80 10

 (d) 4 2 1 1 1 ;
80 10
  

(e) 8 14 22 11
80 80 40

 
2.59 (a) 0.24  0.22  0.46; (b) 0.15 0.03 0.22  0.40
(c) 0.03 0.08  0.11; (d) 0.15 0.03 0.28 0.22  0.68
2.60
16
2 120 20
52 1326 221
2
 
   
 
 
   
2.61 Let P(A)  4 p, ( P B)  2 p, ( P C)  2 p, and ( P D)  p .
Then 9 p  1 and 1
9
p  ;
(a) 2 ;
9
(b) 1 4 5
9 9
 
2.62 (a)
13 4 4
44
2 2 2 13 12 6 6 44 120
                 | 198
52
5
 
   
2 52 51 50 49 48 4165
  
    0.0475
     
(b) 13 48 13 48 120 1
52 51 51 50 49 48 4165
 5
 
 
     
   
16 Mathematical Statistics, 8E
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2.63 (a) 5 | 6  | 6 6 6 6 6 108
 
   6 5 3
4
2 2 2 15 10 3 4 25
     
               
(b) 5
5
6 5 4
3 6 10 5 4 25 4 25
6 6 6 6 6 6 648 162
 
         
  
   
(c) 5
6 5
3 2
          | 6 5 10
 
  | 25
6  | 6 6 6 6 6 648
   5 2
   
(d) 5 | 6  | 6 6 6 6 6 1296
 
   5
6 5
4 6 5 5 25
 
      
2.65 78 [64 36 34] 12 2
78 78 13
  
 
2.64 (a) P(A B) is less than P(A) .
(b) P(A B) exceeds P(A) .
(c) P(A B)  0.72  0.84  0.52  1.04 exceeds 1
2.66 2 ,0
3
2.67 The area of the triangle is 4 3 6;
 2
 If the point is a distance x from the vertex on the longer leg,
then is will be 3
x 4
units from the vertex on the other leg. The area of the required triangle is
3 3 2
4 2 8
x x
x  

. For this to be greater than 3, or half the area of the triangle, x2  8, or 2 x  2 .
Thus, the probability of the line segment dividing the area in at least one-half is
4 2 2 2
1
4 2

 
2.68 0.21 0.28  0.15  0.34
2.69 (a) 0.59  0.30  0.21  0.68 ; (b) 0.59  0.21  0.38
(c) 1 0.21  0.79 ; (d) 1 0.68  0.32
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2.70 1
3
b  d 
5 9
c  d 
3
;
4
a  b  c  hence 1
4
d 
1 1 1 5 1 11 1 11 1 13
, , 1
3 4 12 9 4 36 12 36 4 36
b    c    a     
a  P(out of state living on campus)
b  P(out of state not living on campus)
c  P(from Virginia living on campus)
d  P(from Virginia not living on campus)
2.71 (a) (0.08)  0.05 0.02  0.11; (b) 1