Chapter 2
2.1 (a)
[ ] [( ) ( )] ( ) ( ) ( )PA PAB AB PAB PAB PAB
(b)
()()( ) ( )AB AB AB A B A A B
2.9 (a)
() () ( ) 0 ( ) () ()PA PB PA B PA B PA PB
(b)
( ) ( ) ( ) 1 ( ) ( ) ( ) 1PA PB PA B PA B PA PB
2.11
()()[ )
() ( ) ( ) ( )
( ) ( ) ( ) QED
PA B PA PA B
PA PA B PA B PA B
PA PB PA B
Chapter 2 9
2.13
()()()()()()()()
()()( )( )
()()( )
PA PB PC PD PA B PA C PA D PB C
PBD PCD PABC PABD
PACD PBCD PABCD
2.14 For
2,n
12 1212 12
()()()()()()PE E PE PE PE E PE PE
Assume that for some n :
12
()()
n
nj
PE E E PE
, then
2.15
1
pA
pB
,
pb A Ap
,
PA pB A
,
()pA B A
,
A
pAB
2.16 (a) P:ostulate 1
() 0
a
PA ab
2.17 (a)
()
() 0;
()
PA B
PAB PB
(b)
()()
() 1
() ()
PB B PB
PBB PB PB
10 Mathematical Statistics, 8E
2.18 For example
(a) If
()()()PA B PA B PA B
1
()
4
PA B
so that
1
() ,
2
PBA 1
(),
2
PBA and
6
2.19 ()()()
()( )( )
()( )( )( )
PABC D PABCPDABC
PABPCABPDABC
PAPBAPCA BPDA B C
2.22 (a)
() ( ) ( ) ()() ( )PB PA B PA B PAPB PA B
( ) ( ) ( ) ( ) ( )[(1 ( )] ( ) ( ) QEDPABPBPAPBPB PA PBPA
2.23 Assume that A and
B
are independent and show that this leads to contradiction.
2.24
( ) 0.60, ( ) 0.80, ( ) 0.50, ( ) 0.48, ( ) 0.30
PA PB PC PA B PA C
( ) 0.38, ( ) 0.24
PBC PABC
Chapter 2 11
2.25 Refer to 2.21
2.26 (Refer to 2.24 and 2.25) Already showed that A and B independent,.
2.27 (a)
[(( )]( )()()()()( ) QED
P A B C PA B C PAPBPC PAPB C
2.28
()
( ) () ( ) ()
()
PA B
PAB PB PBA PB
PA
2.29 Proof by induction: If n = 2, then
12 1 2 1 2
( ) () () () ()
PA A PA PA PA PA
and
1 [1 ( )] [1 ( )] 1 1 ( ) ( ) ( ) ( ).
PA PA PA PA PA PA
2.30 Two at time
2
k
2.31
( ) () ( ) () (), since ( ) () 0.PA PA P A PA P P A P
12 Mathematical Statistics, 8E
2.32 Since
12 12
, ( ) .
kk
BB BSABB B A
Thus, by the distributive property,
2.33 The probability of no matches on any given trial is
1n
n
; since the trials are independent, the
2.34
()()()()[1()][1()]()
1()()[1( )].
PA B PA PB PA B PA PB PA B
PA PB PA B
Since
1( )0, ( )1()() QEDPA B PA B PA PB
2.36 (a) Los Angeles, Long Beach, Pasadena, Anaheim, Santa Maria, Westwood;
(b) San Diego, Long Beach, Pasadena, Anaheim, 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.
2.39 (a) House has fewer than three baths;
(b) does not have fire place;
(c) does not cost more than $200,000
(d) is not new;
Chapter 2 13
2.41 (a) (H,1), (H,2), (H,3), (H,4), (H,5), (H,6)
2.42 (a) S =
{(0,0,0) (1,1,1)}
2.43
112123
3, 3, 3, 3,xxx xxx
where
1
1, 2, 4, 5, 6,x
for all i
2.45 (a) ( 3 10)xx ; (b) (5 8)xx ; (c) (3 5);xx
(d)
( 0 3 or 5 10)xx x
2.47 (a) A driver has liability insurance.
(b) A driver does not have collision insurance.
(c) A driver has either liability or collision insurance, but not both.
(d) 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
14 Mathematical Statistics, 8E
2.50 500 (308 266) 103 29 59 results are inconsistent
2.52 (a) 12; (b) 6; (c) 20
2.53 (a) permissible;
(b) not permissible because the sum of the probabilities exceeds 1;
2.54 (a)
10.37 0.63
; (b)
1 0.44 0.56
; (c)
0.37 0.44 0.81
;
(d) 0; (e) 0.37,
()()
PA B PA for mutually exclusive events;
(f)
10.81 0.19
Chapter 2 15
2.57
(0,0), (1,0), (2,0), (3,0), (4,0), (5,0), (0,1), (1,1), (2,1), (3,1),
2.58 (a)
20 10 3 4 5 1
;
80 8 80 4
; (c)
24 1
;
80 10
(d)
4211 1
;
80 10
(e)
814 22 11
80 80 40
2.61 Let
() 4, () 2, () 2, and ( )
PA p PB pPC p PD p .
Then
91
p and
1
9
p;
(a)
2;
9
(b)
45
199
16 Mathematical Statistics, 8E
2.63 (a)
5
653
4
222 15 10 3 4 25
66666 108
6
2.65
78 [64 36 34] 12 2
78 78 13
2.66
2,0
3
2.67 The area of the triangle is
43 6;
2
If the point is a distance x from the vertex on the longer leg,
2.69 (a)
0.59 0.30 0.21 0.68
; (b)
0.59 0.21 0.38
(c)
10.21 0.79
; (d)
1 0.68 0.32
Chapter 2 17
2.70
1
3
bd
5
cd
2.71 (a)
(0.08) 0.05 0.02 0.11;
(b)
1 0.02 0.98
(c)
0.08 0.05 2(0.02) 0.09
2.74 (a) The probability is
34 34
34 21 55
that one of the eggs will be cracked.
2.75 (a) The odds are
64
to
10 10
or 3 to 2;
2.76 (a)
18 36 54 3 ;
90 90 5
(b)
36 27 63 7 ;
90 90 10
(c)
18 2 1 ;
90 10 5
18 Mathematical Statistics, 8E
2.77 (a)
11/5
33/5
(b)
33/10
77/10
2.78
34 34 17
34 2 36 18
2.81
25 40
()
100
2
20
99
PR W
2.84
135
;
488
odds are 5 to 3 that either car will win.
2.85 Using MINITAB software, first we generate 1,000 uniformly distributed pseudo-random
numbers, putting them in Column 1 (C1) as follows:
MTB> Random 1000 C1;
Chapter 2 19
2.86 (a) Repeating the work of Exercise 2.59, we found the corresponding probability for the
second set to be 99/1,000 = 0.099. Obtaining
()PA B
is facilitated by using the LET
2.87
0.20 0.20 1
0.20 0.30 0.10 0.60 3
2.90 (0.55)(0.80) = 0.44
2.91 (a)
(0.8)(0.2)(0.6) 0.096
; (b)
(0.20)(0.40)(0.60) 0.048
;
(c)
(0.8)(0.8)(0.2)(0.4) 0.0512
; (d)
(0.8)(0.8) (0.2)(0.6) 0.76
2.94 A even first, B even second, C same number both
1
() ,
2
PA
1
() ,
2
PB
1
() 6
PC
,
1
()
4
PA B
,
31
()
36 12
PA C
20 Mathematical Statistics, 8E
2.95 (a) The required probability is approximately
4
(0.99) 0.9606
(assuming independence).
The exact probability is
990 989 988 987 0.9605
1,000 999 998 997
2.96 (a)
3
(0.52) 0.1406
; (b)
2
(0.48) (0.52) 0.1198
2.97
543 1
10 9 8 12
2.100 (a)
(0.9)(0.9)(0.9) 0.729
(b)
(0.6)(0.6)(0.4) 0.144
2.101
1111 1 1
( ) , ( ) , ( ) , ( ) , ( ) , ( ) ,
2233 4 6
PA PB PC PD PA B PA C
2.102
(0.7)(0.84) (0.3)(0.49) 0.735
2.104
(0.5)(0.68) (0.5)(0.84) 0.76
2.107
(0.6)(0.35) 0.21 0.21 0.3818
(0.6)(0.35) (0.4)(0.85) 0.21 0.34 0.55
2.108 (a)
(0.08)(0.95) (0.92)(0.02)
0.076 0.0814 0.0944
22 Mathematical Statistics, 8E
2.112
( ) 0.6 ( ) 0.4[1 ( )]PY PM PM
(a)
( ) 0.4 0.2 ( )PY PM
(b)
5() 2 ( )PY PM
106
()5 20.12
250
PM
2.113
33
(0.95) (0.99) 0.832
2.117
1 (1 0.8)(1 0.7)(1 0.65) 0.979
2.118
1 (1 0.85)(1 0.80)(1 0.65)(1 0.60)(1 0.70) 0.999