Section 3.3
3.3
Sample Space
w
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
3
1
1
1
-1
-1
-1
-3
3.5
3
(a) 1   c( x 2  4)  30c
x 0
Thus, c=1/30
2
 2  3 
  10c
(b) 1   c 
x  0  x  3  x 
Thus, c=1/10
3.8
P(w=3)=(2/3)(2/3)(2/3)=8/27
P(w=1)=3(2/3)(2/3)(1/3)=4/9
P(w=-1)=3(2/3)(1/3)(1/3)=2/9
P(w=-3)=(1/3)(1/3)(1/3)=1/27
3.10
f ( x)  1 / 6,
1 x  6
3.14
(a) P ( X 
12
 0.2)  F (0.2)  1  e 1.6
60
(b) f ( x)  F ' ( x)  8e 8 x
0.2
Thus, P ( X  0.2)   8e 8 x dx  e 8 x | 00.2  1  e 1.6
0
3.21
1
(a) 1   k x dx 
0
2k 3 / 2 1 2k
x |0 
3
3
Thus, k=3/2
(b) F ( x)  P( X  x)  
x
0
3
t dt  t 3 / 2 |0x  x 3 / 2
2
P (0.3  X  0.6)  F (0.6)  F (0.3)  (0.6) 3 / 2  (0.3) 3 / 2
3.22
Let X be the number of spades in the three draws.
P(X=0)=(39/52)(38/51)(37/50)
P(X=1)=3(13/52)(39/51)(38/50)
P(X=2)=3(13/52)(12/51)(39/50)
P(X=3)=(13/52)(12/51)(11/50)
OR
13  39 

 
x  3  x 

, 0 x3
f ( x) 
 52 
 
3
Section 3.4
3.37
3
3
(a) 1   cxy  36c
x 1 y 1
Thus, c=1/36
(b) 1   c | x  y | 15c
x
y
Thus, c=1/15
3.38
(a)
(b)
(c)
(d)
P( X
P( X
P( X
P( X
 2, Y  1)  f (0,1)  f (1,1)  f (2,1)  1 / 5
 2, Y  1)  f (3,0)  f (3,1)  7 / 30
 Y )  f (1,0)  f (2,0)  f (2,1)  f (3,0)  f (3,1)  f (3,2)  3 / 5
 Y  4)  f (2,2)  f (3,1)  4 / 15
3.42
f ( x | y) 
f ( x, y )

h( y )
e ( x y )


0
e ( x  y ) dx

e ( x y )
 ex
y
e
1
1
0
0
P (0  X  1 | Y  2)   f ( x | y )dx   e  x dx  1  e 1
3.46
2
x  y x 1

,
10
y  0 30
2
(a) g ( x)   f ( x, y )  
y 0
3
3
x 0
x 0
(b) h( y )   f ( x, y )  
x  y 2y  3

,
30
15
x  0,1,2,3
y  0,1,2
3.49
(a)
x
1
2
3
g(x) 0.1 0.35 0.55
(b)
y
1
3
5
h(y) 0.2 0.5 0.3
(c) P(Y  3 | X  2) 
f (2,3)
0.1

g (2)
0.35
3.51
4
 4  4 

  

x
y
x
y
3


,
(a) f ( x, y )    
12 
 
3
0 x y3
(b) P ( X  Y  2)  1  P ( X  Y  2)  1  [ f (0,0)  f (0,1)  f (1,0)] 
3.54
Since f (1,1)  g (1)h(1), the two random variables are dependent
3.56
(a) f ( x | y ) 
f ( x, y )

h( y )
6x
1 y

0
6 xdx

6x
2x

2
3(1  y )
(1  y ) 2
Since f(x|y) involves the variable y, X and Y are not independent
OR
Prove f ( x, y )  g ( x)h( y )
1 y
(b) P ( X  0.3 | Y  0.5)  
0.3
f ( x | y )dx  
0.5
0.3
2x
dx  0.64
(1  0.5) 2
42
55
3.57
(a) 1  
2 1 1
0
  kxy
0 0
2
zdxdydz 
k
3
Thus, k=3
(b) P ( X 
2 1
1/ 4
1
1
21
, Y  ,1  Z  2)     3xy 2 zdxdydz 
1 1/ 2 0
4
2
512
3.60
4 xyz 2
2 yz 2
dx 
, 0  y  1, 0  z  3
(a) g ( y, z )  
0
9
9
1
(b) h( y )  
3
0
2 yz 2
dz  2 y, 0  y  1
9
2
2 1
1 / 2 4 xyz
1
1
1
7
(c) P (  X  , Y  ,1  Z  2)    
dxdydz 
1
1
/
3
1
/
4
4
2
3
9
162
f ( x, y , z )
 2 x, 0  x  1
(d) f ( x | y, z ) 
g ( y, z )
Thus, P (0  X 
1/ 2
1
1
1
| Y  , Z  2)   2 xdx 
0
2
4
4
Review Exercises
3.63

(a) g ( x)   ye  y (1 x ) dy 
0
 ye  y (1 x )
1 x

0

1   y (1 x )
1
e
dy 
, x0

1 x 0
(1  x) 2
Hint: Integration by parts


0
0
Let u  y, dv  e  y (1 x ) dy , then g ( x)   udv uv   vdu

h( y )   ye  y (1 x ) dx  e  y 1 x  |0  e  y , y  0
0
(b) P ( X  2, Y  2)  

2


2
ye
 y (1 x )
dxdy  e
 y 1 x 
e 6

3
3.67
5
f ( x)   (0.1) x (0.9) 5 x , 0  x  5
 x
3.68
(a) g ( x)  
2
1
3x  y
4 2y
dx  
, 1 y  2
1
9
3 9
No. Since f ( x, y )  g ( x)h( y )
h( y )  
(b)
3x  y
x 1
dy   , 1  x  3
9
3 6
3
3 x
1
2
(c) P ( x  2)   (  )dx 
2 3
6
3
3.69
(a) f ( x)  F ' ( x) 
1  x / 50
e
, x0
50
(b) P( x  70)  1  P( X  70)  1  F (70)  1  (1  e 70 / 50 )