Download Probability Theory Homework Solutions: Exercises 1.32, 1.38, 1.39, 1.52, 2.1, and 2.4 and more Assignments Probability and Statistics in PDF only on Docsity!
Stat8001 hw2, due October 1
st
: Solve exercise problems: 1.32, 1.38, 1.39, 1.52, 2.1, and 2.
1.32, p.
This is most easily seen by doing each possibility. Let P(i) = probability that the candidate hired on the i
th
trial is
the best. Then
P P P i P N
N N N i
1.38, p.
a)
Remember that, by Theorem, ( ) ( ) ( )
C
P A P A B P A B
C C
A B B
P B ( ) 1
C
P B P B
Since ( )
C C
A B B
C
A B
C
P A B
C
P A P A B P A B
P A ( B ) 0
P A ( ) P A ( B )
P A B P A
P A B
P B
P A B ( ) P A ( )
b)
A B
A B A
P A B P A
P B A
P A P A
P B A ( ) 1
P A B P A
P A B
P B P B
c)
If A and B are mutually exclusive:
P A ( B ) P A ( ) P B ( )
A ( A B ) A
P A A B P A
P A A B
P A B P A P B
d)
P A ( B C ) P A ( ( B C ))
P A B ( C P B ) ( C )
P A ( B C ) P A B ( C P B C P C ) ( ) ( )
1.39, p.
a)
There are 2 steps to this—1) A and B are mutually exclusive and 2) A and B are THEN ALSO independent.
- If A and B are mutually exclusive:
A B
P A ( B ) 0
- Since
P A ( B ) P A ( )* P B ( )
, if A and B are THEN ALSO independent:
P A ( B ) P A ( )* P B ( )
For this to occur, either
P A ( ) 0
and/or
P B ( ) 0
1/ 3 1/ 3
x
d
f y y
dy
5/ 3 1/ 3 2 / 3
y y y
1/ 3
14 (1 y y )
4 / 3
14 y 14 y , 0 y 1
To check the integral,
1
4 /
0
(14 y 14 y ) dy
7 / 3
2
y
y
1 2 7 / 3
0
7 y 6 y
1
0
b)
7
x
x
f x e x y x
, monotone, and =
, using theorem.
Y x
y
f y f
d y
dy
(7 / 4)( 3)
y
e
(7 / 4)( 3)
y
e
,3 y
To check the integral,
(7 / 4)( 3)
3
y
e dy
(7 / 4)( 3)
3
y
e
c)
Y X
F y P X y F y
Y
f y fx y
y
2 2
Y
f y y y
y
2
15( y )(1 y ) , 0 y 1
To check the integral,
1 1
2 2
0
15 y (1 y ) dy
1 1 1
2 2
0
(15 y 30 y 15 y ) dy
2.4, p.
a)
f(x) is a pdf and it is positive and
f ( ) x dx
0
0
x x
e dx e dx
and for t 0 , P (|X|<t) = P (-t<X<t)
0
0
t
x x
t
e dx e dx
[1 ] [ 1]
t t
e e
t
e
