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Probability Theory Homework Solutions: Exercises 1.32, 1.38, 1.39, 1.52, 2.1, and 2.4, Assignments of Probability and Statistics

Solutions to the given probability theory exercises from a statistics textbook. The exercises cover topics such as mutually exclusive and independent events, conditional probability, and integrals of probability density functions.

Typology: Assignments

Pre 2010

Uploaded on 09/17/2009

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Stat8001 hw2, due October 1st: Solve exercise problems: 1.32, 1.38, 1.39, 1.52, 2.1, and 2.4
1.32, p.41
This is most easily seen by doing each possibility. Let P(i) = probability that the candidate hired on the ith trial is
the best. Then
1 1 1
(1) , (2) ,..., ( ) ,..., ( ) 1
1 1
P P P i P N
N N N i
1.38, p.42
a)
Remember that, by Theorem,
( ) ( ) ( )
C
P A P A B P A B
( )
C C
A B B
( ) 1P B
( ) 1 ( ) 0
C
P B P B
Since
( )
C C
A B B
,
( )
C
A B
( ) 0
C
P A B
( ) ( ) ( )
C
P A P A B P A B
( ) 0P A B
( ) ( )P A P A B
( ) ( )P A B P A
b)
A B
A B A
pf3
pf4
pf5

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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.

  1. If A and B are mutually exclusive:

A  B  

P A (  B )  0

  1. 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 yy )

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 yPXyF 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

 