36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 8
1) Exercise 5.2 (8 points)
a) The possible outcomes of the experiment are cfw_T T T, T T H, T HT, HT T, T HH, HT H, HHT, HH
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 5
1) Exercise 3.38 (8 points)
We are given that there are two glasses with formula A and one with formula B, so the
probabili
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 3
1) Exercise 2.84 (8 points)
The derivation is straightforward, particularly if you split the three events in a helpful
mann
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 2
1) Exercise 2.14 (8 points)
Let N be the event that the adult needs glasses and U be the event that the adult uses
glasses.
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 4
1) DeGroot & Schervish, Exercise 2.4.2 (8 points)
a) We are given that the initial state is c (for cloudy), and we want the
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 6
1) Exercise 3.138 (10 points)
The rst step is to compute E[Y (Y 1)]:
E[Y (Y 1)] =
y(y 1)
y=0
y
e
y!
y(y 1)
= 0+0+
y=2
y
=
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 1
1) Exercise 1.4 (20 points)
a) One can construct the histogram in a manner similar to that shown in the rst set
of lecture
36-225 Introduction to Probability Theory
Fall 2011
Homework 6
Due Wednesday, October 19th at 9:30 AM in DH 2315
Exercise numbers refer to Wackerly, et al., 7th ed.
1) [10 pts] Exercise 3.138
2) [6 pt
36-225 Introduction to Probability Theory
Fall 2011
Homework 7
Due Wednesday, October 26th at 9:30 AM in DH 2315
Exercise numbers refer to Wackerly, et al., 7th ed.
1) [10 pts] Exercise 3.147
2) [10 p
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 11
1) Exercise 7.20 (12 points)
a) If U 2 (), then U Gamma(/2, 2). Thus:
f (u) =
u/21 eu/2
2/2
2
for 0 u < . So:
E[U ] =
duu
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 10
1) Exercise 6.2 (10 points)
We are given that
3 2
2y
f (y) =
0
1 y 1
otherwise
Note that instead of using U1 , U2 , and U3
36-225 INTRODUCTION TO PROBABILITY & STATISTICS I Practice Exam
You must show your work in order to get full or be considered for partial credit.
You may use a one-sided sheet of notes (8.5 by 11 inch
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 7
1) Exercise 3.147 (10 points)
We know that p(y) = (1 p)y1 p for y N. Thus
E[etY ] =
ety (1 p)y1 p
y=1
= pe
t
et(y1) (1 p)y1
36-225 Introduction to Probability Theory
Fall 2011
Solutions for Homework 9
1) Exercise 5.92 (14 points)
We are given that f (y1 , y2 ) = 6(1 y2 ) for 0 y1 y2 1, and we are to nd
Cov(Y1 , Y2 ). First