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
=
(y 2)!
y=2
=
2
y=2
= 2
z=0
=
y
e
y!
e
y2
e
(y 2)!
z
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 transition
probability pcc . This can be read directly
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. Then we can rewrite the interior of the 2 2 table as
P
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
manner:
P (A2 A3 A1 ) = P (A2 ) + P (A3 A1 ) P (A2 (A3 A1 )
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
probability of picking B randomly is p = 1/3.
a) The probability
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, HHH,
i.e., the number of heads are cfw_ 0, 1, 1, 1, 2, 2,
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
0
du
u/2 eu/2
2/2
2
du
=
2u/2 eu/2 + 1
2
2/2+1 + 1
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 below, well just go with the simpler U . It
doesnt cha
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, we do the obligatory sketch:
(1,1)
We have that Cov(Y
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
y=1
= pet
etz (1 p)z
z=0
= pet
[et (1 p)]z
z=0
= pe
t
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 inches) and a calculator. You may not share a
calculator, p
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 notes:
> d = scan("otc.dat")
Read 40 items
> pdf("otc.p