University of Massachusetts - Amherst
Department of Electrical and Computer Engineering
ECE 314 Introduction to Probability and Random Processes
Spring 2016 Syllabus
DESCRIPTION
This course provides an elementary introduction to probability and statistics
HW 9 Solutions
ECE 314 Introduction to Probability and Random Processes Spring 2013
April 23, 2013
Problem 1
(a)
We can easily nd the piecewise constant density of Y
1 , |y | 1
4
1
fY (y ) = 8 , 1 < |y | 3
0, otherwise
The conditional probabilities of X
HW 10 Solutions
ECE 314 Introduction to Probability and Random Processes Spring 2013
May 1, 2013
Problem 1
First we need to nd ( 2 ).
2
Since we want 95% condence, 1 = .95, so = .05.
2
= .975. Using z-function table, (.975) = 1.96.
2
Now you can solve for
Name:
ID:
ECE 314 - Introduction to Probability and Random Processes, Spring 2012
Quiz #3
In class: 02/15/2012
Exercise 1. A box containing 8 red, 3 white, 9 blue balls. If 3 balls are drawn at random without replacement determine the probability
(a) that
1. A coffee shop has 4 different types of coffee. You can order your coffee in a small,
medium, or large cup. You can also choose whether you want to add cream, sugar, or
milk (any combination is possible, for example, you can choose to add all three). In
ECE 314 - Introduction to Probability and Random
Processes, Spring 2016
Sample Exam 1
Name:
Overview:
The exam consists of 6 problems for 100 points. The points for
each part of each problem are given - you should spend your two
hours accordingly.
The e
HW 8 Solutions
ECE 314 Introduction to Probability and Random Processes Spring 2013
April 17, 2013
Problem 1
We can write the joint pdf for X and Y jointly Gaussian as:
fX,Y (x, y ) =
exp([a(x x )2 + b(y Y )2 + c(x x )(y Y )])
2X Y
where a =
1
2,
2(12 )X
HW 6 Solutions
ECE 314 Introduction to Probability and Random Processes Spring 2013
March 29, 2013
Problem 1
We have for x 0:
P (X > x|A) = P (T > t + x|T > t) =
=
P (T > t + x and T > t)
P (T > t)
e(t+x)
P ( T > t + x)
=
= ex
t
P (T > t )
e
Problem 2
If
Tong Huang ECE314 Spring 2007 Assignment 3 1. a) Assume drawing cards without replacements. The probability of drawing all 4 4 3 2 1 24 3.693785 *10 6 . aces with 4 cards is * * * 52 51 50 49 6497400 We are drawing 5 cards, therefore there are 5 choo
Tong Huang ECE314 spring 2997 Homework 4
0.3 y 0 0.05 y 1 0.15 y 2 1. a) PY (y) 0.1 y 3 0.4 y 4 0 otherwise b) According to the CDF given, 35% of students score 1 or less, and 50% of students score 2 or less. We have to made the passing probability
Department of Electrical and Computer Engineering
University of Massachusetts, Amherst
ECE 314: Introduction to Probability and Random Processes
Spring 2011
Course Webiste:
Course offered on SPARK.
Instructors:
Prof. Hossein Pishro-Nik
215I Marcus Hall
Ph
HW 7 Solutions
ECE 314 Introduction to Probability and Random Processes Spring 2013
April 9, 2013
Problem 1
(a)
Region where fX,Y (x, y )is non-zero.
f X ( x) =
Z1
fX,Y (x, y )dy
1
81 x
< R 3(x + y )dy = 3x(1
=
:0
0,
fY (y ) =
x) + 3 y 2
2
1x
0
=
3
2
32
x
Solutions homework 3
Problem 1
We break this into two parts.
The first part we look at is that the 15th flip is the 10th head. That means that the
15th flip is a head (this occurs with probability p) and 9 of the first 14 flips were
heads. Since flipping
HW 4 Solutions
ECE 314 Introduction to Probability and Random Processes Spring 2013
March 7 , 2013
Problem 1
To verify this rule, we let Y = g (x) and use the formula
pY ( y ) =
pX (x)
cfw_x|g (x)=y
E [g (x)] = E [Y ]
=
ypY (y )
y
=
y
y
pX (x)
cfw_x|g (x
HW 5 Solutions
ECE 314 Introduction to Probability and Random Processes Spring 2013
March 12, 2013
Problem 1
(a)
Legitimate PMF because
1
X
k=0
k
e
k!
2
=e
(1 +
+
2!
3
+
3!
+ ) = e
e =1
(b)
E [X ] =
=
=
=
1
X
k=0
1
X
k
ke
k
ke
k=1
1
X
k=1
1
X
m=0
=
k!
k!
Tong Huang ECE 314 spring 2007 Homework 2 1. Given P(A) 0.3; P(B) 0.6; P(C) 0.2; P(A C) 0; P(B | C) 0.5;P(A B C) 0.8. a) Because P(A C) 0 , which means A and C are independent events, so the intersection of A, B and C must be 0. b) P(B | C) P(BC) / P