ECE 340: PROBABILISTIC METHODS IN ENGINEERING
SOLUTIONS TO HOMEWORK #2
2.16
Solution
a) Write an expression for the event overall system is up.
A= (A11 A12) (A21 A22) (A31 A32)
b) We begin by stating
ECE 340: PROBABILISTIC METHODS IN ENGINEERING
H
SOLUTIONS TO HOMEWORK #9
4.94
X is a uniformly distributed in the range of (-1,1), thus X has the following pdf
1
1
,
1
2
0,
Since
tan
, the plot of
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X .^\,. \)r-.1\o*^ (* t -,
/\ Uu^,'.\,r\rv\-\ ( \O, \f)
Y
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\
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ECE 340 (Spring 2016)
Homework#2 Solutions
1.3.14
a)
Note that
= ,
=
(1)
By applying Axiom 3 to the mutually exclusive sets S and , we obtain
( ) = ( ) + ()
(2)
However, from (1) we have
( ) = ( ) =
UNIVERSITY OF NEW MEXICO
Department of Electrical and Computer Engineering
ECE 340 Probability Methods in Engineering (Summer 2016)
Sample Final Exam
Question no 1- Example no 9.2 (page no 313)
Questi
18.440 Practice Midterm: 50 minutes, 100 points
Carefully and clearly show your work on each problem (without
writing anything that is technically not true) and put a box
around each of your final com
UNIVERSITY OF NEW MEXICO
Department of Electrical and Computer Engineering
ECE 340 Probability Methods in Engineering (Summer 2016)
Sample Mid-term Exam no 2
Question no 1- Example no 7.11 (page no 25
UNIVERSITY OF NEW MEXICO
Department of Electrical and Computer Engineering
ECE 340 Probability Methods in Engineering (Summer 2016)
Sample Quiz no 3
Question no 1- Example no 4.17 (page no 142) not su
ECE 340 (Spring 2016)
Solutions to Assignment #1
1.1.3
M
R
W
O
T
1.2.1
a) Sample space: S= cfw_hc, ht, mc, mt, lc, lt
b) A1= cfw_mc, mt
c) A2= cfw_hc, mc, lc
d) A3= cfw_hc, ht, lc, lt
e) No,
=
,
Chapter no 1 and 2 Extra Questions
Example: Show that the following equallity is correct.
( ) = A (
)
)
= A (
= A ( )
= (A )
= ( ) = A
= A (
)
Identity is hold.
Example:
(A ) ( ) = ?
= [(A ) ] [(A
ECE 340 (Spring 2016)
Solutions to Homework#3
2.1.1
a)
The problem statement on any flip the coin turns up heads with probability 1/4 implies (loosely I am
afraid) that the outcomes of each coin is in
ECE 340: PROBABILISTIC METHODS IN ENGINEERING
SOLUTIONS TO HOMEWORK #8
4.54
a) Solution:
1
1
d) Solution:
First, we need to obtain the pdf of X. We do so by first obtaining the cdf of X. Notice that
ECE 340: PROBABILISTIC METHODS IN ENGINEERING
H
SOLUTIONS TO HOMEWORK #7
Y is the difference between the number of heads and the number of tails in the 3 tosses
of a fair coin. Let m be the number of
ECE-340, Spring 2010
Final Examination, May 10th, 2010; 5:30-7:30 PM
1. Suppose that there are two categories of eggs: large eggs and small eggs, occurring with
probabilities 0.7 and 0.3, respectively
ECE340 Spring 2011
Recitation class
April 06, 2011
Problem 1
Two random variables, X and Y, have a joint probability density function of the form:
(+1)
0 , 1
(, ) =
0
Find
a) The values of and for
2.1 Solution
a) The sample space consists of the twelve hours:
S=cfw_1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
b) A=cfw_1, 2, 3, 4
c) A
B
AC
B=cfw_2, 3, 4, 5, 6, 7, 8
D=cfw_1, 3, 5, 7, 9, 11
D = cfw_3
B =
ECE 340: PROBABILISTIC METHODS IN ENGINEERING
SOLUTIONS TO HOMEWORK #3
2.39
Here, we have replacement and we do care about the ordering.
Thus, the number of distinct ordered triples is 60 x 60 x 60 =
ECE 340: PROBABILISTIC METHODS IN ENGINEERING
SOLUTIONS TO HOMEWORK #4
2.54
a) The number of ways of choosing (without replacement nor caring about order) M out of
100
. This is the total number of
ECE 340: PROBABILISTIC METHODS IN ENGINEERING
SOLUTIONS TO HOMEWORK #5
3.10.
Solution
a) Note that the size of the alphabet of each bit in the password is 2. Then there is a total
2 number of possible