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 that signal transmission through two switches in series
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 function
ta n
is shown below:
Notice that for any gi
\"N:u*\* (X,Y),
X
0",.,.e\
Y
Z-vwcfw_,r-y(X, y)
<-
(\,N
i.r".cl*p",^ C.Lt\
,_)
X .^\,. \)r-.1\o*^ (* t -,
/\ Uu^,'.\,r\rv\-\ ( \O, \f)
Y
CPF r
L\NZ [w,b):
\
Y,
: r"
'N
r t,
f
f
W
g 4A
\ny.\-,.,.
/
CX,'Y)
Z
s
<L^:
a.\
81
Ki
u* a-x ( X, Y
:.k\
l
tArn-k cfw_
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
( ) = ( ) = 1
(3)
where the last equality follows from Axiom 2.
Su
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)
Question no 2- Example no 7.11 (page no 258 and page no 253)
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 computations.
1. (30 points) Twenty people in a room each
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 253) and Example no 7.19 (262)
Question no 2- Example no
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 sure yet
Question no 2- Example no 4.19 (page no 149)
Que
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,
=
,
d) Yes,
1.2.3
Sample space = cfw_H1, , HK, S1, , SK
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 ) ]
= ( ) ( ) = ( ) ( )
=
= [ ( )]= ( () = A = A
Or s
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 independent of that of the other coin. Hence, we can writ
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
X=U3 and U is uniformly distributed in [-1,1]. Since U
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 tails 0 m 3. Then 3-m is the number of heads
and the di
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. For a large egg, the probabilities of having 1, 2,
or
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 which the random variables X and Y are statistically in
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 = cfw_5, 6, 7, 8
A U (B
DC) = cfw_1, 2, 3, 4 U (cfw_2, 3
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 = 603
2.43
There are 24 special characters, 10 numbers, 2
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 equiprobable outcomes in the sample space.
M
100 is
W
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 passwords. The sample space S can be expressed as
,
,