EE3150:Probability for Electrical and Computer Engineering
Fall 2015
Instructor: Prof. Shuangqing Wei
Ofce: Room 341, ERAD
Ofce Hours: 2:00-4:00PM (Monday and Wednesday), 2:00-3:00PM (Friday), or by appointment
E-mail: [email protected]
Class meets:
9:3010:
(c) L is a Poisson ( = E[L] = 2) random variable. Thus its PMF is
2l e2 /l! l = 0, 1, 2, . . . ,
0
otherwise.
PL (l) =
(4)
It follows that
P [L 1] = PL (0) + PL (1) = 3e2 = 0.406.
(5)
Problem 3.5.4 Solution
Both coins come up heads with probability p = 1/
EE 3150, Probability for Electrical and Computer Engineering, Fall 2015
Final, Dec. 11 , 2015
Overview
The exam consists of four problems for 100 points. The points for each
part of each problem are given in the brackets - you should spend your
120 minut
ECE302 HW2 Solution
Problem 1.4.2 Solution
2/01/15
Let si denote the outcome that the roll is i. So, for 1 i 6, Ri = cfw_si . Similarly,
Note that P[RiGj] = P[Ri] if i>j and 0 otherwise.
Gj = cfw_sj+1 , . . . , s6 .
P[Ri]=1/6 for all i. P[E]=1/2. P[Gj]=(6
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EE 3150, Probability for Electrical and Computer Engineering, Fall 2015
Midterm 2, Nov. 6, 2015
Overview
The exam consists of two problems for 100 points. The points for each
part of each problem are given in the brackets - you should spend your
50 minut
EE 3150, Probability for Electrical and Computer Engineering, Fall 2015
Midterm 1, Oct. 2
Overview
The exam consists of three problems for 100 points. The points for
each part of each problem are given in the brackets - you should spend
your 50 minutes a
Problem 4.2.4 Solution
In this problem, the CDF of W is
0
(w + 5)/8
FW (w) = 1/4
1/4 + 3(w 3)/8
1
w < 5,
5 w < 3,
3 w < 3,
3 w < 5,
w 5.
(1)
Each question can be answered directly from this CDF.
(a)
P [W 4] = FW (4) = 1/4 + 3/8 = 5/8.
(2)
P [2 < W 2] = FW
(b) The professor counts each M&M either once or twice so that N = 20+R.
This implies
PN (n) = P [20 + R = n] = PR (n 20)
20
=
pn20 (1 p)40n .
n 20
(2)
Problem 3.7.1 Solution
Let Wn equal the number of winning tickets you purchase in n days. Since
each da
Problem 1
(a) The events are cfw_1,2,3,4,5,6, the probability of each event is
P(i)=1/6
The events that 4 or higher are cfw_4,5,6, therefore, the probability of roll 4 or higher is
P(i4)=P(4)+P(5)+P(6)=1/6+1/6+1/6=3/6
(b) The events that the square of an
Probability and Stochastic Processes
A Friendly Introduction for Electrical and Computer Engineers
Third Edition
STUDENTS SOLUTION MANUAL
(Solutions to the odd-numbered problems)
Roy D. Yates, David J. Goodman, David Famolari
August 27, 2014
1
Comments on
EE 3150, Homework 6: Due in class on 11/4/2015
The following problems are taken from the handout for Chapter 3 which can be downloaded from Moodle. To get full credits for each problem, you are required to present all
intermediate steps to reach your nal
Problem 3.2.1 Solution
(a) We wish to nd the value of c that makes the PMF sum up to one.
PN (n) =
Therefore,
2
n=0
c(1/2)n
0
n = 0, 1, 2,
otherwise.
(1)
PN(n) = c + c/2 + c/4 = 1, implying c = 4/7.
(b) The probability that N 1 is
P [N 1] = P [N = 0] + P
EE 3150, Homework 1: Due in class on 9/4/2015
The following problems are taken from the handout for Chapter 1 which can be downloaded from Moodle. To get full credits for each problem, you are required to present all
intermediate steps to reach your nal a
EE 3150, Homework 3: Due in class on 9/25/2015
The following problems are taken from the handout for Chapter 1 which can be downloaded from Moodle. To get full credits for each problem, you are required to present all
intermediate steps to reach your nal
EE 3150, Homework 2: Due in class on 9/16/2015
The following problems are taken from the handout for Chapter 1 which can be downloaded from Moodle. To get full credits for each problem, you are required to present all
intermediate steps to reach your nal
(e) Yes. In terms of the Venn diagram, these pizzas are in the set (T M
O)c .
Problem 1.1.3 Solution
R
At Ricardos, the pizza crust is either Roman (R) or Neapolitan (N ). To draw the Venn diagram on the right, we make
the following observations:
W
N
M
O
Problem 1.3.8 Solution
Let si denote the outcome that the down face has i dots. The sample space
is S = cfw_s1 , . . . , s6 . The probability of each sample outcome is P[si ] = 1/6.
From Theorem 1.1, the probability of the event E that the roll is even is
Problem 1.3.5 Solution
The sample space of the experiment is
S = cfw_LF, BF, LW, BW .
(1)
From the problem statement, we know that P[LF ] = 0.5, P[BF ] = 0.2 and
P[BW ] = 0.2. This implies P[LW ] = 1 0.5 0.2 0.2 = 0.1. The questions
can be answered using
EE 3150, Homework 4: Due in class on 10/7/2015
The following problems are taken from the handout for Chapter 3 which can be downloaded from Moodle. To get full credits for each problem, you are required to present all
intermediate steps to reach your nal
EE 3150, Homework 5: Due in class on 10/19/2015
The following problems are taken from the handout for Chapter 3 which can be downloaded from Moodle. To get full credits for each problem, you are required to present all
intermediate steps to reach your nal
Problem 3.3.9 Solution
(a) K is a Pascal (5, p = 0.1) random variable and has PMF
PK (k) =
k1 5
p (1 p)k5 =
4
k1
(0.1)5 (0.9)k5 .
4
(1)
(b) L is a Pascal (k = 33, p = 1/2) random variable and so its PMF is
PL (l) =
l 1 33
p (1 p)l33 =
32
l1
32
1
2
l
.
(2)