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Exam (elaborations) TEST BANK FOR Probability and Stochastic Processes

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  • February 13, 2022
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  • 2021/2022
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, Probability and Stochastic Processes
A Friendly Introduction for Electrical and Computer Engineers
SECOND EDITION
Problem Solutions
September 28, 2005 Draft
Roy D. Yates, David J. Goodman, David Famolari

September 28, 2005


• This solution manual remains under construction. The current count is that 678 (out of 687)
problems have solutions. The unsolved problems are

12.1.7, 12.1.8, 12.5.8, 12.5.9, 12.11.5 – 12.11.9.

If you volunteer a solution for one of those problems, we’ll be happy to include it . . . and, of
course, “your wildest dreams will come true.”

• Of course, the correctness of every single solution reamins unconfirmed. If you find errors or
have suggestions or comments, please send email: ryates@winlab.rutgers.edu.

• If you need to make solution sets for your class, you might like the Solution Set Constructor
at the instructors site www.winlab.rutgers.edu/probsolns. If you need access, send email:
ryates@winlab.rutgers.edu.

• Matlab functions written as solutions to homework problems can be found in the archive
matsoln.zip (available to instructors) or in the directory matsoln. Other Matlab functions
used in the text or in these homework solutions can be found in the archive matcode.zip
or directory matcode. The .m files in matcode are available for download from the Wiley
website. Two other documents of interest are also available for download:

– A manual probmatlab.pdf describing the matcode .m functions is also available.
– The quiz solutions manual quizsol.pdf.

• A web-based solution set constructor for the second edition is available to instructors at
http://www.winlab.rutgers.edu/probsolns

• The next update of this solution manual is likely to occur in January, 2006.




1

,Problem Solutions – Chapter 1

Problem 1.1.1 Solution
Based on the Venn diagram

M O


T

the answers are fairly straightforward:
(a) Since T ∩ M = φ, T and M are not mutually exclusive.

(b) Every pizza is either Regular (R), or Tuscan (T ). Hence R ∪ T = S so that R and T are
collectively exhaustive. Thus its also (trivially) true that R ∪ T ∪ M = S. That is, R, T and
M are also collectively exhaustive.

(c) From the Venn diagram, T and O are mutually exclusive. In words, this means that Tuscan
pizzas never have onions or pizzas with onions are never Tuscan. As an aside, “Tuscan” is
a fake pizza designation; one shouldn’t conclude that people from Tuscany actually dislike
onions.

(d) From the Venn diagram, M ∩ T and O are mutually exclusive. Thus Gerlanda’s doesn’t make
Tuscan pizza with mushrooms and onions.

(e) Yes. In terms of the Venn diagram, these pizzas are in the set (T ∪ M ∪ O)c .

Problem 1.1.2 Solution
Based on the Venn diagram,

M O


T

the complete Gerlandas pizza menu is
• Regular without toppings
• Regular with mushrooms
• Regular with onions
• Regular with mushrooms and onions
• Tuscan without toppings
• Tuscan with mushrooms

Problem 1.2.1 Solution

(a) An outcome specifies whether the fax is high (h), medium (m), or low (l) speed, and whether
the fax has two (t) pages or four (f ) pages. The sample space is

S = {ht, hf, mt, mf, lt, lf } . (1)

2

, (b) The event that the fax is medium speed is A1 = {mt, mf }.

(c) The event that a fax has two pages is A2 = {ht, mt, lt}.

(d) The event that a fax is either high speed or low speed is A3 = {ht, hf, lt, lf }.

(e) Since A1 ∩ A2 = {mt} and is not empty, A1 , A2 , and A3 are not mutually exclusive.

(f) Since
A1 ∪ A2 ∪ A3 = {ht, hf, mt, mf, lt, lf } = S, (2)
the collection A1 , A2 , A3 is collectively exhaustive.


Problem 1.2.2 Solution

(a) The sample space of the experiment is

S = {aaa, aaf, af a, f aa, f f a, f af, af f, f f f} . (1)

(b) The event that the circuit from Z fails is

ZF = {aaf, af f, f af, f f f } . (2)

The event that the circuit from X is acceptable is

XA = {aaa, aaf, af a, af f } . (3)

(c) Since ZF ∩ XA = {aaf, af f } = φ, ZF and XA are not mutually exclusive.

(d) Since ZF ∪ XA = {aaa, aaf, af a, af f, f af, f f f} = S, ZF and XA are not collectively exhaus-
tive.

(e) The event that more than one circuit is acceptable is

C = {aaa, aaf, af a, f aa} . (4)

The event that at least two circuits fail is

D = {f f a, f af, af f, f f f } . (5)

(f) Inspection shows that C ∩ D = φ so C and D are mutually exclusive.

(g) Since C ∪ D = S, C and D are collectively exhaustive.


Problem 1.2.3 Solution
The sample space is

S = {A♣, . . . , K♣, A♦, . . . , K♦, A♥, . . . , K♥, A♠, . . . , K♠} . (1)

The event H is the set
H = {A♥, . . . , K♥} . (2)


3

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