TEST BANK FOR Probability and Stochastic 2nd Ed By Roy D. Yates

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: [email protected].
• 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:
[email protected].
• 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, afa, faa, ffa, faf, aff, fff} . (1)
(b) The event that the circuit from Z fails is
ZF = {aaf, aff, faf, fff} . (2)
The event that the circuit from X is acceptable is
XA = {aaa, aaf, afa, aff} . (3)
(c) Since ZF ∩ XA = {aaf, aff} = φ, ZF and XA are not mutually exclusive.
(d) Since ZF ∪XA = {aaa, aaf, afa, aff, faf, fff} = S, ZF and XA are not collectively exhaustive.
(e) The event that more than one circuit is acceptable is
C = {aaa, aaf, afa, faa} . (4)
The event that at least two circuits fail is
D = {ffa, faf, aff, fff} . (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
Problem 1.2.4 Solution
The sample space is
S =
⎧⎨
⎩
1/1 . . . 1/31, 2/1 . . . 2/29, 3/1 . . . 3/31, 4/1 . . . 4/30,
5/1 . . . 5/31, 6/1 . . . 6/30, 7/1 . . . 7/31, 8/1 . . . 8/31,
9/1 . . . 9/31, 10/1 . . . 10/31, 11/1 . . . 11/30, 12/1 . . . 12/31
⎫⎬
⎭. (1)
The event H defined by the event of a July birthday is described by following 31 sample points.
H = {7/1, 7/2, . . . , 7/31} . (2)
Problem 1.2.5 Solution
Of course, there are many answers to this problem. Here are four event spaces.
1. We can divide students into engineers or non-engineers. Let A1 equal the set of engineering
students and A2 the non-engineers. The pair {A1,A2} is an event space.
2. We can also separate students by GPA. Let Bi denote the subset of students with GPAs G
satisfying i − 1 ≤ G < i. At Rutgers, {B1,B2, . . . , B5} is an event space. Note that B5 is
the set of all students with perfect 4.0 GPAs. Of course, other schools use different scales for
GPA.
3. We can also divide the students by age. Let Ci denote the subset of students of age i in years.
At most universities, {C10, C11, . . . , C100} would be an event space. Since a university may
have prodigies either under 10 or over 100, we note that {C0, C1, . . .} is always an event space
4. Lastly, we can categorize students by attendance. Let D0 denote the number of students who
have missed zero lectures and let D1 denote all other students. Although it is likely that D0
is an empty set, {D0,D1} is a well defined event space.
Problem 1.2.6 Solution
Let R1 and R2 denote the measured resistances. The pair (R1,R2) is an outcome of the experiment.
Some event spaces include
1. If we need to check that neither resistance is too high, an event space is
A1 = {R1 < 100,R2 < 100}, A2 = {either R1 ≥ 100 or R2 ≥ 100} . (1)
2. If we need to check whether the first resistance exceeds the second resistance, an event space
is
B1 = {R1 > R2} B2 = {R1 ≤ R2} . (2)
3. If we need to check whether each resistance doesn’t fall below a minimum value (in this case
50 ohms for R1 and 100 ohms for R2), an event space is
C1 = {R1 < 50,R2 < 100}, C2 = {R1 < 50,R2 ≥ 100} , (3)
C3 = {R1 ≥ 50,R2 < 100}, C4 = {R1 ≥ 50,R2 ≥ 100} . (4)
4. If we want to check whether the resistors in parallel are within an acceptable range of 90 to
110 ohms, an event space is
D1 =

(1/R1 + 1/R2)−1 < 90

, (5)
D2 =

90 ≤ (1/R1 + 1/R2)−1 ≤ 110

, (6)
D2 =

110 < (1/R1 + 1/R2)−1
. (7)
4
Problem 1.3.1 Solution
The sample space of the experiment is
S = {LF,BF,LW,BW} .