Unformatted text preview: Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions : Yates and Goodman,1.2.1 1.2.2 1.2.3 1.2.5 1.3.1 1.3.3 1.3.5 1.4.1 1.4.4 and
1.4.5
Problem 1.2.1 ¡ ¡ ¡ (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 ¦ ¥ ¥ §¢
£
¦ ¥ §¢
£
¦ ¥ ¥ ¥ ¥ ¥ £¤¢
¡ ht h f mt m f lt l f (b) The event that the fax is medium speed is A1
(c) The event that a fax has two pages is A2 mt m f . ht mt lt . (d) The event that a fax is either high speed or low speed is A3 ¦ ¥ ¥ ¥ §¢
£ ¡ S mt and is not empty, A1 , A2 , and A3 are not mutually exclusive. A1 A2 A3 ¥ ¦ ¥ ¥ ¥ ¥ ¥ §¢
¢
£ ¨ (f) Since ¦ §¢
£ (e) Since A1 A2 ht h f lt l f . ht h f mt m f lt l f S ©© the collection A1 , A2 , A3 is collectively exhaustive.
Problem 1.2.2
(a) The sample space of the experiment is ¥ ¥¥ ¥ ¥ ¥ ¥ §¢
£ aaa aa f a f a f aa f f a f a f a f f f f f ¦ S (b) The event that the circuit from Z fails is ¥¥¥ aa f a f f f a f f f f £
§¢ The event that the circuit from X is acceptable is φ, ZF and XA are not mutually exclusive. ¥ XA aa f a f f aaa aa f a f a a f f aaa aa f a f a a f f f a f f f f 1 ¢
¦ © ¨ (d) Since ZF
tive. XA ¥ ¥ ¥ ¥ ¤¢
£
¢
¦ ¥ ¢
£ (c) Since ZF ¦ ¥ ¥ ¥ §¢
£ XA ¦ ZF S, ZF and XA are not collectively exhaus (e) The event that more than one circuit is acceptable is ¦ ¥ ¥ ¥ §¢
£ C aaa aa f a f a f aa The event that at least two circuits fail is £
§¢ ¢¨ ¢© (g) Since C D ¥¥¥ (f) Inspection shows that C D f fa faf af f f f f ¦ D φ so C and D are mutually exclusive. S, C and D are collectively exhaustive. Problem 1.2.3
The sample space is
A K A K H A K A K ¦ " ¥ " ¢
¥
£ The event H is the set ¦ $ !¥ $ #" #" ¥ !¥ §¢
¥
¥¥¥
¥
¥¥¥£ S A K Problem 1.2.5
this problem. Here are four event spaces. 1. We can divide students into engineers or nonengineers. Let A1 equal the set of engineering
students and A2 the nonengineers. 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 1.3.1
The sample space of the experiment is 20 ) 10 )
¢
¢
¦ ¥ ¥ ¥ §¢
£
LF BF LW BW 20 )
¢ S 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 Theorem 1.5. ¢ % % % 30 )
¢ (a) The probability that a program is slow is ¢ 5 70 ) 60 ) 40 )
¢
5
¢ PW P LW P BW
2 01 02 03 (b) The probability that a program is big is ¢ 5 40 ) 0 ) 70 )
¢
5
¢ PB P BF P BW 02 02 04 (c) The probability that a program is slow or big is ¢ % 5 70 ) 60 ) 0 ) 30 © )
¢
%5
¢ PW B PW PB P BW 03 04 02 05 Problem 1.3.3
A reasonable probability model that is consistent with the notion of a shufﬂed deck is that each card
in the deck is equally likely to be the ﬁrst card. Let Hi denote the event that the ﬁrst card drawn is
the ith heart where the ﬁrst heart is the ace, the second heart is the deuce and so on. In that case,
P Hi
1 52 for 1 i 13. The event H that the ﬁrst card is a heart can be written as the disjoint
union
H13
H H1 H2 ©9
@99© && ¢ © 8 30 )
¢ Using Theorem 1.1, we have 13 ∑ P Hi A i1 8 70 )
¢ ¢
70 ) PH 13 52 This is the answer you would expect since 13 out of 52 cards are hearts. The point to keep in mind
is that this is not just the common sense answer but is the result of a probability model for a shufﬂed
deck and the axioms of probability.
Problem 1.3.5
Let si equal the outcome of the student’s quiz. The sample space is then composed of all the
possible grades that she can receive. ¦ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ §¢
£
0 1 2 3 4 5 6 7 8 9 10 S Since each of the 11 possible outcomes is equally likely, the probability of receiving a grade of i, for
each i 0 1
10 is P si 1 11. The probability that the student gets an A is the probability that
she gets a score of 9 or higher. That is 8 ¢ 8 5 8 70 ) 0 ) ¢70
¢
5
)
8 C0 )
¢ ¥
B¥ ¥ ¢ P Grade of A P9 P 10 1 11 1 11 2 11 The probability of failing requires the student to get a grade less than 4. 8 ¢ 8 5 8 5 8 5 8 30 ) 0 ) 60 ) !0 ) 70
¢555¢ ) P Failing P3 P2 P1 P0 1 11 1 11 1 11 1 11 4 11 Problem 1.4.1
From the table we look to add all the disjoint events that contain H0 to express the probability
that a caller makes no handoffs as ¢ 5 40 ) 0 ) 70 )
¢
5
¢ P H0 P LH0 P BH0 01 04 05 In a similar fashion we can express the probability that a call is brief by ¢ 5 5 70 ) 60 ) 0 ) 70 )
¢
5
5
¢ PB P BH0 P BH1 P BH2 04 01 01 The probability that a call is long or makes at least two handoffs is ¢ 5 5 5
0 ) 0 ) 60 ) 0 )
5
5
5 ¢
D0 © ) P LH0 ¢ P L H2 01 P LH1 01 02 3 P LH2 01 05 P BH2 06 Problem 1.4.4
consequence of part 4 of Theorem 1.4. ¨
¨
&70
&
70 ) ©E A B, P A ) E¨
E¨
©E PA B. A B, P B ) (b) Since B 0 ) 70
&
0 ) 70
&
0 ©)
0 ©) (a) Since A PA B. A, P A B PA. (d) Since A B B, P A B ) PB. Problem 1.4.5 0)¢ (a) For convenience, let pi P F Hi and qi
p0 p1 p2 q0 q1 q2 ﬁll the table as 0)¢ (c) Since A B P V Hi . Using this shorthand, the six unknowns ¥¥¥¥¥ H0 H1 H2
F p0 p1 p2
V q0 q1 q2 However, we are given a number of facts: 5 8¢ 5 5
8¢ 5
13 p0 5 q2 p1 5 p1 q1
p2 13 8¢ 8¢ 13 5 p2 q0 8¢ p0 5 12 Other facts, such as q0 q1 q2 7 12, can be derived from these facts. Thus, we have four
equations and six unknowns, choosing p0 and p1 will specify the other unknowns. Unfortunately, arbitrary choices for either p0 or p1 will lead to negative values for the other probabilities. In terms of p0 and p1 , the other unknowns are %8 ¢
%8 ¢ q1 13 p0 p2 13 p1 q2 8% 5 ¢
¡ 5 % 8 ¢ q0 5 12
p0 p0 p1 p1 1 12 Because the probabilities must be nonnegative, we see that
13 p1 8& 5 &8 & 1 12 p1 13 8& & p0 0 8& p0 0 5 12 Although there are an inﬁnite number of solutions, three possible solutions are: 8¢ 4 8¢ q1 13 p2 1 12
3 12 q2 3 12 13 ¢ ¢ p1 8¢ 0 q2 3 12 1 12 8¢ 8¢ q0 q1 8¢ p0 1 12 p2 1 12 13 8¢ and 8¢ q0 p1 14 8¢ p0 q2 14 0 8¢ ¢ and 8¢ 8¢ q1 p2 1 12 8¢ 0 q0 8¢ p1 ¢ 13 p0 0 8¢ 5 16
16 p2
q2 0 8¢ q1 8¢ 8¢
8¢ 1 12 p1 1 6. These extra facts ¢ ¥ q0 14 8¢ p0 1 4 and q1 8¢ (b) In terms of the pi qi notation, the new facts are p0
uniquely specify the probabilities. In this case, 13 ...
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This note was uploaded on 02/11/2012 for the course EEE 352 taught by Professor Ferry during the Spring '08 term at ASU.
 Spring '08
 Ferry

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