ECE 863 – Analysis of Stochastic Systems

Fall 2001

Solutions for Homework Set #1

Michigan State University

Department of Electrical & Computer Engineering

Problem 1

2.4 (a)

The sample space S = set of ordered pairs which can be expressed as:
(i, 1) (i, 2) …… (i, i), for i = 1, 2, 3, 4, 5, 6

Therefore:
mb g
S = 1, 1
b2,1g, b2,2g
b3,1g, b3,2g, b3,3g


b6,1g, b6,2g, b6,3g, b6,4 g, b6, 5g, b6,6gr
2.4 (b)

mb4,1g, b4,2g, b4, 3g, b4,4 gr

2.4 (c)

mb3,3g, b4, 3g, b5, 3g, b6,3gr

2.4 (d)

mb6, 6gr

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Problem 2

2.5 (a)

Each testing of a pen has two possible outcomes: “pen is good” (g) or “pen is
bad” (b). The experiment consists of testing the pens until a good pen is
found. Therefore, each outcome of the experiment consists of a string of
“b”s ended by a “g.” Assuming that each tested pen is not put back in the
l
drawer, the sample space S = g, bg, bbg, bbbg . q
2.5 (b)

In this case, we simply record the number of pens that have been tested.
l q
Therefore, S = 1, 2, 3, 4 .

2.5 (c)

In this case,

l
S = gg, bgg, gbg, bbgg, bgbg, gbbg, bbbgg, bbgbg, bgbbg, gbbbg q
2.5 (d)

l
S = 2, 3, 4, 5 q
Problem 3

2.9

b g
A = 5, • lifetime is greater than 5
B = b7, •g lifetime is greater than 7
C = b0, 3g lifetime is not greater than 3

b g
A « B = 7, • lifetime is greater than 5 and 7
A « C = ∆ lifetime is greater than 5 and not greater than 3
b g
A » B = 5, • lifetime is greater than 5 or 7

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Problem 4

2.13 (a)

dA « B c
i d i d
« C c » Ac « B « Cc » Ac « B c « C i
2.13 (b)

dA « B « C i » dA « B
c c
i d
« C » Ac « B « C i
2.13 (c)

A»B » C

2.13 (d)

nThe events in bbg aboves » bA « B « Cg

2.13 (e)

Ac « B c « Cc

Problem 5

2.18

l q
Since the event a, c can be decomposed into two mutually exclusive events:
laq and lcq , fi P la, cq = P alq + P clq = 5 /8

Similarly P b, c l q = P blq +P c lq = 7 /8

We also have P s = 1 = P a, b, c l q
= P laq + P lbq + P lcq

lq
fi P a = 1 /8 P blq = 3 /8 P c lq = 4 /8

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Problem 6

2.21

P A »B » C = P A»B » C b g
b
= P A»B + P Cg - P bA » Bg « C
= P A + P B - P A«B + P C - P A« C » B « C b g b g
= P A + P B + P C - P A«B - P bA « Cg - P bB « Cg b g b
+ P A« C « B « C g
= P A + P B + P C - P A«B - P A« C - P B « C + P A«B « C

Problem 7

2.24

l
We have A = k > 5 and B = k > 10q l q
FG 1 IJ 6

PA =
F 1 I F 1 I R 1 1 U H 2K = FG 1 IJ
j
 G J = G J S1 + +
V =
•
6 5

H 2K H 2K T 2 4 W 1 - 1 H 2K
j=6
2

Similarly PB
F 1I
= G J
10

H 2K
PB c F 1I
= 1-G J
10

H 2K
P A«B = P B =
FG 1 IJ 10

H 2K
P A »B = P A
F 1I
= G J
5

H 2K
(You can also solve this using P A » B = P A + P B - P A « B )

Problem 8

2.48 (a)

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 P A«B
Using the definition P A /B = ,
PB

when A « B = ∆ fi P A « B = 0 fi P A /B = 0

PA
when A Ã B fi A « B = A fi P A /B =
PB

when B Ã A fi A « B = B fi P A /B = 1

2.48 (b)

If P A /B > P A fi P A « B > P A P B

fi P B / A P A > P A P B

fi P B / A > P B

Therefore, P A /B > P A (which is also equivalent to P B / A > P B as we

just showed) implies that A & B tend to occur jointly.

Problem 9

2.50

b g b g
P A«B « C = P A/ B « C P B « C
= P A / bB « Cg P B / C P C

Problem 10

2.51

Let A = total number of dots is even

B = both tosses are even

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 P A «B PB
P A /B = = = 1
PB PB
P A«B PB 1
P B /A = = =
PA PA 2

Problem 11

Let A = the event that 2 or more students have the same birthday
fi Ac = the event that all students have different birthdays

fi The desired probability is P A = 1 - P Ac

P Ac = ’
19 b365 - ig = b365gb365 - 1gb365 - 2g
b365 - 19g
i=0 365 b365g20

= 0.588

\ P A = 1 - 0.588 = 0.412

Problem 12

2.56

Let A1 = The event that the professor will arrive between 8:30 and 9:00.
B = The event that the professor will arrive within the next minute.

Therefore, the desired probability is:

P A1 « B P A1 /B P B
P B / A1 = =
P A1 P A1

Notice that P A1 /B = 1 (i.e. given that the professor will arrive within the
next minute, we are certain that he/she will arrive within the next 30
minutes).

fi P B / A1 =
b
1 1 / 60 g= 1
b30 / 60g 30

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 Similarly, let A2 = The event that the professor will arrive between 8:50 and
9:00.

fi P B / A2 =
P A2 /B P B
=
b
1 1 / 60g= 1
P A2 b10 / 60g 10
Problem 13

2.57

Let’s use the following diagram (i.e. instead of the one in the book)

Input Output
1 - e0
0 0
p0 e0

e1
1 1
p1 1 – e1

2.57 (a)

Find the probability that the output is O.

Let: O0 represents the event that the output is 0.

O1 represents the event that the output is 1.
I0 represents the event that the output is 0.
I1 represents the event that the output is 1.

Since I0 and I1 represent a Partition (i.e. I0 « I1 = ∆ & I0 » I1 = S )

fi P O 0 = P O 0 / I0 P I0 + P O 0 / I1 P I1
b g
= 1 - e0 P0 + e1 p1

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2.57 (b)

Find the probability that the input is 0 given that the output is 1.

P O 1 / I0 P I0
P I0 / O 1 =
P O1 / I0 P I0 + P O1 / I1 P I1
e0 p 0
=
e 0p 0 b
+ 1 - e1 p 1 g
2.57 (c)

Find the probability that the input is 1 given that the output is 1.

P O1 / I1 P I1
P I1 / O1 =
P O1 / I0 P I0 + P O1 / I1 P I1

=
b1 - e g p 1 1
e0 p + b1 - e g p
0 1 1

2.57 (d)

Under what conditions it is more likely that the input is 1 given that the
output is 1.

P I1 / O1 > P I0 / O1
b g
fi 1 - e1 p1 > e0 p 0
b1 - e g p
1 1 b g
> e 0 1 - p1

fi - e1 >
e0 b1 - p g - 11
p1

e1 < 1 - e 0
b1 - p g 1
p1

Problem 14

2.63

8
 First, we express A = A « S
d
= A « B » Bc i
b g d
= A « B » A « Bc i
fi P A = P A « B + P A « B c

= P A /B P B + P A /B c P B c
Since, P A /B = P A /Bc

{
fi P A = P A /B P B + P Bc }
fi P A = P A /B ¤ P A P B = P A « B

Therefore, A & B are independent.

Problem 15

2.64 (a)

P A »B = P A + P B - P A «B
= PA + PB - PA PB

2.64 (b)

P A »B = P A + P B

Problem 16

2.66 (a)

P A P B c P Cc + P Ac P B P C c + P Ac P B c P C

2.66 (b)

P A P B P Cc + P Ac P B P C + P A P B c P C

2.66 (c)

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 1 - P Ac P Bc P Cc

I think you got the idea (d) and (e) can be solved the same way.

Problem 17

2.68

In terms of relative frequencies, we expect

bg bg bg
fA « B n = fA n fB n

bg
where fA « B n is the relative frequency of the joint occurrence of A and B.

Problem 18

l q
A transmitter sends four possible symbols s 0 , s1, s2, s3 with probabilities
p 0 , p1 , p2 & p 3 , respectively. The receiver detects the transmitted signal
and makes a decision that one of the four symbols has been sent. The
l q
output of the receiver can be donated by: r0 , r1, r2 , r3 . Each of the
transmitted symbols consists of two bits: s0 = 00; s1 = 01; s2 = 10; s 3 = 11 . If
the probability of a bit-error is e, what is the probability that the
transmitter has sent s0 given that the receiver detects r3?

P s0 « r3 P r3 / s 0 P s0
P s0 / r3 = =
P r3 n
P r3 / s0 P s0 + P r3 / s1 P s1 + P r3 / s2 P s2 + P r3 / s3 P s3 s

e2 p 0
=
b g b g b
e2 p 0 + e 1 - e p1 + 1 - e e p2 + 1 - e g
2
p3

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Probability of a symbol error

= P s0 « r1 + P s0 « r2 + P s0 « r3
+ P s1 « r0 + P s1 « r2 + P s 0 « r3
+ P s2 « r0 + P s2 « r1 + P s2 « r3
+ P s3 « r0 + P s 3 « r1 + P s3 « r2

b g b g
= p 0 1 - e e + p 0 1 - e e + p 0 e2
p b1 - ege + p b1 - ege + p e
1 1 1
2

p b1 - ege + p b1 - ege + p e
2 2 2
2

p b1 - ege + p b1 - ege + p e
3 3 3
2

= b1 - ege + b1 - ege + e = 2b1 - ege + e

2 2

= 2e - 2e2 + e2 = 2e - e2

ECE.863.Homework.Set.1.Solutions.Fall2000.slh

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