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Probability and random processes for electrical and computer
engineers second edition solution
확률론 (Sungkyunkwan University)
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Probability and random processes for
electrical and computer engineers
second edition solution
Problem
2.1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
2.2
(a) I)
II)
III)
(b) I)
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II)
III)
2.3
I) area always
II) normalize the area of the Venn diagram. So Pr[S] = 1
III)
of the areas
the sum
The area of the union of two events is
(no intersection)
IV) cannot be shown by Venn diagram
Corollaries (table 2.4)
area of A
always > 0
If , then
Null set has area size 0
Intersection is included twice if we sum the
areas
2.4 This solution does not contain the process.
2.5
I)
II)
III)
2.6
00
00
01
01
10
10
11
11
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a)
b)
c)
2.7
a)
1
5
1
1.35
b)
c)
2.8
a)sample spaces (not to scale)
-1
0
1
2
0.65
79.25
b)
2.9
a)
b)
c) this is not possible
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2.10
a)
b) P
c)
2.11
2.12
a)
b)
2. 13
a) probability =
b)
2.14
a)
b) Pr[one or more good] = 0.947
2.15
a) Pr[all good] = 0.512
b) Pr[all bad] =0.008
c) Pr[one bad] = 0.384
d)
2.16
a) total number = 65,536
b) possible commands = 43,680
2. 17
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a) different ways two errors occur = 28
b) probability =
c)
d) Notice to the log scale
Pr
0.0001
0.001
0.01
0.1
2.79832041994400e
-07
2.78324194404198e
-05
0.002636144418322
80
0.148803480000000
2.18
a)
b)
c) Pr[retrains within the first 5 packet] = 0.049
d) N > 10
2. 19
a)
b)
2.20
F = C ( A+ B)(D + E)
2.21
F = A(D + CE) + B(E + CD) or (A + B(C+E))(D+E(B+C))
2.22
2.23
2.24
a) F = AB+ AC
b) F = A + BC
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C) net has lower probability.
2.25
a) F = A + BCD + E
b)
2.26
a)
b)
c) Pr[communicate] = 0.6875
d) b failed (explain why)
2.27
a) p < 0.001
b) p <0.046
2.28
a) X = (C+D)(A+B+E)
b)
c) P[Y] = 0.9994
2.29
a)
b)
c) Pr[1] = 0.5
d) Pr[0] = 0.5
2.30
a) Not independent
b) = 0.07
2.31
a) Independent
b) = 0.02
2.32
a) Yes
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b)
2.33
2.34
Pr[hardware failure|software failure] = 0.004
2.35
a)
111
111
1
111
110
0
110
1
110
101
0
101
1
101
100
0
100
1
100
011
0
011
1
011
010
0
010
1
010
001
0
001
1
001
000
0
000
1
000
11
1
10
11
0
10
b)
0
c)
d)
2.36
2.37
GB
a)
G
GBF Probability:0.81Success
Probability:0.09Success
Probability:0.02 Success
5
Probability:0.02 Fail
5
Probability:0.02 Fail
5
Probability:0.02 Fail
5
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b) I) 0.075
II) 0.025
c) I) 0
II)
III)
d) Pr[mission succeeded | beer cooler failed] = 0
2.38
a)
b)
2.39
a) Pr[at least 1 IC is good] = 0.975
b) Pr[IC is good] = 0.8125
2NB3
a) Pr[defective component select] = 0.1167
b) Pr[Manufacturer A |defective component] = 0.2857
2.40
a)
b)
c)
2.41
a)
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b)
c)
d)
2.42
a)
b)
c)
2.43
a)
S
0
0
1
1
R Probability
0
1
0
1
b)
c)
d)
2.44
a)
b)
c) I)
II)
III)
IV)
2.45
a)
b)
c) I)
II)
III)
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IV)
2.46
a)
b)
c)
d)
2.47
Pr[under 3 error] = 0.9777
2.48
a)
1
0.
10.
9
0.95
0.0
5 1
b)
c)
2.49
a)
b) for received, choose A, for received, choose B
c) for received, choose A, for received, choose A,
2.50
a)
b) H = 3.3661
2.51
a)
b) Pr[error] = 0.255
c) No (explain why)
2.52
a) Pr[A] = Pr[T] = , Pr[C] = Pr[I] = , Pr[E] = Pr[N] = , Pr[M] = Pr[SPACE] =
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b) H = 2.8050
c)
LETTE
R
M
SPACE
A
T
E
N
C
I
d)
CODE
10
11
001
010
011
0000
0001
0
0001
1
2.875
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3.1
a)
10
9
8
7
6
5
4
3
2
b)
1
1
0
0
0.9
1
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
3.2
0
a)
0
1
9
8
7
6
5
4
3
2
1
0
0.5
-1
0
1
2
3
4
b)
5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
0
1
2
3
4
5
3.3
a) k -> 3, 0, 2, 3, 2, 1, 3, 1, 3, 0, 0 ,3, 3, 2, 3, 0, 2, 2, 0, 0
(b) i
(b) k
8
8
6
6
4
4
2
2
0
-1
0
1
i
(c) i
2
3
4
0
-1
0
1
k
(c) k
2
3
4
-1
0
1
k
2
3
4
1
0.4
0.3
0.5
0.2
0.1
0
-1
0
1
i
2
3
4
0
3.4
a)
b) Pr[I = 0] = Pr[I = 8] =
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70
60
50
40
30
20
10
0
-1
0
1
2
3
1
2
3
4
5
6
7
8
9
3.5
a)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
0
4
b)
3.6
a)
0.2
0.18
0.16
0.14
0.12
0.1
0.08
b)
0.06
0.04
1
0.02
0.9
0.80-1
0
1
2
3
4
5
6
7
8
9
10
11
0.7
0.6
0.5
0.4
0.3
0.2
0.1
c)
0
-1
0
1
2
3
4
5
0.4
X=1
Y = 0.387
0.35
0.3
X =0
Y = 0.349
0.25
X=2
Y = 0.194
0.2
0.15
0.1
d)
X= 3
Y = 0.0574
0.05
X=4
Y = 0.0112
0.12
0
0
1
2
0
1
5
X3= 5 4
Y = 0.00149
6
7
4
5
8
9
10
0.1
0.08
0.06
0.04
0.02
0
2
3
6
7
8
9
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3.7
Pr[less than 5 errors] = 0.000968
3.8
3.9
a)
b)
3.10
a)
geometric
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
5
6
7
8
9
5
6
7
8
9
uniform
0.2
0.1
0
0
1
2
3
4
Poisson
0.4
0.2
0
0
1
2
3
4
b)
type
Geometr
ic
Uniform
Poisson
c)
Probability
type
Geometr
ic
Uniform
Poisson
Probability
N = 12
No solution
3.11
a)
b) l =2
3.12
a) C =
b)
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c) choose 1.5918
3. 13
a) C = 4
b)
3.14
a)
b)
I)
II)
III)
c)
d) b) 와 같음
3.15
a)
b) I)
II)
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3.16
a) C =
b)
c)
for
for
for
3.17
a) C =
b)
c) M = 5
3.18
증명 문제 생략
3.19
증명 문제 생략
3.20
NOT valid because
3.21
a) It is invalid because by forcing B=A
b) It is valid. Subject is A =
c) It is valid, if
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3.22
a)
A=5
b)
3.23
a)
b)
3.24
a) c =
b)
3.25
a)
b)
c)
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d) I)
II)
III)
3.26
a)
b)
3.27
a) C =
b)
c)
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d)
3.28
a)
b)
3.29
a) C
=1
b) I)
II)
c) I)
II)
Largest probability occurs for
d)
3. 30
a)
I)
III)
II)
IV)
b)
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c)
v) No, if is negative ,
3.31
a)
b)
c) the same. The probabilities is not depends on the value of
3.32
a) M =
b) M =
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c) M = m
3.33
a)
b)
c)
3.34
a) I)
II)
b) I)
c) I)
II)
II)
3.35
a)
b)
c)
3.36
a)
0~1
1~2
2~3
3~4
4~5
5~6
6~7
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7~8
8~
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3.37
a)C =
b)
c)
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3.38
a)
b)
c) I)
II)
3.39
a)
b)
3.40
a) C = 5.5
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b)
c)
3.41
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3.42
a)
b)
3.43
3.44
a)
b)
3.45
x
Y
-x –c
0
x-c
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3.46
3. 47
a)
b)
3.48
a)
Linear increase from 0 to 10
Y = 10
b)
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3.49
a)
b)
3.50
a)
b)
3.51
3.52
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a)
b)
3.53
a)
b)
3.54
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a)
b)
3.55
a)
b)
c)
3.56
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3.57
a)
b)
3.58
a)
b)
3.59
a)
b)
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3.60
a)
b)
c)
d) s = 20
3.61
a)
b)
c)
3.62
증명문제 생략
3.63
a)
b)
3.64
a)
b)
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3.65
a)
b)
3.66
a)
b)
c)
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4.1
a)
b)
c)
d)
4.2
4.3
4.4
4.5
a)
b)
4.6
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a)
Ed
you
1
2
3
4
1
2
3
4
2
3
4
5
3
4
5
6
4
5
6
7
5
6
7
8
b)
c)
4.7
a)
b)
c)
d)
4.8
a)
b)
4.9
4.10
a)
b)
4.11
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a)
b) I)
II)
c) claudia say yes to rolf
claudia say no to rolf
cloudia will say ‘no’ to Rolf
d) Rolf is wrong in his thinking
e) after the full day of the week we have
Rolf should ask cloudia as early as possible on the day.
4.12
a) c = 4
b)
4.13
a) A = 1.0524
b)
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4.14
4.15
4.16
a)
b)
c)
4.17
Value
frequenc
y
1
5
6 9
6 4
a)
b)
4.18
a)
Value
frequency
2
5
5
2
7
7
b)
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8
2
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4.19
4.20
a)
b)
c) is maximized for . Max variance is 1
4.21
a)
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b)
4.22
4.23
a)
b)
c) I)
d)
4.24
a)
b)
c)
4.25
a)
b)
4.26
a)
b)
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4.27
4.28
4.29
4.30
a) A
b)
c)
d)
4.31
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a) c = 0.3012
b)
c) E[Y] = 3.4338
4.32
a)
b)
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c)
약간의 오차가 있습니다.
d)
4.33
a)
b)
c)
4.34
a) A =
b)
4.35
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4.36
4.37
4.38
a)
b) E[X|1 <= X <= 3.5 ] = 2.25
4.39
a)
b) var
c) yes
4.40
a)
b)
c)
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4.41
a)
b)
4.42
a)
b)
c)
4.43
a)
b)
4.44
4.45
in particular, skewness would be non-zero
4.46
4.47
a)
b)
c)
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4.48
a)
b)
c)
4.49
4.50
a)
b) E[X] = 0
c)
d)
e)
4.51
4.52
4.53
4.54
a)
b)
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4.55
a)
b)
4.56
b)
letter
A
E
H
K
N
S
Y
pro
Codeword
01
0
0000
101
10
000
1011
c)
d)
4.57
4.58
a) both game hae equal entropy.
b) the second game would have greater entropy.
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Length
2
1
4
3
2
3
4
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5.1
a) 0.1357
b)
b)
5.2
a) c= 0.908
b)
5.3
a)
b)
d)
5.4
a) = C =
b)
Not independent
c)
5.5
5.6
증명 생략
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5.7
a)
b) i)
ii)
iii)
5.8
a)
b)
5.9
a)
b)
c) not independent
5.10
a) sketch!!!!!
b) C = 1
c)
d)
e) No
5.11
a)
b) i)
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ii)
c
5.12
a) c = 2
b)
5.13
a)
b)
c)
d) yes
5.14
a)
b)
5.15
a) C =
b)
c)
5.16
5.17
a)
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b) they are independent
5.18
a)
b)
c)
5.19
a)
b)
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c)
d)
e)
5.20
a)
b)
c) thet are not independent
d)
5.21
a)
b)
c)
5.22
a)
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b)
c)
5.23
a)
b)
c)
d)
e)
5.24
증명문제 생략
5.25
a)
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b)
c)
d)
e)
f)
g)
h)
i)
j)
5.26
a)
b)
5.27
a)
b)
c)
d)
5.28
In
5.29
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a)
b)
c)
5.30
a) 증명문제 생략
b) gaussian form
5.31
a) J(y) = 1
b)
c) if we choose
5.32
5.33
a)
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b)
c)
5.34
증명 생략
5.35
증명 생략
5.36
5.37
a)
b)
c)
d)
5.38
a)
b)
c)
5.39
a)
b)
c)
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d)
e)
f) NOT independent
5.40
a) 증명생략
b)
c)
d)
5.41
증명 생략
5.42
a)
b)
c)
5.43
a)
b)
c)
5.44
a)
b)
c)
5.45
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a)
b)
5.46
a) E[W] = 10.5 S
b)
5.47
a)
b)
c)
5.48
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