Chapter 12
Solutions to Problems
1.2 Independent vs. mutually exclusive
(a) If
E
is an event independent of itself, then
P
[
E
] =
P
[
E
∩
E
] =
P
[
E
]
P
[
E
]. This can happen if
P
[
E
] = 0
.
If
P
[
E
]
6
= 0 then cancelling a factor of
P
[
E
] on each side yields
P
[
E
] = 1. In summary,
either
P
[
E
] = 0 or
P
[
E
] = 1.
(b) In general, we have
P
[
A
∪
B
] =
P
[
A
] +
P
[
B
]

P
[
AB
]. If the events
A
and
B
are independent,
then
P
[
A
∪
B
] =
P
[
A
] +
P
[
B
]

P
[
A
]
P
[
B
] = 0
.
3 + 0
.
4

(0
.
3)(0
.
4) = 0
.
58
.
On the other hand, if the
events
A
and
B
are mutually exclusive, then
P
[
AB
] = 0 and therefore
P
]
A
∪
B
] = 0
.
3 + 0
.
4 = 0
.
7.
(c) If
P
[
A
] = 0
.
6 and
P
[
B
] = 0
.
8, then the two events could be independent. However, if
A
and
B
were mutually exclusive, then
P
[
A
] +
P
[
B
] =
P
[
A
∪
B
]
≤
1, so it would not possible for
A
and
B
to be mutually exclusive if
P
[
A
] = 0
.
6 and
P
[
B
] = 0
.
8.
1.4 Frantic search
Let
D
,
T
,
B
, and
O
denote the events that the glasses are in the drawer, on the table, in the brief
case, or in the office, respectively. These four events partition the probability space.
(a) Let
E
denote the event that the glasses were not found in the first drawer search.
P
[
T

E
] =
P
[
TE
]
P
[
E
]
=
P
[
E

T
]
P
[
T
]
P
[
E

D
]
P
[
D
]+
P
[
E

D
c
]
P
[
D
c
]
=
(1)(0
.
06)
(0
.
1)(0
.
9)+(1)(0
.
1)
=
0
.
06
0
.
19
≈
0
.
315
(b) Let
F
denote the event that the glasses were not found after the first drawer search and first
table search.
P
[
B

F
] =
P
[
BF
]
P
[
F
]
=
P
[
F

B
]
P
[
B
]
P
[
F

D
]
P
[
D
]+
P
[
F

T
]
P
[
T
]+
P
[
F

B
]
P
[
B
]+
P
[
F

O
]
P
[
O
]
=
(1)(0
.
03)
(0
.
1)(0
.
9)+(0
.
1)(0
.
06)+(1)(0
.
03)+(1)(0
.
01)
≈
0
.
22
(c) Let
G
denote the event that the glasses were not found after the two drawer searches, two table
searches, and one briefcase search.
P
[
O

G
] =
P
[
OG
]
P
[
G
]
=
P
[
G

O
]
P
[
O
]
P
[
G

D
]
P
[
D
]+
P
[
G

T
]
P
[
T
]+
P
[
G

B
]
P
[
B
]+
P
[
G

O
]
P
[
O
]
=
(1)(0
.
01)
(0
.
1)
2
(0
.
9)+(0
.
1)
2
(0
.
06)+(0
.
1)(0
.
03)+(1)(0
.
01)
≈
0
.
4225
1.6 Conditional probabilities–basic computations of iterative decoding
(a) Here is one of several approaches to this problem. Note that the
n
pairs (
B
1
, Y
1
)
, . . . ,
(
B
n
, Y
n
)
345
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CHAPTER 12.
SOLUTIONS TO PROBLEMS
are mutually independent, and
λ
i
(
b
i
)
def
=
P
[
B
i
=
b
i

Y
i
=
y
i
] =
q
i
(
y
i

b
i
)
q
i
(
y
i

0)+
q
i
(
y
i

1)
.
Therefore
P
[
B
= 1

Y
1
=
y
1
, . . . , Y
n
=
y
n
]
=
X
b
1
,...,b
n
:
b
1
⊕···⊕
b
n
=1
P
[
B
1
=
b
1
, . . . , B
n
=
b
n

Y
1
=
y
1
, . . . , Y
n
=
y
n
]
=
X
b
1
,...,b
n
:
b
1
⊕···⊕
b
n
=1
n
Y
i
=1
λ
i
(
b
i
)
.
(b) Using the definitions,
P
[
B
= 1

Z
1
=
z
1
, . . . , Z
k
=
z
k
]
=
p
(1
, z
1
, . . . , z
k
)
p
(0
, z
1
, . . . , z
k
) +
p
(1
, z
1
, . . . , z
k
)
=
1
2
Q
k
j
=1
r
j
(1

z
j
)
1
2
Q
k
j
=1
r
j
(0

z
j
) +
1
2
Q
k
j
=1
r
j
(1

z
j
)
=
η
1 +
η
where
η
=
k
Y
j
=1
r
j
(1

z
j
)
r
j
(0

z
j
)
.
1.8 Blue corners
(a) There are 24 ways to color 5 corners so that at least one face has four blue corners (there are 6
choices of the face, and for each face there are four choices for which additional corner to color blue.)
Since there are
(
8
5
)
= 56 ways to select 5 out of 8 corners,
P
[
B

exactly 5 corners colored blue] =
24
/
56 = 3
/
7.
(b) By counting the number of ways that
B
can happen for different numbers of blue corners we
find
P
[
B
] = 6
p
4
(1

p
)
4
+ 24
p
5
(1

p
)
3
+ 24
p
6
(1

p
)
2
+ 8
p
7
(1

p
) +
p
8
.
1.10 Recognizing cumulative distribution functions
(a) Valid (draw a sketch)
P
[
X
2
≤
5] =
P
[
X
≤ 
√
5]+
P
[
X
≥
√
5] =
F
1
(

√
5)+1

F
1
(
√
5) =
e

5
2
.
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 Fall '11
 SerapSavari
 Normal Distribution, Probability theory, CDF, Xn

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