EE 464
Mitra, Spring 2005
Midterm Solutions
1.
Problem 1
(19 points total) Consider a probability space
P
= (Ω
,
F
, P
)
and the events
A, B
∈
Ω
.
For each statement below (except the last), either give a proof that the statement is true, or provide
a specific example of a probability space and events for which the statement is false. You can also
provide a proof of the statement being false as well.
For the first three parts, two points for a correct assessment of the statement (true or false) and
three points for a justification. Note, many did not indicate whether the statement was true or false
and with incorrect answers it was not always clear what was being proved or disproved.
(a) (5 points) If
P
[
A
] =
P
[
B
]
then
A
=
B
FALSE
Consider tossing a fair coin:
A
=
HEADS
, B
=
TAILS.
P
[
A
] =
P
[
B
] =
1
2
but clearly
A
negationslash
=
B
.
(b) (5 points) If
A
and
B
are independent events, they cannot be disjoint.
FALSE
If
A
and
B
are independent then,
P
[
AB
] =
P
[
A
]
P
[
B
]
and if
A
and
B
are disjoint then
AB
= Φ
.
Thus
A
and
B
can be both independent and disjoint if either
P
[
A
] = 0
or
P
[
B
] = 0
, or both
probabilities are equal to zero.
Partial credit given for the correct definitions of independence and disjoint
.
(c) (5 points) If
P
[
A
]
≥
P
[
B
]
then for another arbitrary set
C
,
P
[
A

C
]
≥
P
[
B

C
]
FALSE
Let
AC
= Φ
, but
BC
negationslash
= Φ
; furthermore let
P
[
BC
]
>
0
. Then if
P
[
A
]
≥
P
[
B
]
we also have:
P
[
AC
] =
P
[Φ] = 0
and
P
[
BC
]
>
0
; clearly then we cannot have
P
[
A

C
]
≥
P
[
B

C
]
since
P
[
A

C
] = 0
and
P
[
B

C
]
>
0
.
Partial credit given for the correct definition of conditional probability.
(d) (4 points) Given that
A
⊂
B
and
P
[
A
] =
1
4
and
P
[
B
] =
1
3
, compute the following:
P
[
A

B
]
and
P
[
B

A
]
.
P
[
A

B
]
=
P
[
AB
]
P
[
B
]
=
P
[
A
]
P
[
B
]
=
3
4
P
[
B

A
]
=
P
[
AB
]
P
[
A
]
=
P
[
A
]
P
[
A
]
= 1
2.
Problem 2
(30 points total)
Note that this problem is a generalization of the binary channel example done in lecture; now we
have a
ternary
channel.
(a) (6 points) Given the conditions stated, and letting
T
be the variable for transmission and let
R
be the variable for reception, the table of interest is determined as:
P
[
T
= 0] = 1
−
α
−
β
P
[
R
= 0

T
= 0] = 1
−
β
−
p
P
[
R
= 0

T
=
G
] =
1
3
P
[
R
= 0

T
= 1] = 2
p
P
[
T
=
G
] =
β
P
[
R
=
G

T
= 0] =
β
P
[
R
=
G

T
=
G
] =
1
3
P
[
R
=
G

T
= 1] =
β
P
[
T
= 1] =
α
P
[
R
= 1

T
= 0] =
p
P
[
R
= 1

T
=
G
] =
1
3
P
[
R
= 1

T
= 1] = 1
−
β
−
2
p
Note that the columns all need to sum to 1.
Also note that if you filled out the table
incorrectly, the rest of the problem was graded based on
your
table and not the true table.
So, if you incorrectly used your own table’s values, points were deducted.
1
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(b) (7 points) What is the probability of the received signal being a garbage symbol,
P
[
R
=
G
]
?
Can you minimize your probability of a received garbage symbol through choice of
α
?
We use total probability:
P
[
R
=
G
]
=
P
[
R
=
G

T
= 0]
P
[
T
= 0] +
P
[
R
=
G

T
= 1]
P
[
T
= 1] +
P
[
R
=
G

T
=
G
]
P
[
T
=
G
]
=
β
(1
−
α
−
β
) +
1
3
β
+
βα
=
4
3
β
−
β
2
As
P
[
R
=
G
]
negationslash
=
g
(
α
)
, we cannot affect
P
[
R
=
G
]
through our choice of
α
.
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 Spring '06
 Caire
 Conditional Probability, Probability, Probability theory, Bobcats, partial credit

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