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EE 351K Probability, Statistics, and Random Processes
SPRING 2009
Instructor: Shakkottai/Vishwanath
{
shakkott,sriram
}
@ece.utexas.edu
Homework 1
Problem 1
We are given that
P
(
A
) = 0
.
3
,
P
(
B
c
) = 0
.
45
, and
P
(
A
∪
B
) = 0
.
6
. Determine
P
(
B
)
and
P
(
A
∩
B
)
.
Solution :
We have
P
(
B
) = 1

P
(
B
c
) = 1

0
.
45 = 0
.
55
.
Also, by rearranging the formula
P
(
A
∪
B
) =
P
(
A
) +
P
(
B
)

P
(
A
∩
B
)
,
we obtain
P
(
A
∩
B
) =
P
(
A
) +
P
(
B
)

P
(
A
∪
B
) = 0
.
3 + 0
.
55

0
.
6 = 0
.
25
.
Problem 2
Let
A
and
B
be two sets.
(a) Show that
(
A
c
∩
B
c
)
c
=
A
∪
B
and
(
A
c
∪
B
c
)
c
=
A
∩
B
.
(b) Consider rolling a sixsided die once. Let
A
be the set of outcomes where a
3
or
4
comes up. Let
B
be
the set of outcomes where a prime number comes up. Calculate the sets on both sides of the equalities
in part (a), and verify that the equalities hold.
Solution :
(a) We ﬁrst prove De Morgan’s law. That is to show
(
E
∩
F
)
c
=
E
c
∪
F
c
for any two sets
E
and
F
. Given
ω
∈
E
c
∪
F
c
,
ω
must be either in
E
c
or
F
c
. As
E
∩
F
⊆
E
and
E
∩
F
⊆
F
,
ω
must be
not in
E
∩
F
. That is
ω
∈
(
E
∩
F
)
c
. On the other hand, suppose
ω
∈
(
E
∩
F
)
c
. As
E
∩
F
⊆
E
,
ω
∈
E
c
or
ω
∈
E
\
(
E
∩
F
)
. Further,
E
\
(
E
∩
F
)
⊆
F
c
. Therefore,
ω
∈
E
c
∪
F
c
. Similarly, we can also prove
(
E
∪
F
)
c
=
E
c
∩
F
c
.
To show
(
A
c
∩
B
c
)
c
=
A
∪
B
. Note that by De Morgan’s law
(
A
c
∩
B
c
)
c
= (
A
c
)
c
∪
(
B
c
)
c
=
A
∪
B.
Similarly,
(
A
c
∪
B
c
)
c
= (
A
c
)
c
∩
(
B
c
)
c
=
A
∩
B.
(b) We have
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This note was uploaded on 10/26/2009 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas at Austin.
 Spring '07
 BARD

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