EE 351K Probability and Random Processes
FALL 2010
Instructor: Haris Vikalo
[email protected]
Homework 1 Solutions
Problem 1
We are given that
P
(
A
) = 0
.
55
,
P
(
B
c
) = 0
.
45
, and
P
(
A
∪
B
) = 0
.
25
. 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
.
55 + 0
.
55

0
.
25 = 0
.
85
.
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 an odd number comes
up. Let
B
be the set of outcomes where a
1
or a
2
comes up. Calculate the sets on both sides of the
equalities in part (a), and verify that the equalities hold.
Solution :
(a) Sketching Venn diagram or applying De Morgan’s law directly leads to the desired result.
Let us here prove De Morgan’s law. That is, let us 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
. Therefore,
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
)
. Furthermore,
E
\
(
E
∩
F
)
⊆
F
c
. Therefore,
ω
∈
E
c
∪
F
c
. Similarly, we can also prove
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 Spring '10
 Vikalo
 Conditional Probability, Probability, Probability theory, Equals sign

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