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Solutions to Homework 02
1.
We are interested in the shaded region shown in Figure 1. The shaded are on the left
corresponds to
A
∩
B
C
while that on the right is
A
C
∩
B
.
Also, the unshaded area
common to both
A
and
B
is
A
∩
B
. Now, note that
A
= (
A
∩
B
C
)
∪
(
A
∩
B
)
⇒
P
[
A
] =
P
[
A
∩
B
C
] +
P
[
A
∩
B
]
⇒
P
[
A
∩
B
C
] =
P
[
A
]

P
[
A
∩
B
]
similarly
P
[
A
C
∩
B
] =
P
[
B
]

P
[
A
∩
B
]
Then,
P
[(
A
∩
B
C
)
∪
(
A
C
∩
B
)]
=
P
[
A
∩
B
C
] +
P
[
A
C
∩
B
]
=
P
[
A
]

P
[
A
∩
B
] +
P
[
B
]

P
[
A
∩
B
]
=
P
[
A
] +
P
[
B
]

2
P
[
A
∩
B
]
A
B
Figure 1: Venn diagram for question 1.
2.
Refer to Figure 2(a) for the region
A
C
∪
B
C
.
P
[
A
C
∪
B
C
]
=
1

P
[
A
∩
B
]
=
1

z
1
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View Full Document Refer to Figure 2(b) for the region
A
∩
B
C
.
P
[
A
∩
B
C
]
=
P
[
A
]

P
[
A
∩
B
]
=
x

z
Refer to Figure 2(c) for the region
A
C
∪
B
.
P
[
A
C
∪
B
]
=
1

P
[
A
∩
B
C
]
=
1

(
x

z
) = 1

x
+
z
Refer to Figure 2(d) for the region
A
C
∪
B
C
.
P
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This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 GELFAND

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