Solution to EE351K HW assignment 2
1
．
ABC
AB
C
C
ABC
C
(0.04)
AB
C
C
C
(0.16)
A
C
BC
(0.05)
A
C
B
C
C
(0.05)
A
C
BC
C
(0)
A
C
B
C
C
C
(0.2)
Since
(
)
0.1
C
P A
C
∩
=
, we know that
(
)
0.1
(
)
0.05
C
C
C
P A B C
P A BC
=

=
.
Since P(A)=0.7, we know
(
)
0.3
C
P A
=
, hence
(
)
0.3
0.05
0.05
0.2
0
C
C
P A BC
=



=
Since P(C)=0.6, we know
(
)
0.4
C
P C
=
, hence
(
)
0.4
0.04
0
0.2
0.16
C
C
P AB C
=



=
We know then P(ABC)+P(AB
C
C)=0.5=P(AC)
so
(
)
(
)
(
)
(
)
0.7
0.6
0.5
0.8
P A
C
P A
P C
P A
C
∪
=
+

∩
=
+

=
a.
P(A
C
BC
C
) = 0
b.
We can number the 8 cells from left to right, from top to bottom as 1 to 8, then we know that
C
A
C
∪
are the boxes 1,2,5,6,7,8, and
A
C
∪
are the boxes 1,2,3,4,5,6; hence the
conditional probability is
0.5
0.05
0.05
0.75
0.5
0.05
0.05
0
0.2
+
+
=
+
+
+
+
c.
The probability is 0 since
C
A
C
∩
has no common set with
A
C
∩
.
d.
Since
C
B
A
∩
and
A
C
∩
certainly has no intersection, we can easily use the rule
(
)
(
)
( )
(
)
(
)
( )
P X
Y
P X
P Y
P X
Y
P X
P Y
∪
=
+

∩
=
+
to get P(A
C
B)+P(AC)=0.05+0+0.5=0.55
e.
(
)
1
(
)
c
c
c
P A
B
C
P A
B
C
∪
∪
=

∩
∩
=0.8
2.
a. The maximum value of
(
)
P A
B
∪
is 1, when A doesn't overlap with B
b. The minimum value of
(
)
P A
B
∪
is 0.6, when A is completely covered by B
c. The maximum value of
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 BARD
 Conditional Probability, Probability, UCI race classifications

Click to edit the document details