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Probability
1.
Consider two events
A
and
B
such that
(
29
0.4
P A
=
and
(
29
0.7
P B
=
. Determine the maximum and
minimum possible values of
(
29
P A
B
∩
and the conditions under which each of the value is attained.
Solution:
)
(
)
(
)
(
)
(
B
A
P
B
P
A
P
B
A
P
∪

+
=
∩
.
7
.
0
)}
(
),
(
{
max
)
(
and
)
(
),
(
)
(
,
,
=
≥
∪
≥
∪
∴
∪
⊂
B
P
A
P
B
A
P
B
P
A
P
B
A
P
B
A
B
A
B
In addition,
).
(
1
B
A
P
∪
≥
Thus
.
7
.
0
)
(
1
≥
∪
≥
B
A
P
So if
,
1
)
(
=
∪
B
A
P
)
(
B
A
P
∩
will attain the minimum and
1
.
0
1
7
.
0
4
.
0
)
(
=

+
=
∩
B
A
P
If
,
7
.
0
)
(
=
∪
B
A
P
)
(
B
A
P
∩
will attain the maximum and
4
.
0
7
.
0
7
.
0
4
.
0
)
(
=

+
=
∩
B
A
P
2.
Let
, ,
A B C
be events, prove that
(
29
29
(
29
(
(
29
1
c
c
c
c
c
c
P A
B
C
P A B
C
P B C
P C
∪
∪
= 
∩
Proof:
(
29
(
29
(
29
29
(
29
(
(
29
1
1
c
c
c
c
c
c
c
c
c
c
c
c
c
P A
B
C
P
A
B
C
P A
B
C
P A B
C
P B C
P C
∪
∪
=
∩
∩
= 
∩
∩
= 
∩
3.
Let
(
29
P
,
,
Γ
Ω
be a probability space. Let
,
A B
∈Γ
and
(
29
(
29
1
P A
P B
=
=
.
Find
(a)
(
29
c
P A
B
∩
,
(b)
(
29
P A
B
∩
,
(
29
P A
B
∪
,
(c)
(
29
c
P A
B
∪
.
(d)
If
, ,
A B C
∈Γ
are independent, show that
,
A B
C
∩
are independent
Solution:
(a)
(
29
(
29
(
29
1
1
0
c
P B
P B
P B
= ⇒
= 
=
,
(
29
(
29
(
29
0
0
0
c
c
c
c
c
A
B
B
P A
B
P B
P A
B
∩
⊂
⇒
≤
∩
≤
=
⇒
∩
=
.
(b)
(
29
(
29
(
29
1
1
1
A
A
B
P A
P A
B
P A
B
⊂
∪
⇒ =
≤
∪
≤ ⇒
∪
=
.
And
(
29
(
29
(
29
(
29
1 1 1 1
P A
B
P A
P B
P A
B
∩
=
+

∪
= +  =
.
(c)
(
29
(
29
(
29
1
1
1
c
c
c
A
A
B
P A
P A
B
P A
B
⊂
∪
⇒ =
≤
∪
≤ ⇒
∪
=
(d)
(
29
(
29
(
29
(
29
(
29
(
29
(
29
, , are independent
, are independent
A B C
B C
P A
B
C
P A
B
C
P A P B P C
P A P B
C
∩
∩
=
∩
∩
=
=
∩
4.
Suppose events
A
and
1
B
, events
A
and
2
B
are independent , the events
1
B
and
2
B
being disjoint.
Prove that the events
A
and
1
2
B
B
∪
are independent.
Proof:
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
1
2
1
2
1
2
1
2
1
2
1
2
P A
B
B
P
A
B
A
B
P A
B
P A
B
P A P B
P A P B
P A
P B
P B
P A P B
B
∩
∪
=
∩
∪
∩
=
∩
+
∩
=
+
=
+
=
∪
5.
Let events
A
and
B
be independent and
A
B
⊂
. Prove that either
(
29
0
P A
=
or
(
29
1
P B
=
.
Proof:
A
B
A
A
B
⊂
⇒
=
∩
. Thus
(
29
(
29
(
29
(
29
P A
P A
B
P A P B
=
∩
=
, events
A
and
B
are independent. So
(
29
(
29
1
0
P A
P B

=
. It follows that
(
29
(
29
0 or
1
0
P A
P B
=

=
, that is,
(
29
0
P A
=
or
(
29
1
P B
=
.
6.
(a)
If
A
is independent of itself, show that
(
29
P A
is 0 or 1.
(b)
If
(
29
P A
is 0 or 1, show that
A
is independent of all events
B
.
1
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This note was uploaded on 01/11/2011 for the course MA 3160 taught by Professor Cwwoo during the Spring '07 term at City University of Hong Kong.
 Spring '07
 CWWoo
 Probability

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