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1. Let
E
be the event that at least one die lands on 6, and let
F
denote the event that the two
dice land on diﬀerent numbers. Then
E
=
{
(6
, j
) :
j
= 1
,
2
,
3
,
4
,
5
,
6
} ∪ {
(
i,
6) :
i
= 1
,
2
,
3
,
4
,
5
,
6
}
,
F
=
{
(
i, j
) :
i
6
=
j, i, j
= 1
,
2
,
3
,
4
,
5
,
6
}
,
and
EF
=
{
(6
, i
)
,
(
i,
6) :
i
= 1
,
2
,
3
,
4
,
5
}
.
It is easy to see that
F
has 30 outcomes and
EF
has 10 outcomes. Therefore
P
(
E

F
) =
10
30
=
1
3
.
6. Let
W
denote the event that exactly 3 white balls are drawn, and let
E
denote the evemnt
that the ﬁrst and third balls drawn are white.
(a) We ﬁrst consider the nonreplacement case. The number of outcomes in
W
is
±
4
3
²
×
8
×
7
×
6
×
4. The number of outcomes in
EW
is 8
×
7
× ×
±
2
1
²
×
6
×
4. Therefore the
desired conditional probability is
P
(
E

W
) =
8
×
7
× ×
±
2
1
²
×
6
×
4
±
4
3
²
×
8
×
7
×
6
×
4
=
±
2
1
²
±
4
3
²
=
1
2
.
(b) We next consider the replacement case. The number of outcomes in
W
is
±
4
3
²
×
8
×
8
×
8
×
4 and the number of outcomes in
EW
is 8
×
8
×
±
2
1
²
×
8
×
4. Thus the desired
probability is
P
(
E

W
) =
8
×
8
×
±
2
1
²
×
8
×
4
±
4
3
²
×
8
×
8
×
8
×
4
=
1
2
.
11. (a) The number of outcomes in
A
s
is 51 and the number of outcomes in
BA
s
is 3. Thus
P
(
B

A
s
) =
3
51
=
1
17
.
(b) The number of outcomes in
A
is
±
4
2
²
+
(
4
)
×
±
48
a
²
= 6 + 192 = 198, and the
number of outcomes in
BA
is
±
4
2
²
= 6. Thus
P
(
B

A
) =
6
198
=
1
33
.
14. Let
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This homework help was uploaded on 04/13/2008 for the course M 362k taught by Professor Berg during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Berg

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