(a) From Total Probability, we have
P
(
L
10
) =
P
(
L
10

A
)
·
P
(
A
) +
P
(
L
10

B
)
·
P
(
B
) =
1
2
·
1
2
+
1
3
·
1
2
=
5
12
.
P
(
L
15
) =
P
(
L
15

A
)
·
P
(
A
) +
P
(
L
15

B
)
·
P
(
B
) =
1
4
·
1
2
+
1
3
·
1
2
=
7
24
.
P
(
L
20
) =
P
(
L
20

A
)
·
P
(
A
) +
P
(
L
20

B
)
·
P
(
B
) =
1
4
·
1
2
+
1
3
·
1
2
=
7
24
.
(b) This corresponds to the event
L
20

((
L
10
∪
L
15
)
c
∩
A
)
, which can be written as
L
20

(
L
20
∩
A
)
.
Thus
P
(
L
20

L
20
∩
A
) =
P
(
L
20
∩
L
20
∩
A
)
P
(
L
20
∩
A
)
=
P
(
L
20
∩
A
)
P
(
L
20
∩
A
)
= 1
.
2
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View Full Document(c) This corresponds to the event
L
20

(
L
c
10
∩
A
)
. Thus
P
(
L
20

L
c
10
∩
A
) =
P
(
L
20
∩
L
c
10
∩
A
)
P
(
L
c
10
∩
A
)
=
P
(
L
20
∩
A
)
P
(
L
c
10
∩
A
)
=
P
(
L
20

A
)
·
P
(
A
)
P
(
L
c
10

A
)
·
P
(
A
)
=
1
/
4
·
1
/
2
(1

1
/
2)
·
1
/
2
=
1
2
.
(d) This corresponds to the event
A

L
c
10
. From part (a),
P
(
L
c
10
) = 1

P
(
L
10
) = 7
/
12
. By Bayes’
Rule,
P
(
A

L
c
10
) =
P
(
L
c
10

A
)
·
P
(
A
)
P
(
L
c
10
)
=
(1

1
/
2)
·
1
/
2
7
/
12
=
3
7
.
(e) This corresponds to the event
L
20

L
c
10
. Thus
P
(
L
20

L
c
10
) =
P
(
L
20
∩
L
c
10
)
P
(
L
c
10
)
=
P
(
L
20
)
P
(
L
c
10
)
=
7
/
24
7
/
12
=
1
2
.
5. Yes. Since
C
is independent from
A
∩
B
we must have
P
(
C
∩
(
A
∩
B
)) =
P
(
C
)
P
(
A
∩
B
)
. However
since
A
and
B
are pairwise independent then
P
(
A
∩
B
) =
P
(
A
)
P
(
B
)
. Hence
P
(
C
∩
(
A
∩
B
)) =
P
(
C
)
P
(
A
)
P
(
B
)
. Finally since
A
,
B
and
C
are pairwise independent, we deduce that
A
,
B
and
C
are independent.
6. Even though in reality play ends as soon as the winning team is determined, for purposes of com
putation we can assume that the teams play all ﬁves games. The probability that the See Knays win
exactly
k
games out of
5
games can be written as
(
5
k
)
(1

p
)
k
p
5

k
. Since the See Knays will win the
series if
k
= 3
,
4
,
or
5
,
P
(
the See Knays win the series
) =
∑
5
k
=3
(
5
k
)
(1

p
)
k
p
5

k
.
7. Let
A
be the event in which the factory A is chosen,
B
the factory B, and
W
be the event in which
the selected jPhone works.
P
(
A
) =
P
(
B
) =
1
2
because we select a factory uniformly at random.
P
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 Fall '05
 HAAS
 Probability theory, Row vector, Quantification, L20, Acahti, 01 row vectors

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