P
(
W
) =
P
(
W

A
)
P
(
A
) +
P
(
W

B
)
P
(
B
) =
p
2
+
q
2
by total probability. Since
P
(
W
∩
A
) =
P
(
W

A
)
P
(
A
) =
p
2
, we want to conditions on
p
and
q
so that
1
2
(
p
2
+
q
2
)
=
p
2
. This results in
p
=
q
.
8. Let
A
=
Top route is working,
B
=
Bottom route is working and
C
=
Bottom left link is working.
Notice that
C
⊂
B
because the bottom route works if and only if both of the bottom links work.
(a) The event that Acahti is connected to the Big City is
A
∪
B
P
(
A
∪
B
) =
P
(
A
)+
P
(
B
)

P
(
A
∩
B
) = (1

p
)
2
+(1

p
)
2

(1

p
)
4
= 2(1

p
)
2

(1

p
)
4
Because
P
(
A
)
is equal to the probability that the two top links are working (
(1

p
)
×
(1

p
)
) and
similarly for
B
and also
P
(
A
∩
B
) =
P
(
A
)
P
(
B
)
because the events
A
and
B
are independent.
(b) The probability that Acahti is connected to the Big City given that the bottom left link is working
is
P
(
A
∪
B

C
) =
P
((
A
∪
B
)
∩
C
)
P
(
C
)
=
P
((
A
∩
C
)
∪
(
B
∩
C
))
1

p
=
P
((
A
∩
C
)
∪
B
)
1

p
.
However
P
((
A
∩
C
)
∪
B
) =
P
(
A
∩
C
) +
P
(
B
)

P
(
A
∩
C
∩
B
) =
P
(
A
)
P
(
C
) +
P
(
B
)

P
(
A
)
P
(
B
) = (1

p
)
3
+ (1

p
)
2

(1

p
)
4
. Consequently, after simpliﬁcations we ﬁnd
P
(
A
∪
B

C
) = 1

2
p
2
+
p
3
.
3
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View Full Document(c) We want to compute
P
(
B

(
A
∪
B
)
∩
C
)
, this is done as follows
P
(
B

(
A
∪
B
)
∩
C
) =
P
(
B
∩
(
A
∪
B
)
∩
C
)
P
((
A
∪
B
)
∩
C
)
=
P
(
B
)
P
((
A
∪
B
)
∩
C
)
,
where the last equality follows from the fact that
B
⊂
C
and
B
⊂
(
A
∪
B
)
. Consequently
P
(
B

(
A
∪
B
)
∩
C
) =
P
(
B
)
P
(
C
)
P
((
A
∪
B
)

C
)
=
(1

p
)
2
(1

p
)(1

2
p
2
+
p
3
)
=
1
1 +
p

p
2
.
(d) We want
P
(
A
∪
B
) = 2(1

p
)
2

(1

p
)
4
>
0
.
99999
. Letting
x
= (1

p
)
2
, the last inequality
becomes
2
x

x
2

0
.
99999
>
0
. So that
x >
0
.
9968
or equivalently
p <
0
.
00158
.
9. Note that for any
k
≥
1
, we have
p
X
(
k
)
p
X
(
k

1)
=
e

λ λ
k
k
!
e

λ
λ
k

1
(
k

1)!
=
λ
k
.
Now if
k
≤
k
*
, then we have
p
X
(
k
)
p
X
(
k

1)
=
λ
k
≥
λ
k
*
≥
λ
λ
= 1
,
so
p
X
(
k
)
is nondecreasing from
0
up to
k
*
. If
k > k
*
, then we must have
k > λ
, so
p
X
(
k
)
p
X
(
k

1)
=
λ
k
<
1
.
Thus
p
X
(
k
)
is strictly decreasing from
k
*
onward.
10.
(a) The hashed region corresponds to the event
{
Z
≤
1
/
3
}
.

6
x
y
1
1
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
±
1
/
3
1
/
3
(b) Note that the probability of any event equals its area. From the picture, we have
F
Z
(
z
) = Pr(
Z
≤
z
) =
z
+ (1

z
)
z
= 2
z

z
2
,
(
z
∈
[0
,
1])
and
F
Z
(
z
) = 0
for
z <
0
, and
F
Z
(
z
) = 1
for
z >
1
.
4
0.5
0
0.5
1
1.5
1
0.5
0
0.5
1
1.5
2
F
Z
(z)
z
(c)
X
takes a continuum of values, so it is not discrete. Its CDF is differentiable, so it is continuous.
Its PDF is
f
Z
(
z
) = 0
for
z
∈
(
∞
,
0)
∪
(1
,
∞
)
, because
F
Z
is constant there. For
z
∈
[0
,
1]
f
Z
(
z
) =
d
dz
F
Z
(
z
) = 2

2
z
(
z
∈
[0
,
1])
.
(d) We have
Pr(
Z >
3
/
4) = 1

Pr(
Z
≤
3
/
4) = 1

F
Z
(3
/
4) = 1
/
16
.
5
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 Fall '05
 HAAS
 Probability theory, Row vector, Quantification, L20, Acahti, 01 row vectors

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