
k
+1
i
=1
A
i
)
≤
N

k
+1
X
i
=1
P
(
A
i
) = (
N

k
+ 1)
P
(
A
1
)
,
where the last equality follows from the fact that the
N

k
+ 1
events
A
i
have
the same probability and this in turn follows from the assumption that the
N
2
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View Full Documentbytes are equally likely to take any value in
{
0
,...,
255
}
. Now
P
(
A
i
) =
(
1
256
)
k
,
consequently
P
(
∪
N

k
+1
i
=1
A
i
)
≤
N

k
+1
X
i
=1
P
(
A
i
) =
N

k
+ 1
256
k
.
For
k
= 4
and
N
= 1024
, the probability of having
k
consecutive bytes some
where on the hard drive is upper bounded by
2
.
38
×
10

7
.
For
k
= 4
and
N
= 1024
2
, the probability of having
k
consecutive bytes some
where on the hard drive is upper bounded by
2
.
44
×
10

4
.
6.
Uncountable
Ω
and the Third Axiom
The Harvard student’s mistake lies in the equality
P
(
∪
x
∈
[0
,
1]
{
x
}
) =
Z
1
0
P
(
{
x
}
)
dx.
For this, the Harvard student seems to have the second axiom in mind (using
the book’s numbering), as s/he is expressing the probability of a disjoint union
as a “sum” (actually, an integral) of the probabilities of the constituent events.
But this is incorrect because the second axiom of probability holds only for a
countable
union of events: if
A
1
,A
2
,...
are disjoint events then
P
(
∪
∞
i
=1
A
i
) =
∑
∞
i
=1
P
(
A
i
)
. In this case the set
{
x
:
x
∈
[0
,
1]
}
is uncountable and as such the
second axiom does not apply here.
7.
Manufacturing Defects
Consider the square
S
whose center is the center of the transistor and with side
length
a
+
b
. If the center of the dust particle falls anywhere inside
S
then it will
touch the transistor and as a result the transistor will stop functioning. Thus we
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 Fall '05
 HAAS
 Probability, Probability theory, randomly chosen person

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