But
[0
,
1]
 {
1
2
}
is not the whole sample space since we have removed the set
{
1
2
}
.
2. (a) The maximum number of transitions that a binary string of length
n
can have is
n

1
. This would correspond to a
string where every two consecutive bits are different, for instance
01010101
.
2. (b) Only 2 strings have the maximum number of transitions (i.e.,
n

1
). If we ﬁx the ﬁrst bit, then the rest of the string
is ﬁxed since every two consecutive bits must be different. Now since the ﬁrst bit can take only two values (
0
or
1
),
we deduce that the number of strings having the maximum number of transitions is 2. For example if
n
= 4
, the
maximum number of transitions is
3
and the two strings having
3
transitions are
0101
or
1010
.
1
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View Full Document2. (c) The number of binary strings with length
n
with exactly
k
(
k
≤
n

1
) transitions is
2
(
n

1
k
)
. Indeed think of the
transitions as separators to be placed between the bits, since there are
n
bits there are
n

1
possible spaces between
these bits, hence there are
(
n

1
k
)
possible conﬁgurations for placing the
k
transitions. Once the positions of the
transitions have been ﬁxed, the binary string will become completely speciﬁed by choosing the ﬁrst bit. Since there
are two possible choices for the ﬁrst bit (0 or 1), we deduce that the total number of binary strings with length
n
with
exactly
k
must be
2
(
n

1
k
)
.
2. (d) The number of binary strings with length
n
that begin with
1
and have exactly
k
transitions is
(
n

1
k
)
. Indeed, note that
by the symmetry of the problem, the number of binary strings with length
n
with exactly
k
transitions that begin with
0 (say
n
0
(
k
)
) is equal to the number of binary strings with length
n
with exactly
k
transitions that begin with 1 (say
n
1
(
k
)
), i.e.,
n
1
(
k
) =
n
0
(
k
)
. Because any string must either start with a 0 or with a 1, we deduce from part (c) that
2
(
n

1
k
)
=
n
0
(
k
) +
n
1
(
k
)
. In other words
n
1
(
k
) =
(
n

1
k
)
.
2. (e) The total number of binary strings of length
n
is
2
n
. Indeed for the ﬁrst bit we have two choices (0 or 1). Once the
ﬁrst bit is ﬁxed we have two possible choices for the second bit (0 or 1), so that at each stage there are two choices,
hence from the counting principle, the total number of strings will be
2
×
...
×
2

{z
}
n
times
= 2
n
.
3. (a) For
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 Fall '05
 HAAS
 Probability theory, CDF, ﬁrst bit

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