Solutions to Homework
#
2
1.
a) Let us consider a sequence of
n
≥
4 outcomes of tosses of a coin. The sequence may
start with
T
, in which case there are no three heads in a row if and only if there are no
three heads in a row in the subsequence of the last
n

1 outcomes, or with
HT
, in which
case there are no three heads in a row if and only if there are no three heads in a row in
the subsequence of the last
n

2 outcomes, or with
HHT
, in which case there are no three
heads in a row if and only if there are no three heads in a row in the subsequence of the
last
n

3 outcomes, or with
HHH
, in which case we get three heads in a row. Hence we
get the formula
a
n
=
a
n

1
+
a
n

2
+
a
n

3
. We observe that
a
1
= 2,
a
2
= 4,
a
3
= 7.
Answer:
we have
a
n
=
a
n

1
+
a
n

2
+
a
n

3
, where
a
1
= 2,
a
2
= 4 and
a
3
= 7.
b) Using the formula of Part a), we compute
a
4
= 13,
a
5
= 24,
a
6
= 44,
a
7
= 81,
a
8
= 149,
a
9
= 274, and
a
10
= 504. Hence the probability that there will be three or more
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 Winter '12
 AlexanderBarvinok
 Probability, #, Mr. Pickwick

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