CS 70
Discrete Mathematics and Probability Theory
Fall 2010
Tse/Wagner
Soln 10
1. (50 pts.)
Expectations
Solve each of the following problems using linearity of expectation. Clearly explain your methods. (Hint:
for each problem, think about what the appropriate random variables should be and define them explicitly.)
(a) A monkey types at a 26letter keyboard with one key corresponding to each of the lowercase English
letters. Each keystroke is chosen independently and uniformly at random from the 26 possibilities.
If the monkey types 1 million letters, what is the expected number of times the sequence “bonbon"
appears?
Answer:
Define the random variable
X
to be the number of times “bonbon” appears in the 1 million
letters the monkey types. We express
X
as the sum
X
=
10
6

5
∑
i
=
1
X
i
where the random variable
X
i
is
X
i
=
(
1
if characters
i
,...,
i
+
5 are “bonbon”
0
otherwise.
Since each letter is independent, the probability that any given string of 6 letters is “bonbon” is
1
26
6
,
and therefore Pr
[
X
i
=
1
] =
1
26
6
. Thus,
E
(
X
i
) =
1
26
6
and so by linearity of expectation
E
(
X
) =
E
(
10
6

5
∑
i
=
1
X
i
) =
10
6

5
∑
i
=
1
E
(
X
i
) =
10
6

5
∑
i
=
1
1
26
6
=
(
10
6

5
)
1
26
6
.
Comment:
This solution is fine, despite the fact that the
X
i
’s are not mutually independent. Linearity
of expectation can be applied even if the underlying random variables are not independent.
(b) A coin with Heads probability
p
is flipped
n
times. A “run” is a maximal sequence of consecutive flips
that are all the same. (Thus, for example, the sequence HTHHHTTH with
n
=
8 has five runs.) Show
that the expected number of runs is 1
+
2
(
n

1
)
p
(
1

p
)
. Justify your answer carefully.
Answer:
Again, we use linearity of expectation. Define the random variable
X
to be the number of
runs in a sequence, and define the indicator random variable
X
i
as
X
i
=
(
1
if the
i
th flip is the beginning of a run
0
otherwise.
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 Fall '08
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 Probability theory, browns, soln

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