Massachusetts
Institute
of
Technology
Department
of
Electrical
Engineering
&
Computer
Science
6.041/6.431:
Probabilistic
Systems
Analysis
(Fall
2010)
Problem
Set
7:
Solutions
1.
(a)
The event
of
the
i
th success occuring before
the
j
th failure
is equivalent to the
i
th success
occurring
within the first (
i
+
j
−
1) trials (since
the
i
th success must occur
no later
than
the trial right
before the
j
th failure).
This is equivalent to event that
i
or
more
successes
occur in the first (
i
+
j
−
1)
trials (where
we
can have,
at most, (
i
+
j
−
1) successes).
Let
S
i
be the time of
the
i
th success,
F
j
be
the
time
of the
j
th failure,
and
N
k
be
the
number
of successes in the first
k
trials (so
N
k
is a binomial
random
variable
over
k
trials).
So
we
have:
i
+
j
−
1
�
P
(
S
i
< F
j
) =
P
(
N
i
+
j
−
1
≥
i
) =
�
�
i
+
j
−
1
p
k
(1
−
p
)
i
+
j
−
1
−
k
k
k
=
i
(b)
Let
K
be the number
of
successes which occur
before
the
j
th failure,
and
L
be
the
number
of trials to
get
to
the
j
th failure.
L
is simply a
j
th order
Pascal,
with probability of
1
−
p
(since we are now interested in the
failures,
not the
successes.) Plugging into the
formula
for
j
th order
Pascal random
variable,
E
[
L
] =
j
, σ
2
p
j
1
−
p
K
=
(1
−
p
)
2
Since
K
=
L
−
j
,
p
p
E
[
K
] =
K
=
j, σ
2
j
1
−
p
(1
−
p
)
2
(c)
This expression is the same as saying we
need at least 42 trials to get the
17th success.
Therefore, it
can be rephrased as having a maximum
of 16 successes in the
first 41 trials.
Hence
b
= 41,
a
= 16.
2. A
successful call occurs with probability
p
=
3
4
2
3
=
1
2
.
·
(a)
Fred will give away his first
sample
on the
third call
if the
first two calls are
failures
and
the third is a
success. Since the
trials are
independent,
the
probability of this sequence
of
events is simply
1
1
1
1
(1
−
p
)(1
−
p
)
p
=
·
·
=
2
2
2
8
(b)
The event
of
interest
requires failures on the
ninth and tenth trials and a success on
the
eleventh trial. For
a
Bernoulli process,
the
outcomes of these
three
trials are
independent
of the results of
any other
trials
and again our
answer
is
1
1
1
1
(1
−
p
)(1
−
p
)
p
=
·
·
=
2
2
2
8
(c)
We desire the probability that
L
2
,
the
time
to the
second arrival
is equal
to five
trials.
We
know
that
p
L
2
(
ℓ
) is a
Pascal PMF of order
2,
and we
have
�
5
−
1
�
�
1
�
5
1
p
L
2
(5) =
p
2
(1
−
p
)
5
−
2
= 4
·
=
2
−
1
2
8
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Massachusetts
Institute
of
Technology
Department
of
Electrical
Engineering
&
Computer
Science
6.041/6.431:
Probabilistic
Systems
Analysis
(Fall
2010)
(d)
Here we require the conditional probability that the
experimental
value
of
L
2
is equal
to
5,
given that
it
is greater
than 2.
p
L
2
(5)
p
L
2
(5)
P
(
L
2
= 5
|
L
2
>
2)
=
=
P
(
L
2
>
2)
1
−
p
L
2
(2)
� �
5
�
5
−
1
�
p
2
(1
−
p
)
5
−
2
4
·
1
2
1
2
−
1
=
=
=
1
−
�
2
−
1
�
p
2
(1
−
p
)
0
�
1
�
2
6
2
−
1
1
−
2
(e)
The probability that
Fred will complete
at least five
calls before
he
needs a new supply is
equal to
the probability that
the
experimental
value
of
L
2
is greater
than or
equal
to
5.
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- Fall '10
- Prof.DimitriBertsekas
- Electrical Engineering, Poisson Distribution, Probability theory, Massachusetts Institute of Technology, Probabilistic Systems Analysis, Department of Electrical Engineering & Computer Science
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