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
Page 1
of 6
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 Prof.DimitriBertsekas
 Electrical Engineering, Poisson Distribution, Probability theory, Massachusetts Institute of Technology, Probabilistic Systems Analysis, Department of Electrical Engineering & Computer Science

Click to edit the document details