Solution to Problem 1.31.
(a) Let
A
be the event that a 0 is transmitted. Using
the total probability theorem, the desired probability is
P
(
A
)(1
−
±
0
)+
(
1
−
P
(
A
)
)
(1
−
±
1
)=
p
(1
−
±
0
)+(1
−
p
)(1
−
±
1
)
.
(b) By independence, the probability that the string 1011 is received correctly is
(1
−
±
0
)(1
−
±
1
)
3
.
(c) In order for a 0 to be decoded correctly, the received string must be 000, 001, 010,
or 100. Given that the string transmitted was 000, the probability of receiving 000 is
(1
−
±
0
)
3
, and the probability of each of the strings 001, 010, and 100 is
±
0
(1
−
±
0
)
2
.
Thus, the probability of correct decoding is
3
±
0
(1
−
±
0
)
2
+(1
−
±
0
)
3
.
(d) When the symbol is 0, the probabilities of correct decoding with and without the
scheme of part (c) are 3
±
0
(1
−
±
0
)
2
−
±
0
)
3
and 1
−
±
0
, respectively.
Thus, the
probability is improved with the scheme of part (c) if
3
±
0
(1
−
±
0
)
2
−
±
0
)
3
>
(1
−
±
0
)
,
or
(1
−
±
0
)(1 + 2
±
0
)
>
1
,
which is equivalent to
±
0
<
1
/
2.
(e) Using Bayes’ rule, we have
P
(0

101) =
P
(0)
P
(101

0)
P
(0)
P
(101

0) +
P
(1)
P
(101

1)
.
The probabilities needed in the above formula are
P
(0) =
p,
P
(1) = 1
−
p,
P
(101

0) =
±
2
0
(1
−
±
0
)
,
P
(101

1) =
±
1
(1
−
±
1
)
2
.
Solution to Problem 1.32.
The answer to this problem is not unique and depends
on the assumptions we make on the reproductive strategy of the king’s parents.
Suppose that the king’s parents had decided to have exactly two children and
then stopped. There are four possible and equally likely outcomes, namely BB, GG,
BG, and GB (B stands for “boy” and G stands for “girl”). Given that at least one
child was a boy (the king), the outcome GG is eliminated and we are left with three
equally likely outcomes (BB, BG, and GB). The probability that the sibling is male
(the conditional probability of BB) is 1/3 .
Suppose on the other hand that the king’s parents had decided to have children
until they would have a male child. In that case, the king is the second child, and the
sibling is female, with certainty.
11
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentSolution to Problem 1.33.
Flip the coin twice.
If the outcome is headstails,
choose the opera. if the outcome is tailsheads, choose the movies. Otherwise, repeat
the process, until a decision can be made. Let
A
k
be the event that a decision was
made at the
k
th round. Conditional on the event
A
k
, the two choices are equally likely,
and we have
P
(opera) =
∞
X
k
=1
P
(opera

A
k
)
P
(
A
k
)=
∞
X
k
=1
1
2
P
(
A
k
1
2
.
We have used here the property
∑
∞
k
=0
P
(
A
k
) = 1, which is true as long as
P
(heads)
>
0
and
P
(tails)
>
0.
Solution to Problem 1.34.
The system may be viewed as a series connection of
three subsystems, denoted 1, 2, and 3 in Fig. 1.19 in the text. The probability that the
entire system is operational is
p
1
p
2
p
3
, where
p
i
is the probability that subsystem
i
is
operational. Using the formulas for the probability of success of a series or a parallel
system given in Example 1.24, we have
p
1
=
p,
p
3
=1
−
(1
−
p
)
2
,
and
p
2
−
(1
−
p
)
(
1
−
p
(
1
−
(1
−
p
)
3
))
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 staff
 Math, Conditional Probability, Probability, Probability theory

Click to edit the document details