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Problem Solutions – Chapter 2
Problem 2.2.1 Solution
(a) We wish to find the value of
c
that makes the PMF sum up to one.
P
N
(
n
)
=
±
c
(
1
/
2
)
n
n
=
0
,
1
,
2
0
otherwise
(1)
Therefore,
∑
2
n
=
0
P
N
(
n
)
=
c
+
c
/
2
+
c
/
4
=
1, implying
c
=
4
/
7.
(b) The probability that
N
≤
1is
P
[
N
≤
1]
=
P
[
N
=
0]
+
P
[
N
=
1]
=
4
/
7
+
2
/
7
=
6
/
7
(2)
Problem 2.2.2 Solution
From Example 2.5, we can write the PMF of
X
and the PMF of
R
as
P
X
(
x
)
=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
1
/
8
x
=
0
3
/
8
x
=
1
3
/
8
x
=
2
1
/
8
x
=
3
0
otherwise
P
R
(
r
)
=
⎧
⎨
⎩
1
/
4
r
=
0
3
/
4
r
=
2
0
otherwise
(1)
From the PMFs
P
X
(
x
)
and
P
R
(
r
)
, we can calculate the requested probabilities
(a)
P
[
X
=
0
]=
P
X
(
0
)
=
1
/
8.
(b)
P
[
X
<
3
P
X
(
0
)
+
P
X
(
1
)
+
P
X
(
2
)
=
7
/
8.
(c)
P
[
R
>
1
P
R
(
2
)
=
3
/
4.
Problem 2.2.3 Solution
(a) We must choose
c
to make the PMF of
V
sum to one.
4
X
v
=
1
P
V
(v)
=
c
(
1
2
+
2
2
+
3
2
+
4
2
)
=
30
c
=
1
(1)
Hence
c
=
1
/
30.
(b) Let
U
={
u
2

u
=
1
,
2
,...
}
so that
P
[
V
∈
U
]
=
P
V
(
1
)
+
P
V
(
4
)
=
1
30
+
4
2
30
=
17
30
(2)
37
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View Full Document (c) The probability that
V
is even is
P
[
V
is even]
=
P
V
(
2
)
+
P
V
(
4
)
=
2
2
30
+
4
2
30
=
2
3
(3)
(d) The probability that
V
>
2is
P
[
V
>
2]
=
P
V
(
3
)
+
P
V
(
4
)
=
3
2
30
+
4
2
30
=
5
6
(4)
Problem 2.2.4 Solution
(a) We choose
c
so that the PMF sums to one.
X
x
P
X
(
x
)
=
c
2
+
c
4
+
c
8
=
7
c
8
=
1
(1)
Thus
c
=
8
/
7.
(b)
P
[
X
=
4]
=
P
X
(
4
)
=
8
7
·
4
=
2
7
(2)
(c)
P
[
X
<
4]
=
P
X
(
2
)
=
8
7
·
2
=
4
7
(3)
(d)
P
[3
≤
X
≤
9]
=
P
X
(
4
)
+
P
X
(
8
)
=
8
7
·
4
+
8
7
·
8
=
3
7
(4)
Problem 2.2.5 Solution
Using
B
(for Bad) to denote a miss and
G
(for Good) to denote a successful free throw, the sample
tree for the number of points scored in the 1 and 1 is
±
±
±
±
±
±
B
1
−
p
G
p
±
±
±
±
±
±
B
1
−
p
G
p
•
Y
=
0
•
Y
=
1
•
Y
=
2
From the tree, the PMF of
Y
is
P
Y
(
y
)
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
1
−
py
=
0
p
(
1
−
p
)
y
=
1
p
2
y
=
2
0
otherwise
(1)
38
Problem 2.2.6 Solution
The probability that a caller fails to get through in three tries is
(
1
−
p
)
3
. To be sure that at least
95% of all callers get through, we need
(
1
−
p
)
3
≤
0
.
05. This implies
p
=
0
.
6316.
Problem 2.2.7 Solution
In Problem 2.2.6, each caller is willing to make 3 attempts to get through. An attempt is a failure
if all
n
operators are busy, which occurs with probability
q
=
(
0
.
8
)
n
. Assuming call attempts are
independent, a caller will suffer three failed attempts with probability
q
3
=
(
0
.
8
)
3
n
. The problem
statement requires that
(
0
.
8
)
3
n
≤
0
.
05. This implies
n
≥
4
.
48 and so we need 5 operators.
Problem 2.2.8 Solution
From the problem statement, a single is twice as likely as a double, which is twice as likely as a
triple, which is twice as likely as a homerun. If
p
is the probability of a home run, then
P
B
(
4
)
=
pP
B
(
3
)
=
2
B
(
2
)
=
4
B
(
1
)
=
8
p
(1)
Since a hit of any kind occurs with probability of .300,
p
+
2
p
+
4
p
+
8
p
=
0
.
300 which implies
p
=
0
.
02. Hence, the PMF of
B
is
P
B
(
b
)
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
0
.
70
b
=
0
0
.
16
b
=
1
0
.
08
b
=
2
0
.
04
b
=
3
0
.
02
b
=
4
0
otherwise
(2)
Problem 2.2.9 Solution
(a) In the setup of a mobile call, the phone will send the “SETUP” message up to six times.
Each time the setup message is sent, we have a Bernoulli trial with success probability
p
.Of
course, the phone stops trying as soon as there is a success. Using
r
to denote a successful
response, and
n
a nonresponse, the sample tree is
±
±
±
±
±
r
p
n
1
−
p
±
±
±
±
±
r
p
n
1
−
p
±
±
±
±
±
r
p
n
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This note was uploaded on 12/12/2011 for the course EE EE980 taught by Professor Ji during the Spring '11 term at Akademia Rolnicza w Wrocławiu.
 Spring '11
 Ji

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