Probability and Stochastic Processes
Homework 3 Solutions
12th Feb, Spring 2010
Problem Solutions
: Yates and Goodman,
2.2.1 2.2.3 2.2.6 2.2.8 2.3.1 2.3.3 2.3.4 2.3.7
2.3.10 2.3.11 2.3.13 2.4.1 2.4.3 and 2.4.5
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
) =
braceleftbigg
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
≤
1 is
P
[
N
≤
1] =
P
[
N
= 0] +
P
[
N
= 1] = 4
/
7 + 2
/
7 = 6
/
7
(2)
Problem 2.2.3 Solution
(a) We must choose
c
to make the PMF of
V
sum to one.
4
summationdisplay
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)
(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 >
2 is
P
[
V >
2] =
P
V
(3) +
P
V
(4) =
3
2
30
+
4
2
30
=
5
6
(4)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 2.2.8 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) =
p
P
B
(3) = 2
p
P
B
(2) = 4
p
P
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 2.3.1 Solution
(a) If it is indeed true that
Y
, the number of yellow M&M’s in a package, is uniformly
distributed between 5 and 15, then the PMF of
Y
, is
P
Y
(
y
) =
braceleftbigg
1
/
11
y
= 5
,
6
,
7
,...,
15
0
otherwise
(1)
(b)
P
[
Y <
10] =
P
Y
(5) +
P
Y
(6) +
· · ·
+
P
Y
(9) = 5
/
11
(2)
(c)
P
[
Y >
12] =
P
Y
(13) +
P
Y
(14) +
P
Y
(15) = 3
/
11
(3)
(d)
P
[8
≤
Y
≤
12] =
P
Y
(8) +
P
Y
(9) +
· · ·
+
P
Y
(12) = 5
/
11
(4)
Problem 2.3.3 Solution
Whether a hook catches a fish is an independent trial with success probability
h
. The the
number of fish hooked,
K
, has the binomial PMF
P
K
(
k
) =
braceleftbigg (
m
k
)
h
k
(1
−
h
)
m
−
k
k
= 0
,
1
,...,m
0
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Probability, Probability theory, CDF, sixers

Click to edit the document details