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Mathematical Systems
Probability
Solutions by Bracket
A First Course in Probability
Chapter 6—Problems
1.
Two fair dice are rolled. Find the join probability mass function of
X
and
Y
when
a.
X
is the largest value obtained on any die and
Y
is the sum of their values.
We have
p
(
x, y
) as given by the table below:
x
y
2
3
4
5
6
7
8
9
10
11
12
1
1
/
36










2

2
/
36
1
/
36








3


2
/
36
2
/
36
1
/
36






4



2
/
36
2
/
36
2
/
36
1
/
36




5




2
/
36
2
/
36
2
/
36
2
/
36
1
/
36


6





2
/
36
2
/
36
2
/
36
2
/
36
2
/
36
1
/
36
b.
X
is the value on the ﬁrst die and
Y
is the larger of the two values.
We have
p
(
x, y
) as given by the table below:
x
y
1
2
3
4
5
6
1
1
/
36
1
/
36
1
/
36
1
/
36
1
/
36
1
/
36
2

2
/
36
1
/
36
1
/
36
1
/
36
1
/
36
3


3
/
36
1
/
36
1
/
36
1
/
36
4



4
/
36
1
/
36
1
/
36
5




5
/
36
1
/
36
6





6
/
36
c.
X
is the smallest and
Y
is the largest value obtained on the dice.
Note that the event corresponding to
p
(
x, y
) occurs precisely when we roll (
x, y
) or (
x, y
) if
x < y
,
in which case
p
(
x, y
) =
2
36
, or
p
(
x, y
) =
1
36
if
x
=
y
.
7.
Consider a sequence of independent Bernoulli trials, each of which is a success with probability
p
. Let
X
1
be the number of failures preceding the ﬁrst success, and let
X
2
be the number of failures between
the ﬁrst two successes. Find the joint mass function of
X
1
and
X
2
.
Note that
X
1
and
X
2
are independent, and thus
P
±
X
1
=
x
1
, X
2
=
x
2
²
=
P
±
X
1
=
x
1
²
P
±
X
2
=
x
2
²
.
Let
X
be a geometrically distributed random variable with parameter
p
.
Then
P
±
X
1
=
x
1
²
=
P
±
X
=
x
1
+ 1
²
= (1

p
)
x
1
p
, and similarly for
X
2
. Thus
p
(
x
1
, x
2
) = (1

p
)
x
1
+
x
2
p
2
.
1
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Probability
8.
The joint probability density function of
X
and
Y
is given by
f
(
x, y
) =
c
(
y
2

x
2
)
e

y

y
≤
x
≤
y,
0
< y <
∞
a.
Find
c
.
We integrate over the entire region to obtain
1 =
Z Z
C
f
(
x, y
)
dx dy
=
c
Z
∞
0
Z
y

y
(
y
2

x
2
)
e

y
dx dy
=
c
Z
∞
0
e

y
Z
y

y
(
y
2

x
2
)
dx dy
=
c
Z
∞
0
(2
y
3

2
3
y
3
)
dy
=
c
Z
∞
0
e

y
(2
y
3

2
3
y
3
)
dy
=
4
3
c
Z
∞
0
y
3
e

y
dy
=
4
3
c
·
Γ(3 + 1)
=
4
3
c
·
3! = 8
c
Thus
c
=
1
8
.
b.
Find the marginal densities of
X
and
Y
.
We have
f
X
(
x
) =
Z
∞
∞
f
(
x, y
)
dy
=
Z
∞

x

c
(
y
2

x
2
)
e

y
dy
=
c
(
x
2

y
2

2
y

2)
e

y

∞
y
=

x

=
1
8
(2

x

+ 2)
e

x

=
1
4
(

x

+ 1)
e

x

.
Similarly
f
Y
(
y
) =
Z
∞
∞
f
(
x, y
)
dx
=
Z
y

y
c
(
y
2

x
2
)
e

y
dx
=
ce

y
Z
y

y
(
y
2

x
2
)
dx
=
ce

y
(2
y
3

2
3
y
4
) =
1
8
·
e

y
·
4
3
y
3
=
1
6
y
3
e

y
.
c.
Find
E
[
X
].
E
[
X
] =
Z
∞
∞
xf
X
(
x
)
dx
=
1
4
Z
∞
∞
x
(

x

+ 1)
e

x

dx
=
1
4
Z
0
∞
x
(

x
+ 1)
e
x
dx
+
1
4
Z
∞
0
x
(
x
+ 1)
e

x
dx
=
1
4
±
e
x
(

x
2
+ 3
x

3)
²
0
x
=
∞
+
1
4
±

e

x
(
x
2
+ 3
x
+ 3)
²
∞
x
=0
=

3
4
+
3
4
= 0
2
Mathematical Systems
Probability
16.
Suppose that
n
points are independently chosen at random on the perimeter of a circle, and we want
the probability that they all lie in some semicircle. (That is, we want the probability that there is
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This note was uploaded on 12/06/2009 for the course MATH 4120 taught by Professor Carlosj.moreno during the Spring '08 term at CUNY Baruch.
 Spring '08
 CARLOSJ.MORENO
 Probability

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