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Statistics 116 - Fall 2004
Theory of Probability
Assignment # 7
Solutions
Q. 1)
(Ross # 6.2)
Suppose that 3 balls are chosen without replacement from
an urn consisting of 5 white and 8 red balls. Let Xi equal 1 if the i-th ball
selected is white, and let it equal 0 otherwise. Give the joint probability
mass function of
(a)
(b)
, X2 ;
X1 , X2 ,
X1
Solution:
(a)
X3
.
5 4
13 12
5 8
P (X1 = 1; X2 = 0) =
13 12
8 5
P (X1 = 0; X2 = 1) =
13 12
8 7
P (X1 = 0; X2 = 0) =
13 12
(
P X1
= 1; X2 = 1) =
1
(b)
(
= 1; X2 = 1; X3 = 1) =
(
= 1; X2 = 0; X3 = 1) =
(
= 0; X2 = 1; X3 = 1) =
(
= 0; X2 = 0; X3 = 1) =
(
= 1; X2 = 1; X3 = 0) =
(
= 1; X2 = 0; X3 = 0) =
(
= 0; X2 = 1; X3 = 0) =
(
= 0; X2 = 0; X3 = 0) =
P X1
P X1
P X1
P X1
P X1
P X1
P X1
P X1
Q. 2)
5 4 3
13 12 11
5 8 4
13 12 11
8 5 4
13 12 11
8 7 5
13 12 11
5 4 8
13 12 11
5 8 7
13 12 11
8 5 7
13 12 11
8 7 6
13 12 11
(Ross # 6.6) A bin of 5 transistors is known to contain 2 that are
defective. The transistors are to be tested, one at a time, until the defective
ones are identied. Denote by N1 the number of tests made until the rst
defective is identied and by N2 the number of additional tests until the
second defective is identied: nd the joint probability mass function of
N1 and N2 .
Solution:
Imagine drawing all 5 transistors and testing each one in order. Each
outcome is equally likely and the number of outcomes is equal to the
number of ways of choosing
two \defective" transistors out of a set of 5,
of which there are 52 = 10 ways. Hence the probability
(
P N1
Q. 3)
= i; N2 = j ) =
1
10
1i<j
<
(Ross # 6.9) The joint probability density function of
by
X
and Y is given
6 2 xy x +
;
0 < x < 1; 0 < y < 2:
7
2
Verify that this is indeed a joint density function.
Compute the density function of X .
Find P (X > Y ).
Find P (Y > 1=2jX < 1=2).
Find E (X ).
Find E (Y ).
(
f x; y
(a)
(b)
(c)
(d)
(e)
(f)
5:
)=
Solution:
2
(a)
Z
1
1
Z
1
1
(
f x; y
Z
Z
6 2 xy x +
dy dx
2
0 7
0
Z 1 Z 2 6 2 xy =
x +
dy
2
0 7
0
Z 1
2 2
6
xy
) dx
=
(x2 y +
7 0
4 0
Z
6 1 2
=
2x + x dx
7 0
6 2x3 x2 1
=
+
7 3
2 0
6 2 1
=
+
7 3 2
= 1:
) dx dy =
1
2
(b) From our above calculations
( )=
Z
1
fX x
Z
1
(
f x; y
) dy
6 2 xy x +
2
0 7
6(2x2 + x)
=
7
=
(c)
(
P X > Y
)=
=
=
=
=
=
=
=
2
ZZ
(
f x; y
Z
(x;y ):x>y
1
1
Z
Z
Z
x
1
(
f x; y
dy
) dy dx
) dy dx
6 1 x 2 xy
x +
dy dx
7 0 0
2
Z 1
2 x
6
xy 2
x y +
dx
7 0
4 0
Z 6 1 3 x3
x +
dx
7 0
4
Z
65 1 3
x dx
74 0
6 5
7 16
15
56
3
dx
(d)
(
P Y >
(
P Y >
1=2jX
1=2; X
<
1=2) =
1=2; X < 1=2)
1=2)
(
P X <
Z
6
< 1=2) =
7
Z
1=2
2
2
x
1=2
0
Z
6
=
7
=
(
P Y >
1=2
2
x y
0
Z
6
7
+
xy
dy dx
2 2
4
15x2 1=2
+
2
32 0
3
x
2
!
dx
1=2
3x2 15
+ x dx
2
16
1=2
0
6
=
7
xy
+
!
6 1
15
+
7 16 128
6 23
=
7 128
Z
6 1=2 2
< 1=2) =
2x + x dx
7 0
=
(
P X
6
=
7
2x3 x2 1=2
+ 3
2 0
6 1
1
=
+
7 12 8
6 5
=
7 24
(
P Y >
1=2jX
<
1=2) =
!
23
128
5
24
' 0 86
:
:
(e)
6
E (X ) =
7
Z
1
0
2x +x
3
2
6
dx =
7
4
4
x
2
+
!
3 1
x
3
0
=
6
7
1 1
+
2 3
=
65 5
= :
76 7
(f)
( )=
E Y
6
7
6
=
7
6
=
7
6
=
7
Z
1
Z
2
2
x y
Z
Z
0
0
1
2 2
x y
2
0
1
0
(
f x; y
Are
X
and
Y
X
(x+y )
xe
)=
3 2
6
0
dy dx
!
dx
dx
!
(Ross # 6.20) The joint density of
(
+
xy
2
2x3 2x2 1
+
3
3 0
6 2 2
=
+
7 3 3
8
= :
7
Q. 4)
xy
4x
3
2x2 +
2
+
0
and
Y
is given by
0; y > 0
otherwise.
x >
independent? What if f (x; y ) were given by
(
(
f x; y
)=
2 0 < x < y; 0 < y < 1
0 otherwise.
Solution:
In the rst case: yes, because
(
f x; y
) = xe
x
1[0;+1) (x) ye y 1[0;+1) (y ):
In the second case: no, because the support of the random vector is
f(
x; y
) : 0 < x < y; 0 < y < 1g
which is a triangle, and is not the Cartesion product of two sets. Hence
X and Y cannot be independent.
Q. 5)
(Ross # 6.29) When a current
I (measured in amperes) ows through
a resistance R (measured in ohms), the power generated is given by I 2 R
(measured in watts). Suppose that I and R are independent random
variables with densities
( ) = 6x(1
fR (x) = 2x
fI x
5
) 0 x 1;
0 x 1:
x
Determine the density function of W .
Solution:
First, we nd the distribution function of W . For any distribution of
and R such that they are independent and non-negative
I
( ) = P (W w)
= P (I 2 R w)
= P (f(i; r) : i2 r wg)
FW w
=
=
1Z
Z
w=i
2
( ) ( ) dr di
fR r fI i
Z0 1
0
()
(
2
) di:
()
(
2
fI i FR w=i
0
Therefore, the density
( )=
fW w
=
d
fI i FR w=i
dw
0
1
Z
1
Z
()
fI i
0
1
i
2
(
) di:
) di:
2
fR w=i
In our particular example this boils down to, for 0 < w < 1
Z
1
( ) = p 6i(1
w
fW w
Z
1
= 12w p
w
= 12w
= 6w
1
i
i
1
i
)
1 2w
i
2 i2
1
3
i
2
di:
di
1 1
2 i2 p w
!
12w1=2 + 6:
The range of integration in the rst line comes from the fact the density
2
2
fR (w=i ) is non-zero only if w=i 1.
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