CS 70 Fall 2006 — Solutions to Homework #9
November 30, 2006
1.
Marbles.
One approach is to explicitly construct the probability space with
•
The sample space Ω =
{
(
A, b
1
)
,
(
A, b
2
)
,
(
A, w
1
)
, . . . ,
(
A, w
5
)
,
(
B, b
1
)
,
(
B, w
1
)
, . . . ,
(
B, w
4
)
}
; and
•
The correct probability on each point—probabilities of
1
14
and
1
10
on each point with box
A
and
B
respectively.
(the events corresponding to box
A
and
B
must have total probability 0.5 each, and
each sample point within each event must have equal probability summing to total probability of
the event).
Then we can calculate the probabilities directly from Frst principles as follows.
•
The probability that the marble is black:
Involves deFning the event
B
=
{
(
A, b
1
)
,
(
A, b
2
)
,
(
B, b
1
)
}
that the marble is black and noticing that
Pr (
B
)
=
X
ω
∈
B
Pr (
ω
)
=
Pr ((
A, b
1
)) + Pr ((
A, b
2
)) + Pr ((
B, b
1
))
=
1
14
+
1
14
+
1
10
=
17
70
.
•
The probability that the marble came from A, given that it is white:
Involves the deFnition of condi
tional probability: let
A
=
{
(
A, b
1
)
,
(
A, b
2
)
,
(
A, w
1
)
, . . . ,
(
A, w
5
)
}
be the event that the marble came
from box A and
W
=
{
(
A, w
1
)
, . . . ,
(
A, w
5
)
,
(
B, w
1
)
, . . . ,
(
B, w
4
)
}
be the event that the marble is
white; then
Pr (
A

W
)
=
Pr (
A
∩
W
)
Pr (
W
)
=
∑
ω
∈
A
∩
W
Pr (
ω
)
∑
ω
∈
W
Pr (
ω
)
=
Pr ((
A, w
1
)) +
. . .
+ Pr ((
A, w
5
))
Pr ((
A, w
1
)) +
. . .
+ Pr ((
A, w
5
)) + Pr ((
B, w
1
)) +
. . .
+ Pr ((
B, w
4
))
=
5
/
14
(5
/
14) + (4
/
10)
=
25
53
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentAnother approach is to acknowledge the existence of the underlying sample space but not bother with the
probabilities on each sample point, rather write down the ‘interesting’ events and the various probabilities
given to us directly:
•
The sample space Ω =
{
(
A, b
1
)
,
(
A, b
2
)
,
(
A, w
1
)
, . . . ,
(
A, w
5
)
,
(
B, b
1
)
,
(
B, w
1
)
, . . . ,
(
B, w
4
)
}
;
•
Let
A
=
{
(
A, b
1
)
,
(
A, b
2
)
,
(
A, w
1
)
, . . . ,
(
A, w
5
)
}
be the event that the marble came from box A and
let
W
=
{
(
A, w
1
)
, . . . ,
(
A, w
5
)
,
(
B, w
1
)
, . . . ,
(
B, w
4
)
}
be the event that the marble is white (NB: the
complement
¬
A
corresponds to box B and
¬
W
corresponds to black marble); then
•
We are told that Pr (
A
) =
1
2
, Pr (
W

A
) =
5
7
and Pr (
W
¬
A
) =
4
5
.
We can then transform the probabilities we know to get those we want:
•
The probability that the marble is black:
Use the fact that Pr (
¬
W
) = 1

Pr (
W
) and that the sets
A
and
¬
A
partition
1
Ω:
Pr (
W
)
=
Pr (
W

A
) Pr (
A
) + Pr (
W
¬
A
) Pr (
¬
A
)
=
5
7
·
1
2
+
4
5
·
1
2
=
53
70
.
•
The probability that the marble came from A, given that it is white:
The same ideas apply here.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 HAULLGREN
 Probability theory

Click to edit the document details