AMS 311 (Fall, 2009)
Joe Mitchell
Homework Set # 1 – Solution Notes
(1).
(13 points)
Problem 5, Ross, Chapter 2. (same problem in 7th or 8th edition)
(a). Since each
x
i
is either 0 or 1 (2 choices), the total number of choices for the outcome vector,
(
x
1
,x
2
,x
3
,x
4
,x
5
), is 2
5
. Thus,

S

= 2
5
= 32.
(b).
W
has 15 elements, namely:
W
=
{
(1,1,0,0,0), (1,1,0,0,1), (1,1,0,1,0), (1,1,0,1,1), (1,1,1,0,0),
(1,1,1,0,1), (1,1,1,1,0), (1,1,1,1,1), (0,0,1,1,0), (0,0,1,1,1), (0,1,1,1,0), (0,1,1,1,1), (1,0,1,1,0), (1,0,1,1,1),
(1,0,1,0,1)
}
.
(c). Since
A
consists of outcomes having
x
4
=
x
5
= 0, we see that there are 2 choices for each
of
x
1
,x
2
,x
3
; thus,
A
has 2
·
2
·
2 = 8 elements. (You should be able to list them.)
(d).
A
∩
W
=
{
(1
,
1
,
0
,
0
,
0)
,
(1
,
1
,
1
,
0
,
0)
}
, since these are the only two elements of
W
having
x
4
=
x
5
= 0.
(2).
(16 points)
Among 33 students in a class, 17 of them earned A’s on the midterm exam, 14 earned A’s
on the Fnal exam, and 11 did not earn A’s on either examination. What is the probability that a
randomly selected student from this class earned an A on both exams? What is the sample space
and event, in your notation?
The “experiment” is the selection of one student from the class of 33; we can record the “out
come” as a pair (
m,f
), where
m
is the grade the student made on the midterm and
f
is the grade
the student made on the ±nal. The sample space is the set
S
of 33 students.
Let
E
be the event that the selected student makes an “A” on the midterm; we know that
P
(
E
) = 17
/
33. Let
F
be the event that the student earns an “A” on the ±nal; we know that
P
(
F
) = 14
/
33. We are also told that
P
(
E
c
∩
F
c
) = 11
/
33 = 1
/
3.
We want to compute the probability
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 Fall '08
 Tucker,A
 Probability theory, class a, maximum possible value

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