SOLUTIONS to Practice Midterm Exam 1
1. You are given the following:
P
(
A
) = 0
.
35
, P
(
B
) = 0
.
45
, P
(
AB
) = 0
.
18
.
(a)
P
(
A
or
B
) = 0
.
35 + 0
.
45

0
.
18 =
0
.
62
.
(b)
P
(
A
c
) = 1

0
.
35 = 0
.
65
.
(c) First,
B
is the disjoint union of
AB
and
A
c
B
. Thus,
0
.
45 =
P
(
B
) =
P
(
AB
)+
P
(
A
c
B
) =
0
.
18 + +
P
(
A
c
B
)
.
Thus,
P
(
A
c
B
) = 0
.
45

0
.
18 = 0
.
27
.
2.
(a) Draw the
5
×
5
array that represents
all possible combinations of
X
1
and
X
2
. These 25 possibilities are equally
likely.
Five entries correspond to
X
1
=
X
2
;
thus its probability is
5
/
25 = 0
.
20
.
(b) Even without drawing the
5
×
5
×
5
array, we know that it has 125 cells
and all cells are equally likely.
Ex
actly three of these cells give
X
1
+
X
2
+
X
3
= 4
, namely (1,1,2), (1,2,1)
and (2,1,1). Thus, the desired proba
bility is
3
/
125
.
3.
(a)
P
(
SFS
) =
pqp
= (0
.
85)
2
(0
.
15) =
0
.
1084
.
(b) The probability of exactly three suc
cesses is
4!
3!1!
(0
.
85)
3
(0
.
15) = 0
.
3685
.
(c) The event
(
Y
= 3)
can occur only
one way: Two out of two successes
on Friday followed by one out of two
successes on Saturday. These proba
bilities are multiplied b/c trials are in
dependent, giving:
P
(
Y
= 3) = (0
.
85)
2
2(0
.
85)(0
.
15) =
0
.
1842
.
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 Fall '09
 ProfessorWardrop
 Statistics, Null hypothesis, Statistical hypothesis testing, Type I and type II errors, AC B, lower right box

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