Statistics 427: Homework #1
Solutions to Assigned Problems
2.3
[8 pts (2,2,2,2) ]
(a)
The sample space for this experiment is
S
=
{
(
SSS
)
,
(
SSF
)
,
(
SFS
)
,
(
SFF
)
,
(
FSS
)
,
(
FSF
)
,
(
FFS
)
,
(
FFF
)
}
where
S
denotes success and
F
denotes failure. Event
A
contains all outcomes with exactly two
S
’s:
A
=
{
(
SSF
)
,
(
SFS
)
,
(
FSS
)
}
.
(b)
The event where at least two of the components function contains the outcomes
B
=
{
(
SSS
)
,
(
SSF
)
,
(
SFS
)
,
(
FSS
)
}
.
(c)
The system will function if component one functions and at least one of components two and
three functions. So
C
=
{
(
SSS
)
,
(
SSF
)
,
(
SFS
)
}
.
(d)
Event
Outcomes
C
0
{
(
SFF
)
,
(
FSS
)
,
(
FSF
)
,
(
FFS
)
,
(
FFF
)
}
A
∪
C
{
(
SSS
)
,
(
SSF
)
,
(
SFS
)
,
(
FSS
)
}
A
∩
C
{
(
SSF
)
,
(
SFS
)
}
B
∪
C
{
(
SSS
)
,
(
SSF
)
,
(
SFS
)
,
(
FSS
)
}
B
∩
C
{
(
SSS
)
,
(
SSF
)
,
(
SFS
)
}
1
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View Full Document2.11
[6 pts (2,2,2)]
(a)
You can read this right oﬀ the table: 0
.
07.
(b)
There are three bond funds, so we compute 0
.
15 + 0
.
10 + 0
.
05 = 0
.
30.
(c)
Let event
A
be the event where the selected customer does not own stocks. Then we can just
sum up all the possibilities 0
.
20+0
.
15+0
.
10+0
.
05+0
.
07 = 0
.
57. We could have done less arithmetic
by ﬁnding the probability
A
0
by summing 0
.
18 + 0
.
25 = 0
.
43 and then subtracting 1

0
.
42 = 0
.
57.
2.22
[9 pts (3,3,3)]
(a)
Let
B
1
be the event that the person stops at the ﬁrst light and
B
2
be the event that the
person stops at the second light. To ﬁnd the probability that the person must stop at both signals,
we can use
P
(
B
1
∩
B
2
) =
P
(
B
1
) +
P
(
B
2
)

P
(
B
1
∪
B
2
) = 0
.
4 + 0
.
5

0
.
6 = 0
.
3
.
(b)
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 Spring '08
 Staff
 Statistics, 6 pts, 8 pts, 7 pts, 0.2%

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