STAT 350: Probability and Random Processes for Engineers
Spring, 2008
ANSWERS TO EXAM #1, VERSION #2
You will need to follow along with your exam.
1.
(a) Combinations Rule:
C
(32
,
4) = 35,960
(b) Partitions Rule:
32!
4!3!25!
= 117,804,960, or
C
(32
,
4)
×
C
(28
,
3) = 117,804,960
(c) Order matters, so Permutations Rule:
P
(32
,
4) = 863,040
(d) The chance of getting exactly 3 EEs and 1 CPE is
C
(23
,
3)
×
C
(9
,
1)
C
(32
,
4)
=
1771
×
9
35960
=
.
4432
2. If you hit black 0 times, you lose 5($10) = $50. If you hit black 1 time, you lose $30 (four losses,
one win). Otherwise, you lose less than $30. So, you want the chance of 0 or 1 black spins in 5 tries.
This requires the binomial model, with
n
= 5 and
p
= 18
/
38:
P
(
k
≤
1) =
5
0
¶
(18
/
38)
0
(20
/
38)
5
+
5
1
¶
(18
/
38)
1
(20
/
38)
4
=
.
2221
3. For each component, let
A
i
means that component #
i
is working properly;
P
(
A
i
) =
.
88 for each
i
.
The reliability of the 12 subsystem is
R
12
=
P
(
A
1
∪
A
2
); by deMorgan’s Law and independence,
P
(
A
1
∪
A
2
) = 1

P
(
¯
A
1
∩
¯
A
2
) = 1

P
(
¯
A
1
)
×
P
(
¯
A
2
)

1

(
.
12)(
.
12) =
.
9856. Likewise,
R
34
=
.
9856,
and of course
R
5
=
.
88.
Since these subsystems are connected in series, by independence
R
sys
=
R
12
×
R
34
×
R
5
= (
.
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 Spring '07
 Carlton
 Probability, DeMorgan, GA

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