Section 37
381
.
a)
5
.
0
5
.
0
)
5
.
0
1
(
)
1
(
0
=
−
=
=
X
P
b)
0625
.
0
5
.
0
5
.
0
)
5
.
0
1
(
)
4
(
4
3
=
=
−
=
=
X
P
c)
0039
.
0
5
.
0
5
.
0
)
5
.
0
1
(
)
8
(
8
7
=
=
−
=
=
X
P
d)
5
.
0
)
5
.
0
1
(
5
.
0
)
5
.
0
1
(
)
2
(
)
1
(
)
2
(
1
0
−
+
−
=
=
+
=
=
≤
X
P
X
P
X
P
75
.
0
5
.
0
5
.
0
2
=
+
=
e.)
25
.
0
75
.
0
1
)
2
(
1
)
2
(
=
−
=
≤
−
=
>
X
P
X
P
382.
E(X) = 2.5 = 1/p giving p = 0.4
a)
4
.
0
4
.
0
)
4
.
0
1
(
)
1
(
0
=
−
=
=
X
P
b)
0864
.
0
4
.
0
)
4
.
0
1
(
)
4
(
3
=
−
=
=
X
P
c)
05184
.
0
5
.
0
)
5
.
0
1
(
)
5
(
4
=
−
=
=
X
P
d)
)
3
(
)
2
(
)
1
(
)
3
(
=
+
=
+
=
=
≤
X
P
X
P
X
P
X
P
7840
.
0
4
.
0
)
4
.
0
1
(
4
.
0
)
4
.
0
1
(
4
.
0
)
4
.
0
1
(
2
1
0
=
−
+
−
+
−
=
e)
2160
.
0
7840
.
0
1
)
3
(
1
)
3
(
=
−
=
≤
−
=
>
X
P
X
P
383.
Let X denote the number of trials to obtain the first success.
a) E(X) = 1/0.2 = 5
b) Because of the lack of memory property, the expected value is still 5.
384.
a) E(X) = 4/0.2 = 20
b) P(X=20) =
0436
.
0
2
.
0
)
80
.
0
(
3
19
4
16
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
315
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c) P(X=19) =
0459
.
0
2
.
0
)
80
.
0
(
3
18
4
15
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
d) P(X=21) =
0411
.
0
2
.
0
)
80
.
0
(
3
20
4
17
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
e) The most likely value for X should be near
μ
X
. By trying several cases, the most likely value is x = 19.
385.
Let X denote the number of trials to obtain the first successful alignment.
Then X is a geometric random
variable with p = 0.8
a)
0064
.
0
8
.
0
2
.
0
8
.
0
)
8
.
0
1
(
)
4
(
3
3
=
=
−
=
=
X
P
b)
)
4
(
)
3
(
)
2
(
)
1
(
)
4
(
=
+
=
+
=
+
=
=
≤
X
P
X
P
X
P
X
P
X
P
8
.
0
)
8
.
0
1
(
8
.
0
)
8
.
0
1
(
8
.
0
)
8
.
0
1
(
8
.
0
)
8
.
0
1
(
3
2
1
0
−
+
−
+
−
+
−
=
9984
.
0
8
.
0
2
.
0
)
8
.
0
(
2
.
0
)
8
.
0
(
2
.
0
8
.
0
3
2
=
+
+
+
=
c)
)]
3
(
)
2
(
)
1
(
[
1
)
3
(
1
)
4
(
=
+
=
+
=
−
=
≤
−
=
≥
X
P
X
P
X
P
X
P
X
P
]
8
.
0
)
8
.
0
1
(
8
.
0
)
8
.
0
1
(
8
.
0
)
8
.
0
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 Spring '10
 Feng
 Probability theory, #, µ, 200 days, µY, p=0.8

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