Solutions to Homework 6
4.78
Let
Y
be the number of people who refuse to be interviewed before she obtains five
people, so
5,
Y
X
=

Y
can reasonably be assumed to have the negative binomial distribution
with
0.6
p
=
,
5
r
=
a.
(
29
(
29
2
5
0
(
7)
(
2)
4
0.6
1
0.6
4
0.419904
i
i
P X
P Y
i
=
≤
=
≤
+
=

=
∑
b.
2
2
5
0.4
10
5
0.4
50
(
)
,
(
)
0.6
3
0.6
9
rq
rq
E Y
V Y
p
p
×
×
=
=
=
=
=
=
Because
5,
Y
X
=

so
5
X
Y
=
+
, then
25
50
(
)
(
)
5
,
(
)
(
)
3
9
E X
E Y
V X
V Y
=
+
=
=
=
4.79
Let
X
be the number of customers that the salesman has to visit to make three sales,
let
Y
denote the number of customers who refuse to purchase a car before the salesman
makes three sales, so
3
Y
X
=

. Suppose that trials are independent,
Y
can be
modeled as negative binomial distribution with
0.2,
3
p
r
=
=
a.
(
29
(
29
1
3
0
(
5)
(
2)
1
(
1)
2
1
0.2
1
0.2
2
0.9728
i
i
P X
P Y
P Y
i
=
≥
=
≥
=

≤
+
=


=
∑
b.
3
0.8
(
)
12
0.2
rq
E Y
p
×
=
=
=
Because
3,
Y
X
=

so
3
X
Y
=
+
, then
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(
)
(
)
3
15
E X
E Y
=
+
=
4.83
a.
Let
i
Y
denote the event that John wins in the
i
th game,
(
1)
0.6
i
P Y
p
=
=
=
. Because the
outcome of each game is independent of the outcomes of the other games, then
3
1
2
3
(
3)
(
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 Spring '08
 Staff
 Binomial, Probability, Binomial distribution, Discrete probability distribution, Negative binomial distribution, Geometric distribution, Hypergeometric Distribution

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