STAT 230
Solutions To Extra Problems
1. (a) Let
X
= the number of tickets required to produce 2 winners. Then
X
has a negative
binomial distribution with parameters
r
= 2 and
p
=0
.
1. Therefore
E
(
X
)=
r
p
=
2
0
.
1
= 20 and
Var(
X
)=
r
(1

p
)
p
2
=
2(0
.
9)
(0
.
1)
2
= 180.
Now
W
= (2)(5
.
00)

X
(1
.
00) = 10
.
00

X
. Therefore
E
(
W
)=

10
.
00 and Var(
W
) = 180.
(b) Let
Y
1
= number of outlets requiring 14 or fewer tickets to get 2 winners, let
Y
2
= number
of outlets requiring 30 or more tickets to get 2 winners, and let
Y
3
=15

Y
1

Y
2
= the remaining
number of outlets. This divides the outlets into three boxes, and the result is a multinomial
(trinomial) distribution. The desired probability is
P
(
Y
1
=3
,Y
2
=2
,Y
3
= 10) =
15!
3! 2! 10!
p
3
1
p
2
2
(1

p
1

p
2
)
10
where
p
1
= probability that 14 or fewer tickets will be required to obtain 2 winners, and
p
2
=
probability that 30 or more tickets will be required to obtain 2 winners. The probabilities
p
1
and
p
2
can be solved for explicitly from the negative binomial distribution. For example,
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 Fall '06
 various
 Binomial, Normal Distribution, Poisson Distribution, Probability theory, Binomial distribution, Negative binomial distribution, Xk Xn

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