MSPE PROGRAM
ECON 506 FALL 2010
SOLUTION TO HW2
(The problem numbers refer to numbers in 7
th
edition)
3.11
There is a 1/3 chance a person has O
+
blood and 2/3 they do not.
Similarly, there is a 1/15 chance a person
has O
–
blood and 14/15 chance they do not.
Assuming the donors are randomly selected, if
X
= # of O
+
blood donors and
Y
= # of O
–
blood donors, the probability distributions are
0
1
2
3
p
(
x
)
(2/3)
3
= 8/27
3(2/3)
2
(1/3) = 12/27
3(2/3)(1/3)
2
=6/27
(1/3)
3
= 1/27
p
(
y
)
2744/3375
3*(196/3375)
3*(14/3375)
1/3375
Note that
Z
=
X
+
Y
= # will type O blood.
The probability a donor will have type O blood is 1/3 + 1/15 =
6/15 = 2/5.
The probability distribution for
Z
is
0
1
2
3
p
(
z
)
(2/5)
3
= 27/125
3(2/5)
2
(3/5) = 54/27
3(2/5)(3/5)
2
=36/125
(3/5)
3
= 27/125
3.12
E
(
Y
) = 1(.4) + 2(.3) + 3(.2) + 4(.1) = 2.0
E
(1/
Y
) = 1(.4) + 1/2(.3) + 1/3(.2) + 1/4(.1) = 0.6417
E
(
Y
2
– 1) =
E
(
Y
2
) – 1 = [1(.4) + 2
2
(.3) + 3
2
(.2) + 4
2
(.1)] – 1 = 5 – 1 = 4.
V
(
Y
) =
E
(
Y
2
) = [
E
(
Y
)]
2
= 5 – 22 = 1.
3.23
Define
G
to be the gain to a person in drawing one card.
The possible values for
G
are $15, $5, or $–4 with
probabilities 3/13, 2/13, and 9/13 respectively.
So,
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 Fall '08
 Staff
 Normal Distribution, Poisson Distribution, Probability theory, Binomial distribution, Discrete probability distribution, High school students

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