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IEOR 3658
Assignment #2 Solutions
Probability
September 23, 2011
Prof. Mariana OlveraCravioto
Page 1 of
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Assignment #2 Solutions
1.
P
(doubleheaded coin

head obtained) =
P
(
doubleheaded coin
∩
head obtained
)
P
(
head obtained
)
=
P
(
doubleheaded coin
)
P
(
head obtained
)
=
1
3
4
6
=
1
2
2.
(a)
P
(green ball) =
3+4+6
3+9+4+8+6+6
=
13
36
(b)
P
(from urn 1

green ball) =
P
(
from urn 1
∩
green ball
)
P
(
green ball
)
=
3
13
3.
(a) There’re six doubles, (1, 1), (2, 2), .
.. , (6, 6). Therefore the probability of getting
double is 6/36 = 1/6
(b)
P
(double

S
≤
4) =
P
(double
∩
S
≤
4)
P
(
S
≤
4)
=
P
(
{
(1
,
1)
,
(2
,
2)
}
)
P
(
{
(1
,
1)
,
(1
,
2)
,
(2
,
1)
,
(1
,
3)
,
(2
,
2)
,
(3
,
1)
}
)
=
1
3
(c) There’re 11 cases where at least one die is 6. Therefore the answer is 11/36
(d) Again we can use conditional probability. Since we’re conditioning on ’Not Double’, the
size of sample space becomes 30 (36 outcomes  6 doubles). In this sample space there’re
10 cases where at least one die is 6. (Since we’re excluding (6, 6) case) Therefore the
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