above. The expectations of
I
C
,I
D
and
I
S
are the same. Hence
E
(
X
) = 4
E
(
I
H
) = 4
1

(
39
5
)
(
52
5
)

(
13
1
)(
39
4
)
(
52
5
)
!
≈
1
.
468
.
4. [8 points] A hat contains two black and two white cards. Pick two cards and let
X
be
the number of black cards. Find
E
(
X
) and
SD
(
X
).
Solution:
Note that
P
(
X
= 0) =
2
4
×
1
3
=
1
6
,
P
(
X
= 1) = 2
±
2
4
×
2
3
²
=
2
3
, and
P
(
X
= 2) =
2
4
×
1
3
=
1
6
.
Therefore,
E
(
X
) = 0
×
1
6
+ 1
×
2
3
+ 2
×
1
6
= 1
,
E
(
X
2
) = 0
×
1
6
+ 1
×
2
3
+ 4
×
1
6
=
4
3
,
V ar
(
X
) =
E
(
X
2
)

(
E
(
X
))
2
=
4
3

1
2
=
1
3
, and ﬁnally
SD
(
X
) =
p
V ar
(
X
) =
1
√
3
.
Page 3
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View Full DocumentMath 431: Exam 1
5. [10 points] Ten people sit randomly around a circular table. Six are men and four are
women. How many men do you expect are not sitting next to a woman?
Solution:
Let
I
n
be the indicator that the person in the
i
th chair is a man and he
has men sitting on both sides. Then, for each
n
∈ {
1
,...,
10
}
E
(
I
n
) =
P
(
I
n
= 1) =
6
10
×
5
9
×
4
8
=
1
6
.
The number of men not sitting next to a woman is
X
=
I
1
···
I
10
and
E
(
X
) =
10
E
(
I
1
) =
10
6
=
5
3
.
6. Let
X
and
Y
be independent random variables with
E
(
X
) = 4,
SD
(
X
) = 3,
E
(
Y
) = 6,
and
SD
(
Y
) = 4.
(a) [4 points] Find
E
(
X
+
Y
) and
SD
(
X
+
Y
).
Solution:
E
(
X
+
Y
) =
E
(
X
) +
E
(
Y
) = 10. Because
X,Y
are independent
V ar
(
X
+
Y
) =
V ar
(
X
) +
V ar
(
Y
) = 3
2
+ 4
2
= 25, so
SD
(
X
+
Y
) =
√
25 = 5.
(b) [8 points] Find an upper bound for
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 Spring '12
 Miller
 Binomial, Probability, Probability theory, Binomial distribution, 1%, 5.7%, 14.38%

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