UCSD ECE153
Handout #24
Prof. YoungHan Kim
Tuesday, May 3, 2011
Solutions to Midterm Examination (Spring 2010)
(Prepared by TA Lele Wang)
1. Let
X
i
denote the color of the
i
th ball.
(a) By symmetry,
P
{
X
1
=
R
}
= 1
/
2.
(b) Again by symmetry,
P
{
X
i
=
R
}
= 1
/
2 for all
i
.
Alternatively, by the law of total
probability, we have
P
{
X
2
=
R
}
=
P
{
X
1
=
R
}
P
{
X
2
=
R

X
1
=
R
}
+
P
{
X
1
=
B
}
P
{
X
2
=
R

X
1
=
B
}
=
1
2
×
2
3
+
1
2
×
1
3
=
1
2
.
(c) By the chain rule, we have
P
{
X
1
=
R, X
2
=
R, X
3
=
R
}
=
P
{
X
1
=
R
}
P
{
X
2
=
R

X
1
=
R
}
P
{
X
3
=
R

X
2
=
R, X
1
=
R
}
=
1
2
×
2
3
×
3
4
=
1
4
.
(d) Let
N
denote the number of red balls in the first three draws. From part (c), we know
that
P
{
N
= 3
}
= 1
/
4 =
P
{
N
= 0
}
, where the latter identity follows by symmetry. Also
we have
P
{
N
= 2
}
=
P
{
N
= 1
}
. Thus,
P
{
N
= 2
}
must be 1
/
4.
Alternatively, we have
P
{
N
= 2
}
=
P
{
X
1
=
B, X
2
=
R, X
3
=
R
}
+
P
{
X
1
=
R, X
2
=
B, X
3
=
R
}
+
P
{
X
1
=
R, X
2
=
R, X
3
=
B
}
=
P
{
X
1
=
B
}
P
{
X
2
=
R

X
1
=
B
}
P
{
X
3
=
R

X
2
=
R, X
1
=
B
}
+
P
{
X
1
=
R
}
P
{
X
2
=
B

X
1
=
R
}
P
{
X
3
=
R

X
2
=
B, X
1
=
R
}
+
P
{
X
1
=
R
}
P
{
X
2
=
R

X
1
=
R
}
P
{
X
3
=
B

X
2
=
R, X
1
=
R
}
=
1
2
×
1
3
×
2
4
+
1
2
×
1
3
×
2
4
+
1
2
×
2
3
×
1
4
=
1
4
.
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 Spring '11
 YoungHanKim
 Derivative, Conditional Probability, Trigraph, X1, dy dx

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