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University of Illinois
Fall 2010
ECE 313:
Problem Set 5: Solutions
Bayes’ Formula and binary hypothesis testing
1.
[Dissecting a vote]
(a) The number of judges voting yes has the binomial distribution with parameters
n
= 3 and
p,
and
M
is the event that either two or all three judges vote yes. So
P
(
M
) =
(
3
2
)
p
2
(1

p
)+
p
3
= 3
p
2

2
p
3
.
(b) Given
A
is true,
M
is equivalent to the event that at least one of the two other judges votes yes,
so by the same reasoning as in part (a),
P
(
M

A
) =
P
(at least one of judge two or judge three votes yes

A
)
=
P
(at least one of judge two or judge three votes yes) =
±
2
1
²
p
(1

p
) +
p
2
= 2
p

p
2
.
Another way to do this problem is to refer to the sample space: Ω =
{
x
1
x
2
x
3
:
x
i
∈ {
0
,
1
}}
,
where
x
i
= 1 means judge
i
votes yes. Then
MA
=
{
110
,
101
,
111
}
and
P
(
) = 2
p
2
(1

p
) +
p
3
.
So,
by the deﬁnition of conditional probability,
P
(
M

A
) =
P
(
)
P
(
A
)
=
2
p
2
(1

p
)+
p
3
p
= 2
p

p
2
.
(c) By the deﬁnition of conditional probability,
P
(
A

M
) =
P
(
AM
)
P
(
M
)
=
P
(
A
)
P
(
M

A
)
P
(
M
)
=
p
(2
p

p
2
)
3
p
2

2
p
3
=
2

p
3

2
p
.
This answer is shown in the following plot:
The limit as
p
→
0 is 2/3. This makes sense, because if
p
is small, by far the most likely way
for the committee to reach a yes decision is to have two judges vote yes and one judge vote no,
so by symmetry the conditional probability the ﬁrst judge votes yes is close to 2/3. The limit as
p
→
1 is one. This makes sense because if
p
is close to one, with probability close to one, all three
judges vote yes. Conditioning on
M
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 Fall '08
 Milenkovic,O

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