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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 Milenkovic,O
 Conditional Probability, Probability theory, right steps, ML rule, MAP rule

Click to edit the document details