Chapter 3 lecture notes
Math 431, Spring 2011
Instructor: David F. Anderson
Chapter 3: Conditional Probability and Independence
Section 3.2: Conditional Probabilities
Example 1.
Consider rolling a die. We know that
S
=
{
1
,
2
,
3
,
4
,
5
,
6
}
. Let
E
be the event
that the outcome is in
{
2
,
3
,
6
}
. We know that
P
(
E
) = 1
/
2. Now suppose I tell you that the
roll was an even number. Given this knowledge, what is the probability that
E
occurred?
Solution:
Let
F
=
{
2
,
4
,
6
}
. Since we know that each of 2
,
4
,
6 is equally likely, the proba
bility of
E
now appears to be
Cond. Prob. of 2 + Cond. Prob. of 3 + Cond. Prob. of 6 = 1
/
3 + 0 + 1
/
3 = 2
/
3
.
Example 2.
Now consider rolling an unfair die.
We know that
S
=
{
1
,
2
,
3
,
4
,
5
,
6
}
, but
assume that
P
(
{
i
}
) =
.
1 for
i
∈ {
1
,
2
,
3
,
4
,
5
}
and
P
(
{
6
}
) =
.
5. Let
E
be the event that the
outcome is in
{
2
,
3
,
6
}
. We know that
P
(
E
) =
P
(
{
2
}
) +
P
(
{
3
}
) +
P
(
{
6
}
) =
.
7
.
Now suppose I tell you that the roll was an even number. Given this knowledge, what is the
probability that
E
occurred?
Solution:
Let
F
=
{
2
,
4
,
6
}
. We have that
P
(
F
) =
P
(
{
2
}
) +
P
(
{
4
}
) +
P
(
{
6
}
) =
.
1 +
.
1 +
.
5 =
.
7
.
Clearly, the outcome
{
3
}
did not occur, so that has zero probability now. What about
{
2
}
and
{
6
}
? We know that
P
(
F
) =
.
7 and that
P
(
{
2
}
) =
.
1
, P
(
{
4
}
) =
.
1
, P
(
{
6
}
) =
.
5
.
Thus,
if seems like
Cond. Prob. of 2 + Cond. Prob. of 3 + Cond. Prob. of 6 =
.
1
/.
7 + 0 +
.
5
/.
7 = 6
/
7
.
Venn Diagram:
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
How did we intuitively compute that?
•
For
E
to occur given that
F
already has, the outcome
must
be in
EF
(so you intuitively
ignored the outcome “3” in
E
).
•
Because we know that
F
⊂
S
has occurred, we make
F
our new, or
reduced
sample
space.
•
Next, the probability of
EF
occurring given that
F
has should be the probability of
EF
relative to the probability of
F
.
Definition 1.
Let
E, F
⊂
S
with
P
(
F
)
>
0, then
P
(
E

F
) =
P
(
EF
)
P
(
F
)
.
In the first example above,
EF
=
{
2
,
6
}
and so
P
(
E

F
) =
P
(
EF
)
P
(
F
)
=
P
(
{
2
,
6
}
)
P
(
{
2
,
4
,
6
}
)
=
2
/
6
3
/
6
=
2
3
.
The above is taken to be the definition of conditional probability in all cases.
Example 3.
The probability that a person lives to 80 is .64. The probability that a person
lives until 90 is .51. What is the probability that a person will live to 90, given that that
person is 80.
Solution:
Let
A
be the event of living until 90 and
B
be the event of living until 80. Note
that
A
⊂
AB
. Then,
P
(
A

B
) =
P
(
AB
)
P
(
B
)
=
P
(
A
)
P
(
B
)
=
.
51
.
64
≈
.
797
.
Example 4.
A coin is flipped twice. Assuming all points of
S
=
{
(
H, H
)
,
(
T, H
)
,
(
H, T
)
,
(
T, T
)
}
are equally likely, what is the conditional probability that both flips are heads, given that
(
a
) the first flip is a head? (
b
) at least one flip is lands on heads.
Solution:
Let
E
be the event that both flips are heads. Let
A
be the event that the first
flip is a head. Then
P
(
E

A
) =
P
(
EA
)
P
(
A
)
=
P
(
{
(
H, H
)
}
)
P
(
{
(
H, T
)
,
(
H, H
)
}
)
=
1
/
4
2
/
4
=
1
2
.
Next, let
B
be the event that at least one flip is a head. Then
P
(
E

B
) =
P
(
EB
)
P
(
B
)
=
P
(
{
(
H, H
)
}
)
P
(
{
(
H, T
)
,
(
H, H
)
,
(
T, H
)
}
)
=
1
/
4
3
/
4
=
1
3
.
This seems counterintuitive as “the other could be heads or tails.” So it seems that answer
should be 1/2. However, the information “at least one coin landed on heads” only tells us
that (
T, T
) didn’t happen. The remaining three options are equally likely.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '05
 BALAZS
 Conditional Probability, Probability, Probability theory, E1 E2

Click to edit the document details