Conditional Probability
MATH 130, Elements of Statistics I
J. Robert Buchanan
Department of Mathematics
Spring 2014
J. Robert Buchanan
Conditional Probability
Background
When events A and B are independent the Multiplication Rule
states that
P(A and B) = P(A) · P(B).
Question: what if A and B are dependent? Can the
Multiplication Rule be generalized?
J. Robert Buchanan
Conditional Probability
Outcomes of Rolling Dice
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
A: sum of the dice is a seven, P(A) = 1/6,
B: doubles, P(B) = 1/6,
C: sum of the dice is a ten, P(C) = 1/12.
J. Robert Buchanan
Conditional Probability
Example
Roll a pair of fair dice and observe the outcomes.
A: sum of the dice is a seven, P(A) = 1/6,
B: doubles, P(B) = 1/6,
C: sum of the dice is a ten, P(C) = 1/12.
1
What is P(A and B)?
2
What is P(B and C)?
J. Robert Buchanan
Conditional Probability
Example
Roll a pair of fair dice and observe the outcomes.
A: sum of the dice is a seven, P(A) = 1/6,
B: doubles, P(B) = 1/6,
C: sum of the dice is a ten, P(C) = 1/12.
1
What is P(A and B)?
P(A and B) = 0
2
(impossible event)
What is P(B and C)?
J. Robert Buchanan
Conditional Probability
Example
Roll a pair of fair dice and observe the outcomes.
A: sum of the dice is a seven, P(A) = 1/6,
B: doubles, P(B) = 1/6,
C: sum of the dice is a ten, P(C) = 1/12.
1
What is P(A and B)?
P(A and B) = 0
2
(impossible event)
What is P(B and C)?
P(B and C) = 1/36
J. Robert Buchanan
(only the (5,5) combination)
Conditional Probability
Conditional Probability
Definition
The notation P(B|A) is read as the probability that event B
occurs given that event A has occurred.
J. Robert Buchanan
Conditional Probability
Conditional Probability
Definition
The notation P(B|A) is read as the probability that event B
occurs given that event A has occurred.
Example
Roll a pair of fair dice and observe the outcomes.
B: doubles, P(B) = 1/6,
C: sum of the dice is a ten, P(C) = 1/12.
1
What is P(B|C)?
2
What is P(C|B)?
J. Robert Buchanan
Conditional Probability
Conditional Probability
Definition
The notation P(B|A) is read as the probability that event B
occurs given that event A has occurred.
Example
Roll a pair of fair dice and observe the outcomes.
B: doubles, P(B) = 1/6,
C: sum of the dice is a ten, P(C) = 1/12.
1
What is P(B|C)?
2
What is P(C|B)?
1/3
J. Robert Buchanan
Conditional Probability
Conditional Probability
Definition
The notation P(B|A) is read as the probability that event B
occurs given that event A has occurred.
Example
Roll a pair of fair dice and observe the outcomes.
B: doubles, P(B) = 1/6,
C: sum of the dice is a ten, P(C) = 1/12.
1
What is P(B|C)?
1/3
2
What is P(C|B)?
1/6
J. Robert Buchanan
Conditional Probability
Example
Pollsters asked 300 TV viewers if they were satisfied with the
TV coverage of a recent natural disaster. The results are listed
below.
Satisfied
Not Satisfied
Female
80
120
Male
55
45
1
Find P(satisfied).
2
Find P(satisfied | female).
3
Find P(satisfied | male).
4
Is the event “satisfied” independent of gender?
J. Robert Buchanan
Conditional Probability
Example
Pollsters asked 300 TV viewers if they were satisfied with the
TV coverage of a recent natural disaster. The results are listed
below.
Satisfied
Not Satisfied
Female
80
120
Male
55
45
135/300 = 0.450
1
Find P(satisfied).
2
Find P(satisfied | female).
3
Find P(satisfied | male).
4
Is the event “satisfied” independent of gender?
J. Robert Buchanan
Conditional Probability
Example
Pollsters asked 300 TV viewers if they were satisfied with the
TV coverage of a recent natural disaster. The results are listed
below.
Satisfied
Not Satisfied
Female
80
120
Male
55
45
135/300 = 0.450
1
Find P(satisfied).
2
Find P(satisfied | female).
3
Find P(satisfied | male).
4
Is the event “satisfied” independent of gender?
J. Robert Buchanan
80/200 = 0.400
Conditional Probability
Example
Pollsters asked 300 TV viewers if they were satisfied with the
TV coverage of a recent natural disaster. The results are listed
below.
Satisfied
Not Satisfied
Female
80
120
Male
55
45
135/300 = 0.450
1
Find P(satisfied).
2
Find P(satisfied | female).
3
Find P(satisfied | male).
4
Is the event “satisfied” independent of gender?
J. Robert Buchanan
80/200 = 0.400
55/100 = 0.550
Conditional Probability
Conditional Probability Rule
Theorem (Conditional Probability Rule)
If A and B are any two events, then
P(B|A) =
J. Robert Buchanan
P(A and B)
P(A)
Conditional Probability
Example (1 of 2)
In a sample of 150 residents, each person was asked whether
he or she was in favor of a change in the county charter. The
county is composed of one large city and many suburban
townships. The results are listed in the table below.
In City
Outside City
J. Robert Buchanan
Favor
80
20
Oppose
40
10
Conditional Probability
Example (2 of 2)
1
What is P(oppose)?
2
What is P(favor | in city)?
3
What is P(favor | outside city)?
J. Robert Buchanan
Conditional Probability
Example (2 of 2)
50/150 = 0.333
1
What is P(oppose)?
2
What is P(favor | in city)?
3
What is P(favor | outside city)?
J. Robert Buchanan
Conditional Probability
Example (2 of 2)
50/150 = 0.333
1
What is P(oppose)?
2
What is P(favor | in city)?
P(favor | in city) =
=
3
P(favor and in city)
P(in city)
80/150
80
=
= 0.667
120/150
120
What is P(favor | outside city)?
J. Robert Buchanan
Conditional Probability
Example (2 of 2)
50/150 = 0.333
1
What is P(oppose)?
2
What is P(favor | in city)?
P(favor | in city) =
=
3
P(favor and in city)
P(in city)
80/150
80
=
= 0.667
120/150
120
What is P(favor | outside city)?
P(favor | outside city) =
=
J. Robert Buchanan
P(favor and outside city)
P(outside city)
20/150
20
=
= 0.667
30/150
30
Conditional Probability
General Multiplication Rule
Theorem (General Multiplication Rule)
The probability that two events A and B both occur is
P(A and B) = P(A) · P(B|A).
J. Robert Buchanan
Conditional Probability
General Multiplication Rule
Theorem (General Multiplication Rule)
The probability that two events A and B both occur is
P(A and B) = P(A) · P(B|A).
Note: if A and B are independent then
P(A and B)
P(A)
P(A) · P(B)
=
P(A)
= P(B).
P(B|A) =
Likewise if A and B are independent then P(A|B) = P(A).
J. Robert Buchanan
Conditional Probability
Other Forms
All of the following are equivalent ways of stating the General
Multiplication Rule. We choose the one to use depending on
which probabilities we know or can easily determine.
P(A and B) = P(B) · P(A|B)
P(A and B)
P(A|B) =
P(B)
P(A and B) = P(A) · P(B|A)
P(A and B)
P(B|A) =
P(A)
J. Robert Buchanan
Conditional Probability
Example
A box contains 25 parts, of which 3 are defective and 22 are
non-defective. If 2 parts are randomly selected without
replacement, the probability that
1
both are defective,
2
neither is defective,
3
exactly one is defective.
J. Robert Buchanan
Conditional Probability
Example
A box contains 25 parts, of which 3 are defective and 22 are
non-defective. If 2 parts are randomly selected without
replacement, the probability that
1
both are defective,
P(both defective) =
2
neither is defective,
3
exactly one is defective.
J. Robert Buchanan
3 2
= 0.010
25 24
Conditional Probability
Example
A box contains 25 parts, of which 3 are defective and 22 are
non-defective. If 2 parts are randomly selected without
replacement, the probability that
1
both are defective,
P(both defective) =
2
3 2
= 0.010
25 24
neither is defective,
P(neither defective) =
3
22 21
= 0.770
25 24
exactly one is defective.
J. Robert Buchanan
Conditional Probability
Example
A box contains 25 parts, of which 3 are defective and 22 are
non-defective. If 2 parts are randomly selected without
replacement, the probability that
1
both are defective,
P(both defective) =
2
3 2
= 0.010
25 24
neither is defective,
P(neither defective) =
3
22 21
= 0.770
25 24
exactly one is defective.
P(exactly one defective) = 1 − 0.010 − 0.770 = 0.220
J. Robert Buchanan
Conditional Probability