Download Conditional Probability and Bayes` Theorem.

Conditional Probability
and Independent Events
Conditional Probability
• if we have some information about the
result…use it to adjust the probability
• likelihood an event E occurs under the
condition that some event F occurs
• notation: P(E | F )
"the probability of E, given F ".
• called a “conditional probability”
Given They’re Male
• If an individual is selected at random, what is the
probability a sedan owner is selected, given that
the owner is male?
• P( sedan owner | male ) = _______?
Smaller Sample Space
• Given the owner is male reduces the total
possible outcomes to 115.
n  sedan  male 
P( sedan | male) 
n(male)
In general...
• In terms of the probabilities, we define
P( A  B)
P( A | B) 
P( B)
sedan mini-van truck totals
male
female
.16
.24
.40
.10
.22
.32
.20
.08
.28
• P( sedan owner | male ) = _______?
.46
.54
1.00
Compute the probability
sedan mini-van truck totals
male
female
.16
.24
.40
.10
.22
.32
.20
.08
.28
.46
.54
1.00
P(van  female)
P(van | female) 
?
P( female)
P( female  van)
P( female | van) 
?
P(van)
Compare
• NOT conditional:
P( truck ) =
• Are Conditional:
P( truck | male ) =
Dependent Events?
• probability of owning a truck…
• ...was affected by the knowledge the owner is
male
• events "owns a truck" and "owner is male" are
called dependent events.
Independent Events
• Two events E and F , are called
independent if
or simply
the probability of E is unaffected by event F
Roll the Dice
• Using the elements of the sample space:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
• Compute the conditional probability:
P( sum = 6 | a “4 was rolled” ) = ?
• are the events “sum = 6" and
“a 4 was rolled" independent events?
“Affected”
• The events are NOT independent
• the given condition does have an effect.
• That is,
P(sum = 6 | 4 is rolled ) = 2/11 = 0.1818
but
P(sum = 6) = 5/36 = 0.1389
• These are dependent events.
Not Independent
• These are dependent events.
• As a result,
P(sum = 6 and a 4 was rolled)
does not equal
P(sum = 6)P(a 4 was rolled) ?
2
 0.0555
36
 5   11 
     0.0424
 36   36 
Probability of “A and B”
P( A  B)
P( A | B) 
P( B)
may be written equivalently as
P( A  B)  P( A | B) P( B)
• Draw two cards in succession,
without replacing the first card.
• P(drawing two spades) = ________?
Multiplication Rule
(spade, spade)
P(1st is spade  2nd is spade)
 P(2
nd
st
st
spade |1 spade) P(1 spade)
Compare with “combinations approach”, ( 13C2 )( 52C2 ).
Multiplicative Law for Probability
• For two events A and B,
P( A  B)  P( A | B) P( B)
 P( B | A) P( A)
• But when A and B are independent events,
this identity simplifies to
P( A  B)  P( A) P( B)
Example
• In a factory, 40% of items produced come
from Line 1 and others from Line 2.
• Line 1 has a defect rate of 8%.
Line 2 has a defect rate of 10%.
• For randomly selected item, find probability
the item is not defective.
D: the selected item is defective
(i.e., ~D means not defective)
The Decision Tree
not defective
Line 1
defective
not defective
Line 2
defective
P(~ D)  P(~ D | L1 ) P( L1 )  P(~ D | L2 ) P( L2 )
P(~ D)  (0.92)(0.40)  (0.90)(0.60)  0.908
The Two Lines
• ~D: the selected item is not defective.
S
~D
L1
L2
P(~ D)  P(~ D  L1 )  P(~ D  L2 )
 P(~ D | L1 ) P( L1 )  P(~ D | L2 ) P( L2 )
Total Probability
S
A
B1
For the partition {B1 , B2 ,
B2
…
Bk
, Bk } of the sample space S ,
we may write A  ( A  B1 )  ( A  B2 ) 
and so
P(A)  P( A  B1 )  P( A  B2 ) 
or equivalently,
P(A)  P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 ) 
 ( A  Bk )
 P( A  Bk ).
 P( A | Bk ) P( Bk ).
Total Probability
B1
P(A|B1)P(B1)
A
P(A|B2)P(B2)
A
P(A|B3)P(B3)
A
B2
B3
A
A
A
P(A)
 P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )  P( A | B3 ) P( B3 ).
Bayes’ Theorem follows…
P( B j | A) 
P( A  B j )
P( A)

P( A | B j ) P( B j )
Since P(A)  P( A | B1 ) P( B1 ) 
P( A)
 P( A | Bk ) P( Bk ),
we also have

P( A | B j ) P( B j )
P( A | B1 ) P( B1 ) 
 P( A | Bk ) P( Bk )
Bayes’
B1
P(A|B1)P(B1)
A
P(A|B2)P(B2)
A
P(A|B3)P(B3)
A
B2
B3
A
A
A
P( A | B2 ) P( B2 )
P( B2 | A) 
P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )  P( A | B3 ) P( B3 )
Back at the factory…
• For randomly selected item, find probability
it came from Line 1, given the item is not
defective.
P( L1 | W ) =
not defective
Line 1
defective
not defective
Line 2
defective
The 3 Urns
• Three urns contain colored balls.
Urn
1
2
3
Red
3
1
4
White
4
2
3
Blue
1
3
2
• An urn is selected at random and one ball is
randomly selected from the urn.
• Given that the ball is red, what is the
probability it came from urn #2 ?