Download Total Probability and Bayes` Rule

Law of Total Probability
and Bayes’ Rule
“Event-composition method”
• Understand the experiment and sample
points.
• Using set notation, express the event of
interest in terms of events for which the
probability is known.
• Applying probability rules, combine the
known probabilities to determine the
probability of the specified event.
Problem 2.86
• 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.
A: the selected item is not defective
Problem 2.86
• A: the selected item is not defective.
S
A
B1
B2
A  ( A  B1 )  ( A  B2 )
• B1: item came from Line 1.
B2: item came from Line 2.
Problem 2.86
• So we may write A  ( A  B1 )  ( A  B2 )
• Since this is the union of disjoint sets,
the Additive Law yields
P( A)  P( A  B1 )  P( A  B2 )
• Or, in terms of conditional probabilities
P( A)  P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )
P( A)  (0.08)(0.40)  (0.10)(0.60)  0.092
The Decision Tree
defective
Line 1
not defective
Line 2
defective
not defective
Problem 2.94
• Must find blood donor for an accident victim in
the next 8 minutes or else…
• Checking blood types of potential donors requires
2 minutes each and may only be tested one at a
time.
• 40% of the potential donors have the required
blood type.
• What is the probability a satisfactory blood donor
is identified in time to save the victim?
Finding a Donor
• A: blood donor is found within 8 minutes
• Some sample points: “B bad, G good”
A = { (G), (B,G), (B,B,G), (B,B,B,G) }
• Let Ai: ith donor has correct blood type
A  ( A1 )  ( A1  A2 )  ( A1  A2  A3 )
 ( A1  A2  A3  A4 )
4 mutually exclusive events
Finding a Donor
A  ( A1 )  ( A1  A2 )  ( A1  A2  A3 )  ( A1  A2  A3  A4 )
• Trials are independent and each P(Ai) = 0.40,
and so
P( A)  P( A1 )  P( A1 ) P( A2 )  P( A1 ) P( A2 ) P( A3 )
 P( A1 ) P( A2 ) P( A3 ) P( A4 )
P( A)  0.4  (0.6)(0.4)  (0.6)(0.6)(0.4)
 (0.6)(0.6)(0.6)(0.4)
 .8704
Finding a Donor
A1
saved!
A2
saved!
A1
A3
A2
Or, more simply,
P ( A)  1  P (donor is not found)
 1  P( A1 ) P ( A2 ) P ( A3 ) P ( A4 )
 1  (0.6) 4
saved!
A4
saved!
A3
A4
too
late!
Problem 2.96
• Of 6 refrigerators, 2 don’t work.
• The refrigerators are tested one at a time.
• When tested, it’s clear whether it works!
A. What is the probability the last defective
refrigerator is found on the 4th test?
B. What is the probability no more than 4 need
to be tested to identify both defective
refrigerators?
Problem 2.96
C. Given that exactly one defective
refrigerator was found during the first 2
tests, what is the probability the other one
is found on the 3rd or 4th test?
Partition of the Sample Space
S
B1
B2
…
A collection of sets {B1 , B2 ,
1. S  B1  B2 
Bk
, Bk } such that
 Bk , and
2. Bi  B j  , for i  j ,
is called a partition of S .
Union of Disjoint Sets
S
A
B1
B2
For the partition {B1 , 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 ) 
 ( A  Bk )
 P( A  Bk ).
Recall Problem 2.86
• A: the selected item is not defective.
S
A
Not defective
B1
B2
A  ( A  B1 )  ( A  B2 )
• B1: item came from Line 1.
B2: item came from Line 2.
Law of Total Probability
S
A
B1
B2
…
If P( Bi )  0, then
Bk
P( A  Bi )  P( A | Bi ) P( Bi )
Hence, for the partition {B1, B2 , , Bk } of the sample space S ,
we have P(A)  P( A  B1 )  P( A  B2 ) 
 P( A  Bk )
or equivalently,
P(A)  P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 ) 
 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…
Since P(A)  P( A | B1 ) P( B1 ) 
we also have
P( B j | A) 

 P( A | Bk ) P( Bk ),
P( A  B j )
P( A)
P( A  B j )
P( A | B1 ) P( B1 ) 
 P( A | Bk ) P( Bk )
or simply,
P( B j | A) 
P( A  B j )
k
 P( A | B ) P( B )
i 1
i
i
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 | A) 
P( A | B1 ) P( B1 )  P( A | B2 ) P( B2 )  P( A | B3 ) P( B3 )
Making Resistors
• Three machines M1, M2, and M3 produce “1000ohm” resistors.
• M1 produces 80% of resistors accurate to within
50 ohms, M2 produces 90% to within 50 ohms,
and M3 produces 60% to within 50 ohms.
• Each hour, M1 produces 3000 resistors, M2
produces 4000, and M3 produces 3000.
• If all of the resistors are mixed together and
shipped in a single container,
what is the probability a selected resistor is
accurate to within 50 ohms?
Making Resistors
• Define A: resistor is accurate to within 50 ohms.
• M1 produces 80% of resistors accurate to within
50 ohms, M2 produces 90% to within 50 ohms,
and M3 produces 60% to within 50 ohms.
P( A | M1 )  0.80, P( A | M 2 )  0.90,
and P( A | M 3 )  0.60
• Each hour, M1 produces 3000 resistors, M2
produces 4000, and M3 produces 3000.
P( M 1 )  0.3, P( M 2 )  0.40, and P( M 3 )  0.30.
Using Total Probability
Since
P(A)  P( A | M 1 ) P( M 1 )  P( A | M 2 ) P( M 2 )  P( A | M 3 ) P( M 3 )
we have
P(A)  (0.8)(0.3)  (0.9)(0.4)  (0.6)(0.3)
 0.78
That is,
78 % are expected to be accurate
to within 50 ohms.
The Tree
A
M1
P (A)  (0.8)(0.3)
A
A
M2
M3
(0.8)(0.3)
(0.9)(0.4)
 (0.6)(0.3)
 0.78
A
A
A
 (0.9)(0.4)
(0.6)(0.3)
…given it’s within 50 ohms…
• Determine the probability that,
given a selected resistor is accurate to within 50
ohms, it was produced by M1. P( M1 | A) = ?
• Determine the probability that,
given a selected resistor is accurate to within 50
ohms, it was produced by M3. P( M3 | A) = ?
Given A…
A
M1
P(M1 | A)
A
M2
M3
(0.8)(0.3)
A
(0.9)(0.4)
A
(0.6)(0.3)
A
A
P(M1  A)

P( A)
(0.8)(0.3)

0.78
Arthritis
• A test detects a particular type of arthritis for
individuals over 50 years old.
• 10% of this age group suffers from this arthritis.
• For individuals in this age group known to have
the arthritis, the test is correct 85% of the time.
• For individuals in this age group known to NOT
have the arthritis, the test indicates arthritis
(incorrectly!) 4% of the time.
• P( has arthritis | tests positive ) = ?
Arthritis
• 10% of this age group suffers from this arthritis.
P(have arthritis) = 0.10
• For individuals in this age group known to have
the arthritis, the test is correct 85% of the time.
P( tests positive | have arthritis ) = 0.85
• For individuals in this age group known to NOT
have the arthritis, the test indicates arthritis
(incorrectly!) 4% of the time.
P(tests positive | no arthritis ) = 0.04
• P( have arthritis | tests positive ) = ?
P(has arthritis | tests positive ) = ?
positive
Has
arthritis
0.1
0.9
No
arthritis
0.85
negative
positive
negative
0.04
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 ?
The Lost Labels
• A large stockpile of cases of light bulbs, 100
bulbs to a box, have lost their labels.
• The boxes of bulbs come in 3 levels of
quality: high, medium, and low.
• It’s known 50% of the boxes were high
quality, 25% medium, and 25% low.
• Two bulbs will be tested from a box to
check if they’re defective.
Lost Labels…
• The likelihood of finding defective bulbs is
dependent on the bulb quality:
Number of defects
0
1
2
Low
Medium
High
.49
.42
.09
.64
.32
.04
.81
.18
.01
• Given neither bulb is found to be defective,
what is the probability the bulbs came from
a box of high quality bulbs?