Upload
stephany-blake
View
214
Download
0
Embed Size (px)
Citation preview
Law of Total Probabilityand 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
B1 B2
A
• B1: item came from Line 1. B2: item came from Line 2.
1 2( ) ( )A A B A B
Problem 2.86
• Since this is the union of disjoint sets,the Additive Law yields
1 2( ) ( )A A B A B
1 2( ) ( ) ( )P A P A B P A B
• Or, in terms of conditional probabilities
1 1 2 2( ) ( | ) ( ) ( | ) ( )P A P A B P B P A B P B
( ) (0.08)(0.40) (0.10)(0.60) 0.092P A
• So we may write
The Decision Tree
Line 1
Line 2
defective
defective
not 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
1 1 2 1 2 3
1 2 3 4
( ) ( ) ( )
( )
A A A A A A A
A A A A
4 mutually exclusive events
Finding a Donor
1 1 2 1 2 3 1 2 3 4( ) ( ) ( ) ( )A A A A A A A A A A A
• Trials are independent and each P(Ai) = 0.40,and so
1 1 2 1 2 3
1 2 3 4
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
P A P A P A P A P A P A P A
P A P A P A P A
( ) 0.4 (0.6)(0.4) (0.6)(0.6)(0.4)
(0.6)(0.6)(0.6)(0.4)
.8704
P A
Finding a Donor
1A
1A
2A
3A
4A2A
3A
4A
1 2 3 4
4
( ) 1 (donor is not found)
1 ( ) ( ) ( ) ( )
1 (0.
Or, more simply,
6)
P A P
P A P A P A P A
saved!
saved!
saved!
saved!
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 Bk…
1 2
1 2
A collection of sets { , , , } such that
1. , and
2. , for ,
is called a partition of .
k
k
i j
B B B
S B B B
B B i j
S
Union of Disjoint Sets
S
B1 B2 Bk…
1 2
1 2
1 2
For the partition { , , , } of the sample space ,
we may write ( ) ( ) ( )
and so
( ) ( ) ( ) ( ).
k
k
k
B B B S
A A B A B A B
P A P A B P A B P A B
A
Recall Problem 2.86
• A: the selected item is not defective.
S
B1 B2
A
• B1: item came from Line 1. B2: item came from Line 2.
1 2( ) ( )A A B A B Not defective
Law of Total Probability
AS
B1 B2 Bk…
1 2
1 2
Hence, for the partition { , , , } of the sample space ,
we have ( ) ( ) ( ) ( )k
k
B B B S
P A P A B P A B P A B
If ( ) 0, then
( ) ( | ) ( )i
i i i
P B
P A B P A B P B
1 1 2 2
or equivalently,
( ) ( | ) ( ) ( | ) ( ) ( | ) ( ).k kP A P A B P B P A B P B P A B P B
Total Probability
B1
B2
B3
A
A
A
A
A
A
P(A|B1)P(B1)
P(A|B2)P(B2)
P(A|B3)P(B3)
1 1 2 2 3 3
( )
( | ) ( ) ( | ) ( ) ( | ) ( ).
P A
P A B P B P A B P B P A B P B
Bayes’ Theorem follows…1 1Since ( ) ( | ) ( ) ( | ) ( ),
we also havek kP A P A B P B P A B P B
1 1
( )( | )
( )
( )
( | ) ( ) ( | ) ( )
jj
j
k k
P A BP B A
P A
P A B
P A B P B P A B P B
1
or simply,
( ) ( | )
( | ) ( )
jj k
i ii
P A BP B A
P A B P B
Bayes’
B1
B2
B3
A
A
A
A
A
A
P(A|B1)P(B1)
P(A|B2)P(B2)
P(A|B3)P(B3)
22
1 1 2 2 3 3
( )( | )
( | ) ( ) ( | ) ( ) ( | ) ( )
P A BP B A
P A B P B P A B P B P A B P B
Making Resistors• Three machines M1, M2, and M3 produce “1000-
ohm” 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.
• Each hour, M1 produces 3000 resistors, M2 produces 4000, and M3 produces 3000.
1 2
3
( | ) 0.80, ( | ) 0.90,
and ( | ) 0.60
P A M P A M
P A M
1 2 3( ) 0.3, ( ) 0.40, and ( ) 0.30.P M P M P M
Using Total Probability
1 1 2 2 3 3
Since
( ) ( | ) ( ) ( | ) ( ) ( | ) ( )
we have
P A P A M P M P A M P M P A M P M
( ) (0.8)(0.3) (0.9)(0.4) (0.6)(0.3)
0.78
P A
That is, 78 % are expected to be accurate to within 50 ohms.
The Tree
M1
M2
M3
A
A
A
A
A
A
(0.8)(0.3)
(0.9)(0.4)
(0.6)(0.3)
( ) (0.8)(0.3)
(0.9)(0.4)
(0.6)(0.3)
0.78
P A
…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…
M1
M2
M3
A
A
A
A
A
A
(0.8)(0.3)
(0.9)(0.4)
(0.6)(0.3)
1
1
(M | )
P(M )
( )
(0.8)(0.3)
0.78
P A
A
P A
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 ) = ?
Hasarthritis
(has arthritis | tests positive ) = ?P
Noarthritis
positive
negative
positive
negative
0.1
0.90.04
0.85
The 3 Urns
• Three urns contain colored balls. Urn Red White Blue 1 3 4 1 2 1 2 3 3 4 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 Low Medium High
0 .49 .64 .811 .42 .32 .182 .09 .04 .01
• Given neither bulb is found to be defective, what is the probability the bulbs came from a box of high quality bulbs?