Law of Total Probability and Bayes’ Rule

• 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.

• 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

•
A : the selected item is not defective.
S
A
B
1
B
2
A

( A

B
1
)

( A

B
2
)
•
B
1
: item came from Line 1.
B
2
: item came from Line 2.

• So we may write
A

( A

B
1
)

( A

B
2
)
• Since this is the union of disjoint sets, the Additive Law yields
( )

(

B
1
)

(

B
2
)
• Or, in terms of conditional probabilities
P A

P A B P B
1 1
)

P A B P B
2
) (
2
)
( )

(0.08)(0.40)

(0.10)(0.60)

0.092

Line 1
defective not defective
Line 2 defective not defective

• 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?

•
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
A i
: i th donor has correct blood type
A

( A
1
)

( A
1

A
2
)

( A
1

A
2

A
3
)

( A
1

A
2

A
3

A
4
)
4 mutually exclusive events

A

( )
1

( A
1

A
2
)

( A
1

A
2

A
3
)

( A
1

A
2

A
3

A
4
)
• Trials are independent and each
P ( A i
) = 0.40, and so
( )

(
1
)

(
1
) (
2
)

(
1
) (
2
) (
3
)

P A P A P A P A
1 2 3 4
)
( )

0.4

(0.6)(0.4)

(0.6)(0.6)(0.4)

(0.6)(0.6)(0.6)(0.4)

.8704

A
1 saved!
A
2 saved!
A
1
A
3
A
2
Or, more simply,
 
P (donor is not found)
 
P A P A P A P A
1 2 3 4
)
 
6)
4
A
3 saved!
A
4
A
4 saved!
too late!

• 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 4 th test?
B. What is the probability no more than 4 need to be tested to identify both defective refrigerators?

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 3 rd or 4 th test?

S
B
1
B
2
…
B k
A collection of sets { ,
1 2
, , B k
} such that
1. S

B
1

B
2
 
B k
, and
2. B i

B j
 
, for i
 j ,

S
A
B
1
B
2
…
B k
B B
1 2
, , B k
S we may write A

( A

B
1
)

( A

B
2
)
 
( A

B k
) and so
( )

(

B
1
)

(

B
2
)
 
(

B k
).

•
A : the selected item is not defective.
B
1
A
Not defective
B
2
S
A

( A

B
1
)

( A

B
2
)
•
B
1
: item came from Line 1.
B
2
: item came from Line 2.

B
1
A
B
2
…
B k
S
P B i

0, then
(

B i
)

( | i
) ( ) i
B B
1 2
, , B k
S we have ( )

(

B
1
)

P A

B
2
)
 
(

B k
) or equivalently,
P A

P A B P B
1 1
)

P A B P B
2
) (
2
)
 
( | k
) ( k
).

A P ( A | B
1
) P ( B
1
)
B
1
A
A P ( A | B
2
) P ( B
2
)
B
2
B
3
A
A P ( A | B
3
) P ( B
3
)
A

( |
1
) (
1
)

( |
2
) (
2
)

( |
3
) (
3
).

Since ( )

( |
1
) (
1
)
 
( | k
) ( k
), we also have
P B j
A

(

B j
)

P A B P B
1 1
)
(

B j
 
)
( | k
) ( k
) or simply,
P B j
A
 i k 

1
(

B j
)
( | i
) ( i
)

A P ( A | B
1
) P ( B
1
)
B
1
A
A P ( A | B
2
) P ( B
2
)
B
2
B
3
A
A P ( A | B
3
) P ( B
3
)
A
P B
2
A

(

B
2
)
P A B P B
1
)

P A B P B
2
)

P A B P B
3
)

• Three machines M
1 ohm” resistors.
, M
2
, and M
3 produce “1000-
• M
1 produces 80% of resistors accurate to within
50 ohms, M
2 produces 90% to within 50 ohms, and M
3 produces 60% to within 50 ohms.
• Each hour, M
1 produces 3000 resistors, M
2 produces 4000, and M
3 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?

• Define A: resistor is accurate to within 50 ohms.
• M
1 produces 80% of resistors accurate to within
50 ohms, M
2 produces 90% to within 50 ohms, and M
3 produces 60% to within 50 ohms.
P A M
1
)

0.80, P A M
2
)

0.90,
P A M
3
)

0.60
• Each hour, M
1 produces 3000 resistors, M
2 produces 4000, and M
3 produces 3000.
P M
1
)

0.3, P M
2
)

P M
3
)

0.30.

Since
P A

P A M P M
1
)

P A M P M
2
)

P A M P M
3
) we have
( )

(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.

M
1
M
2
M
3
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

• Determine the probability that, given a selected resistor is accurate to within 50 ohms, it was produced by M
1
. P( M
1
| A) = ?
• Determine the probability that, given a selected resistor is accurate to within 50 ohms, it was produced by M
3
. P( M
3
| A) = ?

M
1
M
2
M
3
A
A
A
A
A
A
(0.8)(0.3)
(0.9)(0.4)
P (M | )
1
A

P(M
1

A )
(0.6)(0.3)

(0.8)(0.3)
0.78

• 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 ) = ?

• 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 ) = ?
Has arthritis
0.1
No arthritis
0.9
positive
0.85
negative positive
0.04
negative

• Three urns contain colored balls.
Urn
1
2
3
Red White Blue
3 4 1
1 2
4 3
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 ?

• 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.

• The likelihood of finding defective bulbs is dependent on the bulb quality:
Number of defects Low Medium High
0
1
2
.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?