051 probability

7/27/2019 051 Probability

1/311

ProbabilityProbabilityProbabilityProbabilityProbabilityProbabilityProbabilityProbability

100 Rolls 1000 Rolls

Outcome Frequency Outcome Frequency0 80 0 690

1 19 1 282

2 1 2 28

The Nature of Probability

Example: Consider an experiment in which weroll two six-sided fair dice and record thenumber of 3s face up. The only possibleoutcomes are zero 3s, one 3, or two 3s. Hereare the results after 100 rolls, and after 1000rolls:


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Using a Histogram

We can express these results (from the 1000rolls) in terms of relative frequencies and displaythe results using a histogram:

0 1 2

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Relative

Frequency

Threes Face Up

Continuing the Experiment

If we continue this experiment for severalthousandmore rolls:

1. The frequencies will have approximately a25:10:1 ratio in totals

2. The relative frequencies will settle down

Use random number tables

Use a computer to randomly generate number values representing the

various experimental outcomes

Key to either method is to maintain the probabilities

Note: We can simulate many probability experiments:


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Probability of Events

Probability that an Event Will Occur: Therelative frequency with which that event canbe expected to occur

The probability of an event may beobtained in three different ways:

Empirically

Theoretically

Subjectively

Experimental or Empirical Probability

Question: What happens to the observedrelative frequency as n increases?

Experimental or Empirical Probability:

1. The observed relative frequency with which an eventoccurs

2. Prime notation is used to denote empirical probabilities:

3. n(A): number of times the event A has occurred

4. n: number of times the experiment is attempted

=Pn

n( )

( )A

A


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Example

Example: Consider tossing a fair coin. Define the eventH as the occurrence of a head. What is the probability ofthe event H, P(H)?

Notes:

This does not mean exactly one head will occur in everytwo tosses of the coin

In the long run, the proportion of times that a head willoccur is approximately 1/2

1. In a single toss of the coin, there are two possible outcomes

2. Since the coin is fair, each outcome (side) should have an equally likely

chance of occurring

3. Intuitively, P(H) = 1/2 (the expected relative frequency)

Long-Run Behavior

To illustrate the long-run behavior:

1.Consider an experiment in which we toss the coinseveral times and record the number of heads

2.A trial is a set of 10 tosses

3.Graph the relative frequency and cumulative relativefrequency of occurrence of a head

4.A cumulative graph demonstrates the idea of long-runbehavior

5.This cumulative graph suggests a stabilizing, or settlingdown, effect on the observed cumulative probability

6.This stabilizing effect, or long-term average value, isoften referred to as the law of large numbers


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Number of Relative Cumulative

Trial Heads Observed Frequency Relat ive Frequency

1 5 5/10 5/10 = 0.5000

2 4 4/10 9/20 = 0.4500

3 4 4/10 13/30 = 0.4333

4 5 5/10 18/40 = 0.4500

5 6 6/10 24/50 = 0.4800

6 7 7/10 28/60 = 0.4667

7 6 6/10 34/70 = 0.4857

8 4 4/10 38/80 = 0.4750

9 7 7/10 45/90 = 0.5000

10 3 3/10 48/100 = 0.4800

11 4 4/10 52/110 = 0.4727

12 6 6/10 58/120 = 0.4838

13 7 7/10 65/130 = 0.5000

14 4 4/10 69/140 = 0.4929

15 3 3/10 72/150 = 0.4800

16 7 7/10 79/160 = 0.4938

17 6 6/10 85/170 = 0.5000

18 3 3/10 88/180 = 0.4889

19 6 6/10 94/190 = 0.4947

20 4 4/10 98/200 = 0.4900

Experiment

Experimental results of tossing a coin 10times each trial:

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 5 10 15 20 25

Expected value = 1/2

Trial

Relative Frequency


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Cumulative Relative Frequency

0.4

0.45

0.5

0.55

0.6

0 5 10 15 20 25

Expected value = 1/2

Trial

Law of Large Numbers

Law of Large Numbers: If the number of times an

experiment is repeated is increased, the ratio of the

number of successful occurrences to the number of

trials will tend to approach the theoretical

probability of the outcome for an individual trial Interpretation: The law of large numbers says: the

larger the number of experimental trials n, the closer

the empirical probability P(A) is expected to be to thetrue probability P(A)

)A()A(', PPn In symbols: As


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Simple Sample Spaces

We need to talk about data collectionand experimentation more precisely

With many activities, like tossing a coin,rolling a die, selecting a card, there isuncertainty as to what will happen

We will study and characterize thisuncertainty

Experiment & Outcome

Experiment: Any process that yields a result or anobservation

Outcome: A particular result of an experiment

Example: Suppose we select two students at randomand ask each if they have a car on campus:

1. A list of possible outcomes: (Y, Y), (Y, N), (N, Y), (N, N)

2. This is called ordered pair notation3. The outcomes may be displayed using a tree diagram


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Tree Diagram

1. This diagram consists of four branches: 2 first generation branches

and 4 second generation branches

2. Each branch shows a possible outcome

Student 1 Student 2 Outcomes

Y Y, Y

Y

N Y, N

Y N, Y

N

N N, N

Sample Space & EventSample Space: The set of all possible outcomes of anexperiment. The sample space is typically called Sandmay take any number of forms: a list, a tree diagram, alattice grid system, etc. The individual outcomes in asample space are called sample points. n(S) is the numberof sample points in the sample space.

Event: Any subset of the sample space. If A is an event,then n(A) is the number of sample points that belong to A

Example: For the student car example above:

S= { (Y, Y), (Y, N), (N, Y), (N, N) }

n(S) = 4


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Examples

Example: An experiment consists of two trials.The first is tossing a penny and observing ahead or a tail; the second is rolling a die andobserving a 1, 2, 3, 4, 5, or 6. Construct thesample space:

Example: Three voters are randomly selected andasked if they favor an increase in property taxesfor road construction in the county. Construct the

sample space:

S= { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 }

S= { NNN, NNY, NYN, NYY, YNN, YNY, YYN, YYY}

Example

Example: An experiment consists of selecting

electronic parts from an assembly line and testing

each to see if it passes inspection (P) or fails (F).

The experiment terminates as soon as one

acceptable part is found or after three parts are

tested. Construct the sample space: Outcome

F FFF

FF P FFP

P FP

P P

S= { FFF, FFP, FP, P }


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Example

Example: The 1200 students at a localuniversity have been cross tabulatedaccording to resident and college status:

Arts and Sciences Business

Resident 600 280

Nonresident 175 145

The experiment consists of selecting onestudent at random from the entire student body

n(S) = 1200

Coffee

Bagel

1832 16

11

Example Example: On the way to work, some employees at a

certain company stop for a bagel and/or a cup ofcoffee. The accompanying Venn diagram summarizesthe behavior of the employees for a randomly selectedwork day:

The experiment consists of selecting one employee at random

n(S) = 77


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Notes

1. The outcomes in a sample space cannever overlap

2. All possible outcomes must berepresented

3. These two characteristics are calledmutually exclusiveand all inclusive

Rules of Probability

Consider the concept of probabilityand relate it to the sample space

Recall: the probability of an event is therelative frequency with which the event

could be expected to occur, the long-termaverage


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Equally Likely Events

1. In a sample space, suppose all sample pointsare equally likely to occur

2. The probability of an event A is the ratio of thenumber of sample points in A to the numberof sample points in S

4. This formula gives a theoretical probability valueof event As occurrence

3. In symbols: Pn

n S( )

( )

( )A

A=

5. The use of this formula requires the existence of

a sample space in which each outcome isequally likely

Example

Example: A fair coin is tossed 5 times, and ahead (H) or a tail (T) is recorded each time.What is the probability of:

A = {exactly one head in 5 tosses}, and

B = {exactly 5 heads}? The outcomes consist of a sequence of 5 Hs and Ts

A typical outcome includes a mixture of Hs and Ts, like: HHTTH

There are 32 possible outcomes, all equally likely

A = {HTTTT, THTTT, TTHTT, TTTHT, TTTTH}

B = {HHHHH}

Pn

n S( )

( )

( )A

A= =

5

32

Pn

n S( )

( )

( )B

B= =

1

32


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Subjective Probability

1. Suppose the sample space elements are notequally likely, and empirical probabilities cannot

be used

2. Only method available for assigning probabilities

may be personal judgment

3. These probability assignments are called

subjective probabilities

4. Personal judgment of the probability is

expressed by comparing the likelihood among

the various outcomes

Basic Probability Ideas

1. Probability represents a relative frequency

2. P(A) is the ratio of the number of times an eventcan be expected to occur divided by the numberof trials

3. The numerator of the probability ratio must be apositive number or zero

4. The denominator of the probability ratio must be

a positive number (greater than zero)5. The number of times an event can be expected

to occur in ntrials is always less than or equal tothe total number of trials, n


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Properties

P( )Aall outcomes

= 1

2. The sum of the probabilities of all outcomesin the sample space is 1:

Notes:

The probability is zero if the event cannot occur

The probability is one if the event occurs every time

(a sure thing)

0 1 P( )A1. The probability of any event A is between

0 and 1:

P P( ) ( )T J E C= 4

P P( ) ( )T J E C+ = 1

4 1

5 1

1

5

4 41

5

4

5

+ =

=

=

= =

=

P P

P

P

P P

( ) ( )

( )

( )

( ) ( )

E C E C

E C

E C

T J EC

Example Example: On the way to work Bobs personal

judgment is that he is four times more likely to getcaught in a traffic jam (TJ) than have an easy commute(EC). What values should be assigned to P(TJ) andP(EC)?


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Odds

Odds: another way of expressingprobabilities

If the odds in favor of an eventA are ato b,then:

Pa

a b( )A =

+2.The probability of event A is:

P ba b

( )A does not occur =+

3.The probability that event A will not occur is

1.The odds against A are bto a

Example

Example: The odds in favor of you passing anintroductory statistics class are 11 to 3. Find theprobability you will pass and the probability you

will fail.

P( )pass =+

=11

11 3

11

14P( )fail =

+

=3

11 3

3

14

Using the preceding notation: a= 11 and b= 3:


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Complement of An Event

Complement of an Event: The set of all samplepoints in the sample space that do not belong toevent A. The complement of event A is denoted by

(read A complement).

Example Example:

1.The complement of the event success is failure

2.The complement of the event rain is no rain

3.The complement of the event at least 3 patientsrecover out of 5 patients is 2 or fewer recover

Notes:

Every event A has a complementary event

Complementary probabilities are very useful when thequestion asks for the probability of at least one.

P P( ) ( )A A for any event A+ = 1

P P( ) ( )A A= 1

A


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Example

Example: A fair coin is tossed 5 times, and a head(H) or a tail(T) is recorded each time. What is the probability of

1) A = {at least one head in 5 tosses}

2) B = {at most 3 heads in 5 tosses}

P P

P

( ) ( )

( )

A A

0 heads in 5 tosses

=

=

= =

1

1

11

32

31

32

Solutions:

1)

P P

P

P P

( ) ( )

( )

( ( ) ( ))

B B

4 or 5 heads

4 heads 5 heads

=

=

= +

= +

= = =

1

1

1

15

32

1

321

6

32

26

32

13

16

2)

Manual Automatic

Transmission Transmission

2-door 75 155

4-door 85 170

Example Example: A local automobile dealer classifies

purchases by number of doors and transmission type.The table below gives the number of eachclassification.

If one customer is selected at random, find the probability

that:1) The selected individual purchased a car with

automatic transmission

2) The selected individual purchased a 2-door car


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Solutions

P( )Automatic Transmission

=+

+ + += =

155 170

75 85 155 170

325

485

65

97

1)

P( )2-door

=+

+ + += =

75 155

75 85 155 170

230

485

46

97

2)

Mutually Exclusive Events & the Addition Rule

Compound Events: formed by combiningseveral simple events:

The probability that either event A or eventB will occur: P(A or B)

The probability that both events A and Bwill occur: P(A and B)

The probability that event A will occurgiven that event B has occurred: P(A | B)


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Mutually Exclusive Events

Mutually Exclusive Events: Events defined insuch a way that the occurrence of one eventprecludes the occurrence of any of the otherevents. (In short, if one of them happens, theothers cannot happen.)

Notes:

Complementary events are also mutuallyexclusive

Mutually exclusive events are not necessarilycomplementary

All-Day Half-Day

Pass Pass Total

Male 1200 800 2000

Female 900 700 1600

Total 2100 1500 3600

Example

Example: The following table summarizesvisitors to a local amusement park:

One visitor from this group is selected at random:1) Define the event A as the visitor purchased an all-day pass

2) Define the event B as the visitor selected purchased a half-day pass

3) Define the event C as the visitor selected is female


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Solutions

1) The events A and B are mutually exclusive

A C

9001200 700

800

3)

2) The events A and C are notmutually exclusive.The intersection of A and C can be seen in thetable above or in the Venn diagram below:

P P P P( ) ( ) ( ) ( )A or B A B A and B= +

General Addition Rule

General Addition Rule: Let A and B be two eventsdefined in a sample space S:

A B Illustration:

Note:If two events A and B are mutually exclusive:

P( )A and B = 0


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Special Addition Rule

Special Addition Rule: Let A and B be two events definedin a sample space. If A and B are mutually exclusiveevents, then:

P P P( ) ( ) ( )A or B A B= +

AB

CD

This can be expanded to consider more than twomutually exclusive events:

P(A or B or C) = P(A) + P(B) + + P(E)

P P( ) ( ) 0.A A= = =1 1 15 0.85

P( )A and B (A and B are mutually exclusive )= 0

P P P( ) ( ) ( )B or C B C= + = + =0.40 0.25 0.65

Example

Example: All employees at a certain companyare classified as only one of the following:manager (A), service (B), sales (C), or staff (D).

It is known that P(A) = 0.15, P(B) = 0.40,P(C) = 0.25, and P(D) = 0.20

P P P P( ) ( ) ( ) ( )A or B or C A B C= + + = 0.15 + 0.40 + 0.25 = 0.80


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Example

Example: A consumer is selected at random. The probability theconsumer has tried a snack food (F) is 0.5, tried a new soft drink(D) is 0.6, and tried both the snack food and the soft drink is 0.2

P

P P P P

( )

( ) ( ) ( ) ( )

Tried the snack food or the soft drink

F or D F D F and D= = + = 0.5 + 0.6 0.2 = 0.9

P P P( ) ( ) ( )Not tried the snack food F F= = = 1 1 0.5 = 0.5

P P P( ) ( ) ( )Tried only the soft drink D F and D= = 0.6 0.2 = 0.4

P

P P

( )

[( )] ( )

Tried neither the snack food nor the soft drink

F or D F or D= = = 1 1 0.9 = 0.1

Independence, the MultiplicationRule, & Conditional Probability

Independent Events: Two events A and Bare independent events if the occurrence(or nonoccurrence) of one does not affectthe probability assigned to the occurrenceof the other.

Note:If two events are notindependent,they are dependent


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Conditional Probability

1. Sometimes two events are related in such a way that theprobability of one depends upon whether the secondevent has occurred

Conditional Probability: The symbol P(A | B) representsthe probability that A will occur given B has occurred. Thisis called conditional probability.

PP

P( | )

( )

( )A B

A and B

B=1.

2. Partial information may be relevant to the probabilityassignment

2. Given B has occurred, the relevant sample space is nolonger S, but B (reduced sample space)

3. A has occurred if and only if the event A and B hasoccurred

Independent Events

Independent Events: Two events A and B areindependent events if:

Notes:

If A and B are independent, the occurrence of Bdoes not affect the occurrence of A

If A and B are independent, then so are:A and B

A and B

A and B

P(A | B) = P(A) or P(B | A) = P(B)


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PP

P( | )

( )

( )

/

/A B

A and B

B= = =

1 6

3 6

1

3

PP

P( | )

( )

( ) /A C

A and C

C= = =

0

3 60

P P

P( | ) ( )

( )//

B A B and AA

= = =1 61 6

1

Example

Example: Consider the experiment in which asingle fair die is rolled: S= {1, 2, 3, 4, 5, 6 }.Define the following events:

A = a 1 occurs, B = an odd number occurs,and C = an even number occurs

Age Favor (F) Oppose Total

Less than 30 (Y) 250 50 300

30 to 50 (M) 600 75 675

More than 50 (O) 100 125 225

Total 950 250 1200If one resident is selected at random, what is the probability the resident will:

1) Favor the new playground?

2) Favor the playground if the person selected is less than 30?

3) Favor the playground if the person selected is more than 50?

4) Are the events F and M independent?

Example Example:In a sample of 1200 residents, each

person was asked if he or she favored building a

new town playground. The responses are

summarized in the table below:


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Solutions

P( )F = =9501200

1924

1)

PP

P( | )

( )

( )

/

/F Y

F and Y

Y 6= = =

250 1200

300 1200

52)

PP

P( | )

( )

( )

/

/F O

F and O

O= = =

100 1200

225 1200

4

93)

PP

PP( | )

( )

( )

/

/( )F M

F and M

MF

F and M are dependent

= = = 600 1200

675 1200

8

94)

General Multiplication Rule

General Multiplication Rule: Let A and B be twoevents defined in sample space S. Then:

Notes:How to recognize situations that result in the compound

event and:

A followed by B

A and B occurred simultaneously The intersection of A and B

Both A and B

A but not B (equivalent to A and not B)

P P P( ) ( ) ( | )A and B A B A=

P P P( ) ( ) ( | )A and B B A B= or


26/3126

P P P( ) ( ) ( )A and B A B=

Special Multiplication Rule

Special Multiplication Rule: Let A and B be two eventsdefined in sample space S. If A and B are independentevents, then:

P P P P P( ) ( ) ( ) ( ) ( )A and B and C and ... and G A B C G= L

This formula can be expanded. If A, B, C, , G areindependent events, then

Example: Suppose the event A is Allen gets a cold this winter, B is

Bob gets a cold this winter, and C is Chris gets a cold this winter.P(A) = 0.15, P(B) = 0.25, P(C) = 0.3, and all three events are

independent. Find the probability that:

1. All three get colds this winter

2. Allen and Bob get a cold but Chris does not

3. None of the three gets a cold this winter

Solutions

P

P P P P

( )

( ) ( ) ( ) ( )

All three get colds this winter

A and B and C A B C= =

1)

= (0.15)(0.25)(0.30) = 0.0113

P

P P P P

(

( ) ( ) ( ) ( )

Allen and Bob get a cold, but Chris does not)


2)

= (0.15)(0.25)(0.70) = 0.0263

P

P P P P

(

( ) ( ) ( ) ( )

None of the three gets a cold this winter)


3)

= (0.85)(0.75)(0.70) = 0.4463


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Notes

1. Independence and mutually exclusive are two very different conceptsa. Mutually exclusive says the two events cannot occur

together, that is, they have no intersection

b. Independence says each event does not affect the otherevents probability

2. P(A and B) = P(A) P(B) when A and B are independent

a. Since P(A) and P(B) are not zero, P(A and B) is nonzero

b. Thus, independent events have an intersection

3. Events cannot be both mutually exclusive and independent

a. If two events are independent, then they are not mutually exclusive

b. If two events are mutually exclusive, then they are not independent

Combining the Rules of Probability

Many probability problems can berepresented by tree diagrams

Using the tree diagram, the addition andmultiplication rules are easy to apply


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Example

Example: A certain company uses three overnight deliveryservices: A, B, and C. The probability of selecting service A is1/2, of selecting B is 3/10, and of selecting C is 1/5. Suppose theevent T is on time delivery. P(T|A) = 9/10, P(T|B) = 7/10, andP(T|C) = 4/5. A service is randomly selected to deliver a packageovernight. Construct a tree diagram representing this experiment.

Notes:

A set of branches that initiate from a single point has a totalprobability of 1

Each outcome for the experiment is represented by a branch thatbegins at the common starting point and ends at the terminal points atthe right

T

T

T

1 2/

3 10/

1 5/

9 10/

1 10/

7 10/

3 10/

4 5/

1 5/

A

B

C

T

T

T

Solution

The resulting tree diagram:


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P P P( ) ( ) ( | )A and T A T A= = =1

2

9

10

9

20

Using the Tree Diagram

1. The probability of selecting service A and having the packagedelivered on time:

P P P P

P P P P P P

( ) ( ) ( ) ( )

( ) ( | ) ( ) ( | ) ( ) ( | )

T A and T B and T C and T

A T A B T B C T C

= + +

= + +

=

+

+

= + +

=

1

2

9

10

3

10

7

10

1

5

4

5

9

20

21

100

4

2541

50

2. The probability of having the package delivered on time:

Good (G) Defective (D) Total

Line 1 (1) 70 40 110

Line 2 (2) 80 25 105

Total 150 65 215

Example Example: A manufacturer is testing the production of a new product

on two assembly lines. A random sample of parts is selected andeach part is inspected for defects. The results are summarized in the

table below:

Suppose a part is selected at random:1) Find the probability the part is defective

2) Find the probability the part is produced on Line 1

3) Find the probability the part is good or produced on Line 2


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Solutions

P( )D

(total defective divided by total number of parts)

= 65215

1)

P( )1

(total produced by Line 1 divided by total number of parts)

=110

2152)

Pn

n S

P P P

( )( )

( )

( ) ( ) ( )

G or 2G or 2

(total good or produced on Line 2 divided by total parts)

G G and 2

= =

= + = +

175

215

2 150215

105215

80215

3)

= 175215

Example Example: This problem involves testing individuals for the presence

of a disease. Suppose the probability of having the disease (D) is0.001. If a person has the disease, the probability of a positive test

result (Pos) is 0.90. If a person does not have the disease, theprobability of a negative test result (Neg) is 0.95. For a person

selected at random:

1) Find the probability of a negative test result given the person has

the disease

2) Find the probability of having the disease and a positive test result

3) Find the probability of a positive test result


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Resulting Tree Diagram

D

D

Pos

Neg

Pos

Neg

Disease

Test

Result

0.001

0.999

0.90

0.10

0.05

0.95

Solutions

P P( ) ( )Neg| D Pos| D= = 1 11) 0.90 = 0.10

P P P( ) ( ) ( )D and Pos D Pos| D= =2) (0.001)(0.90) = 0.0009

P P P

P P P P

( ) ( ) ( )

( ) ( ) ( ) ( | )

Pos D and Pos D and Pos

D Pos|D D Pos D

= +

= +

=

3)

(0.001)(0.90) +(0.0009)(0.05) = 0.5085

051 probability

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