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2

CAP 6635 Artificial Intelligence

Homework 4

Question 1 [2 pts]: Table 1 shows probability values of different events. Using the table to calculate following values and show proof:

· The probability that a persona has cavity [0.25 pt]

· The probability of a toothache event [0.25 pt]

· The joint probability of cavity and toothache [0.25 pt]

· Calculate conditional probability of no cavity, given the patient has toothache [0.25 pt]

· Calculate conditional probability of no cavity, given the patient does not have toothache [0.25 pt]

· Determine whether cavity and toothache are independent or not, why [0.25 pt]

· Given a patient has cavity, determine whether the tooth probe catch is conditionally independent of toothache or not, why [0.25 pt]

· Given a patient does not have cavity, determine whether the tooth probe catch is conditionally independent of toothache or not, why [0.25 pt]

Table 1

Question 2 [2 pts]: A patient takes a lab test and the result comes back positive. Assume the test returns a co

ect positive result in only 95% of the cases in which the disease is actually present, and a co

ect negative result in only 95% of the cases in which the disease is not present. Assume further that 0.001 of the entire population have this cancer.

· Use Bayes Rule to derive the probability of any test results being positive [1 pt]

· Use Bayes Rule to derive the probability of the patient having the cancer given that his/her lab test is positive (list the major steps). [1 pt]

Question 3: [2 pts]: Figure 1 shows a Bayesian network, using first letter to denote each named variable, e.g, using T to denote tampering, and complete following questions:

· Show joint probability of the whole Bayesian network [0.25 pt]

· How many probability values are needed (i.e, should be given), in order to calculate the joint probability of the whole network, why [0.25 pt]

· Prove that “Alarm” and “Report” are conditionally independent, given “Learning” [0.25 pt]

· Prove that “Alarm” and “Smoke” are conditionally independent, given “Fire” [0.25 pt]

· Assume x ( y denotes that x are independent of y, x ( y | z denotes that x and y are conditionally independent, given z. Complete Table 2, and use ( to mark co

ect answers. [1 pt]

Table 2

Figure 1

Question 4: [2 pts]: Figure 2 shows a Bayesian network where r denotes “rain”, s denotes “sprinkler”, and w denotes “wet lawn” (each variable takes binary values 1 or 0). The prior probabilities of rain and sprinkler, and the conditional probabilities values are given as follows:

p(r = 1) = 0.05

p(s = 1) = 0.1

p(w = 1|r = 0, s = 0) = 0.001

p(w = 1|r = 0, s = 1) = 0.97

p(w = 1|r = 1, s = 0) = 0.90

p(w = 1|r = 1, s = 1) = 0.99

· Show joint probability value formula of the whole network, and calculate the joint probability value of P(r=1, s=1, w=1). [0.25 pt]

· Calculate overall probability of lawn is wet, ie., P(w=1) [0.25 pt]

· After observing the law is wet, calculate the probability that the sprinkler was left off (i.e., s=0). [0.5 pt]

· After observing the law is wet, please calculate the probability that there was rain (i.e., r=1). [0.5 pt]

Figure 2

Question 5 [2 pts]: Figure 3 shows a Bayesian network which is similar to Figure 2, but w1 denotes your lawn, and w2 denotes neighbor’s lawn (each variable takes binary values 1 or 0). In this case, rain will cause both yours and your neighbor’s lawn being wet, whereas your sprinkler will only cause your lawn to be wet. The prior probabilities and conditional probability values are given as follows:

p(r = 1) = 0.05

p(s = 1) = 0.1

p(w1 = 1|r = 0, s = 0) = 0.001

p(w1 = 1|r = 0, s = 1) = 0.97

p(w1 = 1|r = 1, s = 0) = 0.90

p(w1 = 1|r = 1, s = 1) = 0.99

p(w2 = 1|r = 1) = 0.90

p(w2 =1|r=0)=0.1

· Show joint probability value formula of the whole network, and calculate the joint probability value of P(r=1, s=1, w1=1, w2=1). [0.25 pt]

· Calculate overall probability of yours and your neighbor’s lawn are wet, i.e., P(w1=1, w2=1) [0.25 pt]

· After observing that yours and your neighbors’ lawn are both wet, calculate the probability that the sprinkler was left on (i.e., s=1). [0.5 pt]

· After observing that yours and your neighbors’ lawn are both wet, calculate the probability that there was rain (i.e., r=1). [0.5 pt]

Figure 3

Question 6 [3 pts]: Given the following toy dataset with 15 Instances

· Please manually construct a Naïve Bayes Classifier (list the major steps, including the values of the priori probability [1.0 pt] and the conditional probabilities [1.0 pt]. Please use m-estimate to calculate the conditional probabilities (m=1, and p equals to 1 divided by the number of attribute values for each attribute).

· Please use your Naïve Bayes classifier to determine whether a person should play tennis or not, under conditions that “Outlook=Overcast & Temperature=Hot & Humidity =Normal& Wind=Weak”. [1 pt]

ID

Outlook

Temperature

Humidity

Wind

Class

1

Sunny

Hot

High

Weak

No

2

Sunny

Hot

High

Strong

No

3

Overcast

Hot

High

Weak

Yes

4

Rain

Mild

High

Weak

Yes

5

Rain

Cool

Normal

Weak

Yes

6

Rain

Cool

Normal

Strong

No

7

Overcast

Cool

Normal

Strong

Yes

8

Sunny

Mild

High

Weak

No

9

Sunny

Cool

Normal

Weak

Yes

10

Rain

Mild

Normal

Weak

Yes

11

Sunny

Mild

Normal

Strong

Yes

12

Overcast

Mild

High

Strong

Yes

13

Overcast

Mild

Normal

Weak

No

14

Rain

Hot

High

Strong

Yes

15

Rain

Mild

High

Strong

No

Logical Agents

Uncertain Knowledge Reasoning

and Learning

Uncertainty, Bayesian Network, and Naïve Bayes Classification

Chapters: 12, 13, 19

Outline

• Uncertainty & Probability

– Distribution

– Independence, conditional independence

• Bayes’ Rule

• Bayesian Network

– Joint distribution

– 3-Way Bayesian Network

– Bayesian Network Construction & Inference

• Naïve Bayes Classification

Uncertainty

• Suppose we are catching a flight at 9AM at FLL

airport? When should you leave Boca for FLL?

Leaving Catch/Miss P(on time)

8:00AM 1/9 1/10=0.1

7:00AM 6/4 6/10=0.6

6:30AM 9/1 9/10=0.9

…

Which action to choose?

Depends on my preferences for missing flight vs. airport waiting

time, etc.

Probability Theory: Beliefs about events

Utility theory: Representation of preferences

Decision about when to leave is determined by two factors:

Decision theory = utility theory + probability theory

What is Probability

• Probability

– Calculus for dealing with nondeterminism and uncertainty

– The world is full of uncertainty

• Deterministic is inadequate and ineffective

• Probabilistic model

– Determine how often we expect different things to occu

• Where do the numbers for probabilities come from?

– Frequentist view

• Numbers from experiments

– Objectivist view

• Numbers inherent properties of universe

– Subjectivist view

• Numbers denote agent’s beliefs

Random Variable

• A random variable x takes on a defined set of

values with different probabilities.

• For example, if you roll a die, the outcome is random (not fixed)

and there are 6 possible outcomes, each of which occur with

probability one-sixth.

– X=1， 3， 2， 2， 1， 6， 5

• For example, if you poll people about their voting preferences,

the percentage of the sample that responds “Yes on Proposition

100” is a random variable (the percentage will be slightly

differently every time you poll).

• Roughly, probability is how frequently we expect

different outcomes to occur if we repeat the

experiment over and over (“frequentist” view)

Discrete or Continuous Random

Variables

• Discrete random variables have a

countable number of outcomes

– Examples: Dead/alive, treatment/placebo, dice,

counts, etc.

– Count frequency by histogram

• Continuous random variables have an

infinite continuum of possible values.

– Examples: blood pressure, weight, the speed

of a car, the real numbers from 1 to 6.

– Count frequency by density

Axioms of probability

• For any propositions A, B

1. 0 ≤ P(A) ≤ 1; 0 ≤ P(B) ≤ 1

2. Probability of whole space S is 1

1. P(S) = 1

3. P(A B) = P(A) + P(B) - P(A B)

Distributions

• Probability distribution

– Probabilities of all possible values of the random variable.

• Random variable:

– A numerical value whose measured value can change from one replicate of

the experiments to anothe

• Weather is one of

– Normalized, i.e., sums to 1. Also note the bold font..

– Bold form normally means a vecto

– If E1, E2, …, Ek are mutually exclusive

• P(X{E1 E2 … Ek})=P(XE1)+P(XE2)+…+P(XEk)

Priors

• Prior or unconditional probabilities of propositions

– Co

esponding to belief prior to a

ival of any (new) evidence.

– Desired outcomes/The total number of outcomes

e.g., P(cavity) =P(Cavity=true)= 0.1

P(Weather=sunny) = 0.72

P(cavity Weather=sunny) = 0.072

Notations: P(AB), P(A and B), P(A, B), and P(AB) denote the

same meanings

Dilemma at the Dentist’s

• What is the probability of a

cavity given a toothache?

• What is the probability of

cavity given the probe

catches?

Joint Probability

The entire table sums to 1.

Toothache Toothache

Cavity XXXXXXXXXX

Cavity XXXXXXXXXX

P(CavityToothache)=0.12

Toothache Toothache

Cavity 12 8

Cavity 8 72

Observed 100

individuals

P(A, B)=P(B, A)

Law of Total Probability

• If B1, B2, B3, .., is a partition of the sample space S, then

for any event A, we have

– P(A)=iP(ABi)=iP(A|Bi)P(Bi)

– For event A=“cavity”, the partition space are {toothache,

toothache}

• P(Cavity)=P(Cavitytoothache)+P(Cavity toothache)

= XXXXXXXXXX=0.2

– For event A=“toothache”, the partition space are {cavity,

cavity}

• P(toothache)=P(toothachecavity)+P(toothache cavity)

= XXXXXXXXXX=0.8

Toothache Toothache

Cavity XXXXXXXXXX

Cavity XXXXXXXXXX

Will discuss late

Inference by Enumeration

Inference by Enumeration

Conditional Probability

P(AB)=P(A|B)P(B)

=P(B|A)P(A)

Inference by Enumeration

Absolute Independence

Absolute Independence

• Absolute independence is powerful but rare

– For n independent biased coins, O(2n) →O(n)

2*2*2*4=32 entries 2*2*2+4=12 entries

Conditional independence

• If I have a cavity, the probability that the probe catches in it

doesn't depend on whether I have a toothache:

– P(catch | toothache, cavity) = P(catch | cavity)

• P(catch|cavity)=P(catch^cavity)/P(cavity)

= XXXXXXXXXX)/ XXXXXXXXXX.008)=0.18/0.2=0.9

• P(catch | toothache,

Answered 2 days AfterApr 12, 2022

Kayes Nfuwork

CAR 63S

1 4 Y( (awi) - 0.1o1 O.02

0.072 to .bo

0.2

P tothache) = o-109 0.02 40.0l +0.

(f

0.2

3 P Cauihy toothvde)z2P (Loa)kP(fosfflaul

0. 12 /P((ity tehnef)

4f0 .L 6.1 t

4PCN, (ovih ootadhe)= Pkavisky 4, toutiach

p(tpstuadhe)

O.o

0.4

5)Pv.(oskz No Toa hu) Y(No cwif

P/no T8kad)

P.12

P (tttne

0.72 0.1

-8

1

6) Simce for

imdlohemdonel, sholl be

PC,T)= PCc.P C)

p CC,T)= 0.1

2

PCO 0.2

PCT)6.2

TCTD )A PCT), tm aru

nat ndalpendernt knt depenket vaiBe

Gic

( Cot oeh, can)P

(catd. Coh)

Cas)=Plcate, a Cost)PCo

0.106-+o.e12 O. = d.

7

o.0)4:0Bp0.ol2

+6.00f

PC th A cav)

= 0.lO3 G.7

Th a olep el

Ts als hrve fY

PLc (aw)¥(Ancanr

Tm nw inupe-dt nuip mcao

PC+I 1 d )-0.0

1dkenoe) 0.1P(-|A)2d.0S

P+diseasi) -1

us)2.oo

P Pltldesx P...

CAR 63S

1 4 Y( (awi) - 0.1o1 O.02

0.072 to .bo

0.2

P tothache) = o-109 0.02 40.0l +0.

(f

0.2

3 P Cauihy toothvde)z2P (Loa)kP(fosfflaul

0. 12 /P((ity tehnef)

4f0 .L 6.1 t

4PCN, (ovih ootadhe)= Pkavisky 4, toutiach

p(tpstuadhe)

O.o

0.4

5)Pv.(oskz No Toa hu) Y(No cwif

P/no T8kad)

P.12

P (tttne

0.72 0.1

-8

1

6) Simce for

imdlohemdonel, sholl be

PC,T)= PCc.P C)

p CC,T)= 0.1

2

PCO 0.2

PCT)6.2

TCTD )A PCT), tm aru

nat ndalpendernt knt depenket vaiBe

Gic

( Cot oeh, can)P

(catd. Coh)

Cas)=Plcate, a Cost)PCo

0.106-+o.e12 O. = d.

7

o.0)4:0Bp0.ol2

+6.00f

PC th A cav)

= 0.lO3 G.7

Th a olep el

Ts als hrve fY

PLc (aw)¥(Ancanr

Tm nw inupe-dt nuip mcao

PC+I 1 d )-0.0

1dkenoe) 0.1P(-|A)2d.0S

P+diseasi) -1

us)2.oo

P Pltldesx P...

SOLUTION.PDF## Answer To This Question Is Available To Download

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