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...

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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 correct positive result in only 95% of the cases in which the disease is actually present, and a correct 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 correct 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 occur • 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 another • Weather is one of • P(Weather) = <0.72, 0.1,="" 0.08,="" 0.1=""> – Normalized, i.e., sums to 1. Also note the bold font.. – Bold form normally means a vector – If E1, E2, …, Ek are mutually exclusive • P(X{E1 E2 …  Ek})=P(XE1)+P(XE2)+…+P(XEk) Priors • Prior or unconditional probabilities of propositions – Corresponding to belief prior to arrival 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(AB), 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 0.12 0.08 Cavity 0.08 0.72 P(CavityToothache)=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(ABi)=iP(A|Bi)P(Bi) – For event A=“cavity”, the partition space are {toothache,  toothache} • P(Cavity)=P(Cavitytoothache)+P(Cavity  toothache) =0.12+0.08=0.2 – For event A=“toothache”, the partition space are {cavity, cavity} • P(toothache)=P(toothachecavity)+P(toothache cavity) =0.08+0.72=0.8 Toothache Toothache Cavity 0.12 0.08 Cavity 0.08 0.72 Will discuss later 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) =(0.108+0.072)/(0.108+0.012+0.072+0.008)=0.18/0.2=0.9 • P(catch | toothache,