AI Tutorial 2 With Answers Harbin/Adelaide Course July 2016

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1 AI Tutorial 2 With Answers Harbin/Adelaide Course July 2016 Note: you should, wherever possible, attempt these problems TWO ways. The first (and mandatory unless specified otherwise) is to work through the examples pen and paper. Explaining each step you take. The second way is to, wherever you can, use the Bayes Net simulator/java applet to check your answers. Of course if you can see a third way (perhaps two different pen and paper ways) you should also try that if you have time. QUESTION 1 Consider the Burglar alarm network. Answer the following queries: a) P(Burglar Alarm=true) ONE way to do this. 1) from model P(A,R,B)=P(A BE)P(B)P(E), and 2) from Bayes rule (or the definition of conditional probability) what we want, P(B A)=P(A,B)/P(A). But P(A,B)=ΣEP(A,B,E) = ΣEP(A BE)P(B)P(E) (from 1)). = P(B) ΣEP(A BE)P(E) so P(B A)= P(B)/P(A) ΣEP(A BE)P(E) so P(B A=true) = P(B)/P(A=true) ΣEP(A=true BE)P(E) So we can observe that the term in blue is formed by firstly selecting the rows for A=true from the conditional table P(A BE) and multiplying by P(E) and then summing out E. You can see that that is what the Java applet does (f5 is selecting, f6 is by multiplying E and summing). We then multiply by P(B) (again the java applet does this and calls the result f7). Of course we can divide by P(A=true) to complete as per the above equation but since it it the same constant dividing the distribution, we can instread just normalize to achieve the same effect which is what the java applet does to get the answer f8. ANOTHER two ways is to apply the standard methods of the lecture slides for inference join and marginalize out unwanted variables using either the interleaved join and marginalization, of the full join then marginalization after. b) P(Alarm Earthquake=true, Burglar=true) c) P(Burglar John=true, Mary=false) QUESTION 2 (only using the Java applet) Consider the CAR NETWORK. Construct the network in the applet and answer: a) P(CarWontStart Dipstick=false, Lights=true, OilLight=true, FuelGauge=true) b) P(AlternatorBroken CarWontStart=true, Dipstick=true, Lights=false, BatteryMeter=false) QUESTION 3

2 Using the table from AIMA (shown below) for the random variables (Catch, Cavity, Toothache) calculate the following (note that the convention is capitals denote a variable, lower case denotes an instance, - denotes the false instance: that is cavity means no cavity, cavity means there is a cavity, and Cavity refers to the random variable). a) P(Toothache) Summing up all of toothache: P(+t)= =0.2. P(-t) is 1- P(+t) of course! Won t keep giving the redundant parts of the answers b) P(Cavity) Summing up all of cavity: P(cavity)= =0.2 c) P(Toothache Cavity) P(+t cavity)=p(+t,cavity)/p(cavity)=( )/0.2=0.6 P(-t cavity)=p(-t,cavity)/p(cavity)=( )/0.2=0.4 P(+t -cavity)=p(+t,-cavity)/p(-cavity)=( )/0.8=0.1 P(-t -cavity)=p(-t,-cavity)/p(-cavity)=( )/0.8=0.9 d) P(cavity toothache catch) Simply select from the table those entries where either toothache or catch are true (first three cols) and cavity is true (so first row, first three cols) and sum: = = Then normalize by P(toothache v catch)= = So <0.4615,0.5385>. e) P(Catch Cavity) P(catch cavity)=p(catch,cavity)/p(cavity)=0.18/0.2=0.9 P(-catch cavity)= P(-catch,cavity)/P(cavity)=0.02/0.2=0.001 f) P(Catch)

3 Summing up all of catch P(catch)= =0.34 g) Are Toothache and Cavity independent? P(cavity,+t)= =0.12 (from joint table). P(cavity)P(+t)=0.2*0.2 so no they are not independent. h) Are Cavity and Catch independent? P(cavity,catch)= =0.18. P(cavity)P(catch)=0.2*0.34 so no they are not independent i) Are Toothache and Catch independent? P(+t,catch)= = P(+t)P(catch)=0.2* so no they are not independent. j) Assume the sensible interpretation that cavity causes both toothache and catch. Draw the corresponding Bayes Net (graph AND conditional probability tables of course). You can do this in the applet! You can then check some if not all of your answers above QUESTION4 Consider the wet grass example given in lectures :

4 Note: that we save space by only saving/displaying the plus or true parts of the conditional distributions as the negative or false parts must sum to one when added to the plus parts. Calculate (and check with the Java applet) a) P(W c) b) P(S -w,+r) QUESTION 5 Given the following Bayes Network of Boolean Variables (ignore the numbers above each node to start with):

5 a) How large is the joint probability table? (2 8-1)=255. (8 variables, then sum to one means one entry redundant). b) Give the sizes of all of the conditional probability tables and compare the storage required with that required for the joint probability table. Can you now give a meaning to the numbers above each node. Numbers give the size of the CPT for the node (since we need not give both +/- Variable the size is (1/2) of 2^(number of Parents+1)= 2^(number of Parents+1. Giving total storage of 20 entries. QUESTION 6 At the nuclear reactor at the Australian Nuclear Science and Technology Organisation, there is an alarm that senses when a temperature gauge exceeds a given threshold. The gauge measures the temperature of the core. Consider the Boolean variables A (alarm sounds), FA (Alarm faulty), and FG (gauge is faulty) and the multi-valued (continuous random variables) nodes G (gauge reading) and T (actual core temperature). (a) Draw a Bayesian network for this domain, given that the gauge is more likely to fail when the core temperature gets too high. Possible answer:

6 (b) Suppose there are just two possible actual and measured temperatures, NORMAL and HIGH; the probability that the gauge gives the correct temperature is x when it is working, but y when it is faulty. Give the conditional probability table associated with G. answer: (c) Suppose the alarm works correctly unless it is faulty, in which case it never sounds. Give the conditional probability table associated with A. answer: (d) Suppose the alarm and gauge are working and the alarm sounds. Calculate an expression for the probability that the temperature of the core is too high, in terms of the various conditional probabilities in the network.

7 Answer

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