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# Assignment 3: Uncertainty and Probability solution

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COMP 307 — Introduction to AI

Assignment 3:
Uncertainty and Probability

Question Description Part 1: Reasoning Under Uncertainty Basics [10
marks]
1. Create the full joint probability table of X and Y , i.e. the table containing the following
four joint probabilities P(X = 0, Y = 0), P(X = 0, Y = 1), P(X = 1, Y = 0), P(X = 1, Y = 1).
Also explain which probability rules you used.
2. If given P(X = 1, Y = 0, Z = 0) = 0.336, P(X = 0, Y = 1, Z = 0) = 0.168, P(X = 0, Y = 0, Z = 1) =
0.036, and P(X = 0, Y = 1, Z = 1) = 0.042, create the full joint probability table of the three
variables X, Y , and Z. Also explain which probability rules you used.
x Y P(X=x, Y=y)

Category:

## Description

COMP 307 — Introduction to AI

Assignment 3:
Uncertainty and Probability

Question Description Part 1: Reasoning Under Uncertainty Basics [10
marks]
1. Create the full joint probability table of X and Y , i.e. the table containing the following
four joint probabilities P(X = 0, Y = 0), P(X = 0, Y = 1), P(X = 1, Y = 0), P(X = 1, Y = 1).
Also explain which probability rules you used.
2. If given P(X = 1, Y = 0, Z = 0) = 0.336, P(X = 0, Y = 1, Z = 0) = 0.168, P(X = 0, Y = 0, Z = 1) =
0.036, and P(X = 0, Y = 1, Z = 1) = 0.042, create the full joint probability table of the three
variables X, Y , and Z. Also explain which probability rules you used.
x Y P(X=x, Y=y)
1 1 0.14
1 0 0.56
0 1 0.21
0 0 0.09
x Y Z P(X=x,Y=y,Z=z)
0 0 0 0.054
0 0 1 0.036
0 1 0 0.168
0 1 1 0.042
1 0 0 0.336
1 0 1 0.224
1 1 0 0.112
1 1 1 0.028
3. From the above joint probability table of X, Y , and Z:
(i) calculate the probability of P(Z = 0) and P(X = 0, Z = 0),
(ii) judge whether X and Z are independent to each other and explain why.
First we need to find P(Y), from the P(X=x,Y=y)table using the sum rule:
THE NORMALISATION RULE : INSERT EQUATION
Then from here we need to construct a P(Z =z ,Y =y) table.
Use the product rule: P(Z, Y) = P(Y) * P(Z | Y)
From here we use the Sum rule again to construct Z probability from the above table:
THE NORMALISATION RULE : INSERT EQUATION
Now we can construct the P(X=x,Z=z) table:
THE NORMALISATION RULE : INSERT EQUATION
If independent: P(X=x, Z=z) = P(X=x) * P(Z=z)
However as illustrated below this is not the case:
From above table: P(X=0, Z=0) = 0.222.
Y P(Y)
0 0.56 + 0.09 = 0.65
1 0.21 + 0.14 = 0.35
Z Y P(Z=z,Y=y)
0 0 0.65 * 0.6 = 0.39
0 1 0.35 * 0.8 = 0.28
1 0 0.65 * 0.4 = 0.26
1 1 0.35 * 0.2 = 0.07
Z P(Z)
0 0.28 +0.39 = 0.67
1 0.26+0.07= 0.33
X Z P(X=x,Z=z)
0 0 0.054 + 0.168 = 0.222
0 1 0.042 + 0.036 =0.078
1 0 0.112 +0.336 = 0.448
1 1 0.224 +0.028 = 0.252
And:
P(X=0) = 0.3
P(Z=0) = 0.67.
0.3 * 0.67 = 0.14874.
Thus they are not independent.
4. From the above joint probability table of X, Y , and Z:
(i) calculate the probability of P(X = 1, Y = 0|Z = 1),
P(A,B) = P(B) * P(A|B)
let A = X,Y & B = Z
P(A|B) = P(A|B) / P(B)
P(A|B) = P(X,Y,Z)/P(Z)
plug in values from probability table for x=1, y=0, z=1
P(X = 1, Y = 0|Z = 1) = 0.224/0.33
P(X = 1, Y = 0|Z = 1) = 0.679
(ii) calculate the probability of P(X = 0|Y = 0, Z = 0).
P(A,B) = P(B) * P(A|B)
let A = X & B = Z,Y
P(A|B) = P(A|B) / P(B)
P(A|B) = P(X,Y,Z)/P(Y, Z)
plug in values from probability table for x=0, y=0, z=0
P(X = 0|Y = 0, Z = 0) = 0.054/0.39
P(X = 0|Y = 0, Z = 0) = 0.138
Part 2: Naive Bayes Method [25 marks]
1. the probabilities P(Fi |c) for each feature I
Spam Not spam
Total 51 149
P(Feature 1 = t) 0.6667 0.3557
P(Feature 1 = f) 0.3333 0.6443
P(Feature 2 = t) 0.5882 0.5772
P(Feature 2 = f) 0.4118 0.4228
P(Feature 3 = t) 0.451 0.3423
P(Feature 3 = f) 0.549 0.6577
P(Feature 4 = t) 0.6078 0.396
P(Feature 4 = f) 0.3922 0.604
P(Feature 5 = t) 0.4902 0.3356
P(Feature 5 = f) 0.5098 0.6644
P(Feature 6 = t) 0.3529 0.4698
P(Feature 6 = f) 0.6471 0.5302
P(Feature 7 = t) 0.7843 0.5034
P(Feature 7 = f) 0.2157 0.4966
P(Feature 8 = t) 0.7647 0.349
P(Feature 8= f) 0.2353 0.651
P(Feature 9 = t) 0.3333 0.2416
P(Feature 9 = f) 0.6667 0.7584
P(Feature 10 = t) 0.6667 0.2886
P(Feature 10 = f) 0.3333 0.7114
P(Feature 11 = t) 0.6667 0.5839
P(Feature 11 = f) 0.3333 0.4161
P(Feature 12 = t) 0.7843 0.3356
P(Feature 12 = f) 0.2157 0.6644
instance 1 : is not Spam
spam prob = 6.040489748774789E-4
not spam prob = 0.03162718597368903
instance 2: is spam
spam prob = 0.011028195352395692
not spam prob = 0.0027968287684020814
instance 3: is spam
spam prob = 0.03728891074351882
not spam prob = 0.008746516559680537
instance 4: is not Spam
spam prob = 0.0010470182231209631
not spam prob = 0.041333650052675405
instance 5: is spam
spam prob = 0.01172796386288092
not spam prob = 0.006253146952268239
instance 6: is spam
spam prob = 0.011186673223055645
not spam prob = 0.003101980890857802
instance 7: is not Spam
spam prob = 6.871057089231324E-4
not spam prob = 0.022497259785629678
instance 8: is not Spam
spam prob = 0.012380507914844192
not spam prob = 0.026974744168660303
instance 9: is spam
spam prob = 0.03728891074351882
not spam prob = 0.0025285418724905998
instance 10: is not Spam
spam prob = 0.004083371070171757
not spam prob = 0.04710694228517442
3. The derivation of the Naive Bayes algorithm assumes that the attributes are conditionally
independent. Why is this like to be an invalid assumption for the spam data? Discuss the
possible effect of two attributes not being independent.
In reality, if one of the features indicates spam, it is more likely that the other features would too.
Thus they are not independent. This causes problems for the Bayes algorithm as it needs
independence of features to use this calculation: P(A, B | C) = P(A | C) * P(B | C
Part 3: Bayesian Networks [30 marks]
1. Construct a Bayesian network to represent the above scenario.
Meeting
(M)
P(M)
0 0.3
1 0.7
Lecture P(LT)
0 0.4
1 0.6
MEETING
LECTURE
OFFICE
COMPUTER
LIGHTS
LECCTUR
E(LT)
MEETING(
M)
OFFICE(O) P(O|LT,M)
0 0 0 0.94
0 0 1 0.06
0 1 0 0.25
0 1 1 0.75
1 0 0 0.2
1 0 1 0.8
1 1 0 0.05
1 1 1 0.95
OFFICE
(O)
LIGHT
(L)
P(L|O)
0 0 0.98
0 1 0.02
1 0 0.5
1 1 0.5
OFFICE
(O)
COMPUT
ER(C)
P(C|O)
0 0 0.8
0 1 0.2
1 0 0.2
1 1 0.8
2. Calculate how many free parameters in your Bayesian network ?
P(M=0) = 0.3
P(L=0) = 0.4
P(O = 0 | M=1, L=1) = 0.05
P(O = 0 | M=0, L=1) = 0.2
P(O = 0 | M=0, L=0) = 0.94
P(L = 1|O = 1) = 0.5
P(L = 0|O = 0) = 0.98
P(C =0 |O =1) = 0.2
P(C =0 |O =0) = 0.8
3. What is the joint probability that Rachel has lectures, has no meetings, she is in her office
and logged on her computer but with lights off.
P(Lt=1, M=0, O=1, C=1, L= 0):
= P(Lt=1) * P(M=0) * P(O =1| Lt =1, M=0) * P(C=1| 0=1) * P(L=0|O=1)
= 0.6 * 0.3 *0.8*0.8*0.5
= 0.0576
4. Calculate the probability that Rachel is in the office.
P(O) = P(O = 1, M=1, Lt =1)+ P( O =1, M=0, Lt =1) + P(O =1, M=1, Lt =0) + P(O =1, M=0, Lt =0)
P(O) = P(O| M=1, Lt =1)*P(M=1, Lt =1) + P(O| M=0, Lt =1)*P(M=0, Lt =1) + P(O| M=1, Lt
=0)*P(M=1, Lt =0) * P(O| M=0, Lt =0)*P(M=0, Lt =0)
P(O) = P(O| M=1, Lt =1)*P(M=1) * P(Lt =1) + P(O| M=0, Lt =1)*P(M=0) * P(Lt =1) + P(O| M=1,
Lt =0)*P(M=1) * P(Lt =0) + P(O| M=0, Lt =0)*P(M=0) * P(Lt =0)
P(O) = (0.95 * 0.7 * 0.6) + (0.8 * 0.3 * 0.6) + (0.75 * 0.7* 0.4) + (0.06 * 0.3 * 0.4)
P(O) = 0.7602
5. If Rachel is in the office, what is the probability that she is logged on, but her light is off.
P(L=0, C=1 | O =1)
= P(L = 0 | O =1 ) * P(C = 1 | O =1)
= 0.5 * 0.8
= 0.4
6. Suppose a student checks Rachel’s login status and sees that she is logged on. What effect
does this have on the students belief that Rachels light is on ?
Light, Logged-on and Office variables have a common cause relationship, where Lights and Logged
on are effected by whether Rachel is or isn’t in her office. This causes Light and logged on to be
dependent on each other. If the student knows the relationship, the student can infer that there is a
higher probability that Rachel is in her office if she is logged on. Given Rachel is in her office theres
a higher probability that her lights are on, than if she wasn’t.
Part 4: Inference in Bayesian Networks [35 marks]
1. Using inference by enumeration to calculate the probability P(P = t|X = t) (i) describe what
are the evidence, hidden and query variables in this inference, (ii) describe how would you use
variable elimination in this inference, i.e. to perform the join operation and the elimination
operation on which variables and in what order, and (iii) report the probability.
i) Evidence variables is X-ray. Hidden variables are smoking, Dyspnoea and cancer. The query
variable is Pollution.
ii) The first merge will be of the Pollution, Smoker and Cancer classes to create a P(P, C, S) table
From here I can make a P(P,C) table by removing the Smoker class . After this removal I will
merge this table with X-ray creating P(P,C,X). After this we can eliminate Cancer to create
P(P,X) table. From this probability table and the P(X) table we can infer a P(P|X) table.
Stage 1:
P(P,C,S)
Stage 2:
P(P,C)
Stag 3:
P(P,C,X) = merge P(X) and P(P,C) table.
Pollution Cancer Smoker P(P=p, C=c, S=s)
0 0 0 0.07
0 0 1 0
0 1 0 0.029
0 1 1 0.001
1 0 0 0.0617
1 0 1 0.013
1 1 0 0.257
1 1 1 0.014
Pollution Cancer P(P=p, C=c)
0 0 0.099
0 1 0.001
1 0 0.874
1 1 0.027
Pollution Cancer Xrays P(P=p, C=c, X=x)
0 0 0 0.0792
0 0 1 0.0198
0 1 0 0.0001
0 1 1 0.0009
Stage 4: Create a P(X,C) by removing Pollution:
Stage 5: Create P(X)
Finally we can create the P(P|X) table:
iii) P(P=1|X=1) = 0.910
2. Given the Bayesian Network , find the variables that are independent of each other or
conditionally independent given another variable. Find at least three pairs or groups of such
variables.
Group 1: Indirect cause. Smoker, Cancer, Dyspnonea are conditionally independent.
Group 2: Indirect cause. Pollution, Cancer, Dyspnonea are conditionally independent.
Group 3: Indirect cause. Pollution, Cancer, Xray are conditionally independent.
1 0 0 0.6992
1 0 1 0.1748
1 1 0 0.0027
1 1 1 0.0243
Cancer Xray P(C=c, X=x)
0 0 0.0793
0 1 0.0207
1 0 0.7019
1 1 0.1991
X-ray P(X=x)
0 0.7812
1 0.2188
Pollution Xray P(P=p|X=x)
0 0 0.102
0 1 0.094
1 0 0.898
1 1 0.910