A4 Module 3 (Version B) Solution

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Total points 20/20

This module is worth 20 points. Your last submission will be used for the Fnal score. You may attempt this module 5 times without penalty. After 5 attempts, each additional attempt will result in a penalty of 5% (e.g., On your 7th attempt, you obtain a score of 18 points. Then, your Fnal score for this module will be 18 – (2*1) = 16 points.)

If you encounter any problems with the assignment, please email zheweisun@cs.toronto.edu with [CSC384 A4] in the subject. Be sure to include the module number and version.

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CSC384 S19 A4 Module 3 (Version B) 2019-07-30, 1*31 AM

M3P1 – D-Separation

Answer the following True/False questions based on the following graph:

Section score 4/4

A and B are independent, given no evidence.

True

False

A and B are conditionally independent, given C.

True

False

1/1

1/1

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CSC384 S19 A4 Module 3 (Version B) 2019-07-30, 1*31 AM

D and E are conditionally independent, given C.

True

False

D and E are conditionally independent, given A and B.

True

False

M3P2 – Variable Elimination I

Answer the questions based on the following Bayes net:

Section score 6/6

1/1

1/1

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CSC384 S19 A4 Module 3 (Version B) 2019-07-30, 1*31 AM

Select all that are true:

2/2

Pr(X2 = T | X3 = F) = Pr(X2 = T, X3 = F) / Pr(X3 = F)

Pr(X2 = T | X3 = F) = Pr(X2 = T, X3 = F) / [ Pr(X2 = T, X3 = F) + Pr(X2 = F, X3 = F)]

Pr(X2 = T | X3 = F) = Pr(X2 = T, X3 = F) / [ Pr(X2 = T, X3 = T) + Pr(X2 = T, X3 = F)]

Pr(X2 = T | X3 = F) = Pr(X2 = T, X3 = F) / Pr(X2 = T)

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CSC384 S19 A4 Module 3 (Version B) 2019-07-30, 1*31 AM

Calculate the following:

1/1

Solve P(X2 = T | X3 = F).

3/3

*Hint: Using variable elimination with elimination ordering X4, X1, X2, X3 will greatly simplify your calculations! The two previous questions should also guide your calculations. Your answer should be between 0 and 1, rounded to 3 digits after the decimal (e.g. 0.120).

M3P3 – Variable Elimination II

Section score 10/10

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CSC384 S19 A4 Module 3 (Version B) 2019-07-30, 1*31 AM

Given the table of probabilities pictured, what is P(C = -c | D = – 5/5 d)?

Your answer should be between 0 and 1, rounded to 3 digits after the decimal (e.g. 0.120).

Given the table of probabilities pictured, what is

5/5

P(E=e|A=a,D=d)?

Your answer should be between 0 and 1, rounded to 3 digits after the decimal (e.g. 0.120).

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