Task 1.
a.
A = P(Battery age < 3 yrs | battery dead && no oil)
B = P(Battery age < 3 yrs | battery dead && have oil)
A == B. 'No oil' is outside the Markov Blanket of both 'battery dead' and 'battery age', thus changing the value on the node does not affect the probability of battery age < 3 yrs.
b.
A = P(Battery age < 3 yrs | battery dead && no oil)
A = P(Battery age < 3 yrs | battery dead && starter broken)
A == B. The difference between knowing starter broken or no oil cannot affect Battery Age, because both are outside the Markov Blanket for Battery Age.
c.
A = P(alternator broken | no charging)
B = P(alternator broken | no charging && fanbelt broken)
A > B would expectely be true because the parent of one child being true usually lowers the probability of the other parents.
d.
A = P("no gas"=true)
B = P("no gas"=true | "car won't start"=true)
Task 2.
a.
P(alternator broken)= 0.02
P(not charging | alternator broken) = 0.95
P(not charging | alternator working) = 0.01
P(charging) = -P(not charging)
P(not charging)
P(AB) = 0.02
P(NC|AB) = 0.95
P(NC|!AB) = 0.01
P(!NC)
P(NC|AB) = P(AB|NC)P(NC) / P(AB)
0.95 = P(AB|NC)P(NC) / 0.02
P(AB|NC)P(NC) = 0.19
P(AB|NC) = P(NC|AB)P(AB) / P(NC|AB)P(AB) + P(NC|!AB)P(!AB)
= 0.95*0.02 / (0.95*0.02 + 0.01*0.98)
P(AB|NC)= 0.6597
P(NC) = 0.19/0.6597 = 0.288
P(!NC) = 1-0.288 = 0.712
b.
P(battery age <= 3 yrs) = 0.7
P(battery dead | battery age <= 3 yrs) = 0.02
P(battery dead | battery age > 3 yrs) = 0.1
P(battery age <= 3 yrs | battery dead=true)
P(B<3|BD) = P(BD|B<3)P(B<3) / P(BD|B<3)P(B<3) + P(BD|B>3)P(B>3)
= 0.02*0.7 / (0.02*0.7 + 0.1*0.3)
= 0.318318
Task 3.
a.
Markov Blanket of Node L: K, P, Q, M, G
b.
P(A,F) = P(A|F)*P(F)
=
c.
P(F,K)
d.
P(M,!C | H)
Task 4.