Ans 11.4 (10 pts)
There are 180 linearizations:  Because there are no preconditions 
for either hat or coat, they can be executed at any time in the 
total order plans.  Therefore, since there are 6 total order plans 
for putting on the shoes, and the hat (or coat) can be introduced 
at any place before, during, or after the steps of putting a shoe on, 
hence 5 in total.  For example, before LeftSock, before LeftShoe, 
before RightSock, before RightShoe, plus one for after the last 
step: total 5.  Therefore there are 6 x 5 = 30 possibilities.  
Using the same reasoning, if we then introduce the coat (or hat),
 then it can go at any place, which this time totals 6.  The 
possibilities therefore number 30 x 6 = 180.

Ans 14.3 (5 pts)
A: TestPositive
B: HaveDisease
the problem gives:
 P(A|B)=0.99
 P(A|~B)=0.01
 P(B)=0.0001
ask for P(B|A)
P(B|A)=P(A|B)*P(B) /[P(A|B)* P(B)  + P(A|~B)*P(~B) ] =
      =0.99 * 0.0001 /(0.99 *0.0001 + 0.01* 0.9999) = 0.0098

Ans 14.11 (15 pts)
Without the last line (9 out of 10 is green) you can't calculate the
possiblity.
Given this fact, you can.
A: TaxiAppearsBlue
B: TaxiIsBlue
the problem gives:
P(B)=0.1
P(A)=0.1*0.75 + 0.9 *0.25 =0.3
ask for P(B|A) =0.1 * 0.75 /0.3 =0.25


Ans 15.1 (20 pts)
See attached figure for solution.
c) 2^8 = 256.
d) There are three independent probability values: There are three: Gas and Radio,  
Gas and IcyWeather, and finally, IcyWeather and Radio.
e) NOISY-AND.

Ans 15.6 (10 pts)
Proabability is relative -- when you learn
someting new, it changes your perceived probabilty
of something else that you dont yet know about.
so in that sense it is nonmonotonic

but, "P(X)" still has the same value, even
if you know Y and P(X) doesnt equal P(X|Y) -- so
in that sense it is monotonic

(therefore, a student gets it right as long as
they defend their answer correctly)

(20 pts)
Titanic Question.

1) P(H) = P(H|C)*P(C) + P(H|~C)*P(~C)
2) P(H|C) = Read from the table.
3) P(S|C) = P(S|H)P(H|C) + P(S|~H)P(~H|C).
4) P(S|H) = P(S|H^w)P(W) + P(S|H^~W)P(~W)
5) P(S|C^H) = P(S|H) = same as above.
6) P(C|S) = P(S|C) (from ans 3) * P(C) / P(S)
7) P(C|S^W) = make a "smaller" net with no W at all, making the table
at S smaller. Now answer P(C|S) for that smaller net.

ps: First question 10 pts, Question 11.1 (10 pts), question 11.5 was not graded.