(a) (7 pts.) Prove the conditionalized version of the general product rule:
P(A, B | E) = P(A | B, E) P(B | E)(b) (8 pts.) Prove the conditionalized version of Bayes' rule:
P(A | B, C) = P(B | A, C) P(A | C) / P(B | C)
In parts (c) through (f), you should simplify the probability equations
by using the following shorthand:
T = temperature is high
~T = temperature is not high (normal)
G = gauge G reads high
~G = gauge G reads normal
A = alarm goes off
~A = alarm doesn't go off
FG = gauge G is faulty
~FG = gauge G is working
FA = alarm is faulty
~FA = alarm is working
(a) (5 pts.) Draw a belief net for this domain, given that the gauge
is more likely to fail when the core temperature gets too high.
(b) (2 pts.) Is your network a polytree? Why or why not?
(c) (4 pts.) Suppose that there are just two possible actual and measured
temperatures, Normal and High, and that the gauge gives the incorrect temperature
x% of the time when it is working, but y% of the time when it is faulty.
Give the conditional probability table associated with G. Please specify
your CPT in the following form:
| P (G | T, FG) | T = Normal | T = Normal | T = High | T = High |
| FG | ~FG | FG | ~FG | |
| ~G | ||||
| G |
(d) (4 pts.) Suppose the alarm works unless it is faulty, in which case it never goes off. Give the conditional probability table associated with A. Use the same CPT table format, this time for P (A | G, FA).
(e) (25 pts.) Suppose the alarm and gauge are working, and the alarm
sounds. Calculate the probability that the core temperature is too high.
For this problem, you will also need to know that P(T) = p, P (FG | T) =
g, and P (FG | ~T) = h. Hint: Because FA and A are conditionally
independent of T given G,
P(T | FG, G, A, FA) = P(T | FG,
G).
Now go back and look carefully at the behavior of the alarm, given
the gauge, and think about what G must be. Next, start applying Bayes'
Rule and/or the definition of conditional probability to move the terms
around until you arrive at known terms (e.g., probabilities that are expressed
explicitly in the CPTs).
(f) (5 pts.) Suppose we add a second temperature gauge H, connected so that the alarm goes off (if it's working) when either gauge reads high. Where do H and FH (the event of H failing) go in the network? What is the new CPT associated with A?
| Outlook | Temp (F) | Humidity (%) | Windy? | Class |
| sunny | 75 | 70 | true | Play |
| sunny | 80 | 90 | true | Don't Play |
| sunny | 85 | 85 | false | Don't Play |
| sunny | 72 | 95 | false | Don't Play |
| sunny | 69 | 70 | false | Play |
| overcast | 72 | 90 | true | Play |
| overcast | 83 | 78 | false | Play |
| overcast | 64 | 65 | true | Play |
| overcast | 81 | 75 | false | Play |
| rain | 71 | 80 | true | Don't Play |
| rain | 65 | 70 | true | Don't Play |
| rain | 75 | 80 | false | Play |
| rain | 68 | 80 | false | Play |
| rain | 70 | 96 | false | Play |
(a) (15 pts.) At the root node for a deecision tree in this domain, what is the information gain associated with each attribute? (Use a threshold of 75 for humidity and temperature (i.e., humidity <= 75 / humidity > 75, temperature <= 75 / temperature > 75.)
(b) (15 pts.) Again at the root node, what is the gain ratio associated with each attribute (using the same thresholds as in (b))?
(c) (10 pts.) Suppose you build a decision tree that splits on the Outlook
attribute at the root node. How many children nodes are there are at the
first level of the decision tree? Which branches require a further split
in order to create leaf nodes with instances belonging to a single class?
For each of these branches, which attribute can you split on to complete
the decision tree building process at the next level (i.e., so that at
level 2, there are only leaf nodes)? Draw the resulting decision tree,
showing the decisions (class predictions) at the leaves.