TRƯỜNG ĐẠI HỌC BÁCH KHOA TP.HCM Khoa Khoa học & Kỹ thuật Máy tính ARTIFICIAL INTELLIGENCE Tutorial 7 Questions UNCERTAINTY and IMPRECISION Question 1... TRƯỜNG ĐẠI HỌC BÁCH KHOA TP.HC
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ARTIFICIAL INTELLIGENCE
Tutorial 7 Questions UNCERTAINTY and IMPRECISION Question 1
1 Prove that:
a P(A | B ˄ A) = 1
b P(X ˄ Y | E) = P(X | Y ˄ E).P(Y | E)
c P(Y | X ˄ E) = P(X | Y ˄ E).P(Y | E)/P(X | E)
(the conditionalized version of Bayes' rule)
Solution:
) (
) ( )
(
)) (
( )
|
A B P
A B P A
B P
A B A P A
B
A
P
b
) (
) (
)
|
(
E P
E Y X P E
Y
X
c
P Y X E P X Y E P Y E P E P X Y E P Y E
P Y X E
P X E P X E P E P X E
2 Show that the statement
P(A ˄ B | C) = P(A | C).P(B | C)
is equivalent to either of the statements
P(A | B ˄ C) = P(A | C) and P(B | A ˄ C) = P(B | C)
Solution:
)
| ( )
| ( )
(
) ( )
| ( )
(
) (
)
|
C P
C B P C B A P C
P
C B A P C
B
A
Therefore, if P(AB|C)P(A|C).P(B|C) then P(A|BC)P(A|C)
Similarly, if P(AB|C)P(A|C).P(B|C) then P(B|AC)P(B|C)
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Question 2
This exercise investigates the way in which conditional independence relationships affect the amount of information needed for probabilistic calculations
a Suppose we wish to calculate P(H | E1, E2) and we have no conditional independence information Which of the following sets of numbers are sufficient for the calculation? (i) P(E1, E2), P(H), P(E1 | H), P(E2 | H)
(ii) P(E1, E2), P(H), P(E1, E2 | H)
(iii) P(H), P(E1 | H), P(E2 | H)
b Suppose we know that P(E1 | H, E2) = P(E1| H) for all values of H, El, E2 Now which
of the three sets are sufficient?
Solution:
a (ii) Bayes rule
b (i) Because P(E1,E2|H) = P(E1|H).P(E2|H) – conditional independence
Question 3
Orville, the robot juggler, drops balls quite often when its battery is low In previous trials, it has been determined that the probability that it will drop a ball when its battery is low is 0.9 On the other hand when its battery is not low, the probability that it drops a ball is only 0.02 The battery was recharged not so long ago, so there is only a 8% chance that the battery is low A robot observer with a slightly unreliable robot observation system sends the information that Orville dropped a ball The reliability of the robot observer is described by the following probabilities:
P(observer reports Orville dropped ball | Orville dropped ball) = 0.8
P(observer reports Orville dropped ball | Orville did not drop ball) = 0.1
a Draw the Bayesian network
b Calculate the probability that the battery is low given the information of the robot observer
Solution:
a
BL: battery is low
D: Orville dropped ball
O: observer reports Orville dropped ball
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b
) (
) (
) (
) (
) (
) (
)
|
(
D O P D O P
D O BL P D O BL P O
P
O BL P
O
BL
P
P BL O D P BL P D BL P O D
P BL O D P BL P D BL P O D
)
| ( )
( )
| ( )
( ) (
) (
)
= 0.08 x 0.9 + 0.92 x 0.02 = 0.0904
( ) ( ) ( | ) 0.0904 0.8 0.07232
P OD P D P O D
P O D P D P O D
Thus, ( | ) 0.0576 0.0008
0.07232 0.09096
P BL O
Question 4
In your local nuclear power station, there is an alarm that senses when a temperature gauge exceeds a given threshold The gauge measures the temperature of the core
Consider the Boolean variables A (alarm sounds), F A (alarm is faulty), and F G (gauge is
faulty) and the multivalued nodes G (gauge reading) and T (actual core temperature)
a Draw a Bayesian network for this domain, given that the gauge is more likely to fail when the core temperature gets too high
b Suppose there are just two possible actual and measured temperatures, normal and
high; the probability that the gauge gives the correct temperature is x when it is working, but y when it is faulty Give the conditional probability table associated with G
c Suppose the alarm works correctly unless it is faulty, in which case it never sounds
Give the conditional probability table associated with A
0.08
0.02
0.8 0.1
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d Suppose the alarm and gauge are working and the alarm sounds Calculate an
expression for the probability that the temperature of the core is too high, in terms of the various conditional probabilities in the network
Solution:
A suitable network is shown in Figure 1 The key aspects are: the failure nodes are parents of the sensor nodes, and the temperature node is a parent of both the gauge and the gauge failure node It is exactly this kind of correlation that makes it difficult for humans to understand what is happening in complex systems with unreliable sensors
Figure 1 A Bayesian network for the nuclear alarm problem
The conditional probability table (CPT) for G is shown below The wording of the question is a little tricky because FG means “not working” and FG means “working”
c The CPT for A is as follows:
d Abbreviating T = High and G = High by T and G respectively, the probability of interest here is P(T | FG ˄ FA ˄ A) Because the alarm's behavior is deterministic, we can reason that if the alarm is working and sounds, G must be High, we need only calculate P(T | FG ˄ G)
) ( )
| (
) ( )
| (
) ( )
| (
)
|
(
T P T G F P T P T G F P
T P T G F P G
F
T
P
G G
G G
= P(G| F G T).P( F G|T).P(T)
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Letting P(T) = p, P(FG | T) = g, and P(FG | T) = h, we get
) 1 )(
1 )(
1 ( ) 1 (
) 1 ( )
|
(
p h x p
g x
p g x G
F
T
Question 5
Given a fuzzy number A as presented in Figure 1 Calculate A/A
Figure 1 The fuzzy number A
Solution:
5 2
0
] 5 , 4 [ 5
] 4 , 2 [ 2
/ ) 2
(
)
(
x or x
if
x if x
x if x
x
A
A = [2 + 2, 5 - ]
A/A = [(2 + 2]
2 / 5 5
/ 2 0
] 2 / 5 , 1 [ )
1 2 /(
) 2 5
(
] 1 , 5 / 2 [ )
2 /(
) 2 5
(
)
(
/
x or x
if
x if x
x
x if x
x x
A
A
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End