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1

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. 1
)(
)(
)(
))((
)|( 




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:
)|().|(
)(
)().|(
)(
)(
)|( CBPCBAP
CP
CBPCBAP
CP
CBAP
CBAP 






Therefore, if )|().|()|( CBPCAPCBAP  then )|()|( CAPCBAP



Similarly, if )|().|()|( CBPCAPCBAP  then )|()|( CBPCABP




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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3



= 0.08 x 0.9 + 0.92 x 0.02 = 0.0904
( ) ( ). ( | ) 0.0904 0.8 0.07232PO D PD PO D  
( ) ( ). ( | ) (1 0.0904) 0.1 0.09096PO D P D PO D       

Thus,
0.0576 0.0008
(|)
0.07232 0.09096
PBLO


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.

The conditional probability table (CPT) for G is shown below. The wording of the
question is a little tricky because F
G
means “not working” and F
G
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 | F
G
˄ F
A
˄ 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 | F
G
˄ G).
)().|()().|(
)().|(
)|(
TPTGFPTPTGFP
TPTGFP
GFTP
GG
G
G

| T) = h, we get
)1)(1)(1()1(
)1(
)|(
phxpgx
pgx
GFTP
G


 Question 5.

Given a fuzzy number A as presented in Figure 1. Calculate A/A. Figure 1. The fuzzy number A

Solution:













2/55/20
]2/5,1[)12/()25(
]1,5/2[)2/()25(
)(/
xorxif
xifxx
xifxx
xAA

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