We consider a probabilistic model for a fault diagnosis problem. The class varia

Question

We consider a probabilistic model for a fault diagnosis problem. The class variable C rep resents the health of a disk drive: C = 0 means it is operating normally, and C = 1 means it is in a failed state. When the drive is running it continuously monitors itself using a temperature and shock sensor, and records two binary features, X and Y. X = 1 if the drive has been subject to shock (eg, dropped), and X = 0 otherwise. Y-1 if the drive temperature has ever been above 70°C, and 0 otherwise. The following table defines the joint probability mass function of these three random variables: 0.2 0.2 0.0 0.05 0.25 When computing probabilities below, provide the numerical value of your answer, as well as equations showing how you calculated that answer a) What is the probability p(C = 1)? b) What is the probability p(C = 0|X = 1, Y = 0)? c) What is the probability p(X = 0,Y = 0)? d) What is the probability p(C = 01X = 0)? e) Are X and Y independent? Justify your answer f) Are X and Y conditionally independent given C? Justify your answer

Explanation / Answer

a) From the given table P(C = 1 ) is computed as the sum of probabilities of the last 4 rows as:

P( C = 1) = 0 + 0.1 + 0.05 + 0.25 = 0.4

Therefore P(C = 1) = 0.4

b) Here the required probability is computed using the bayes theorem as:

P( C = 0 | X = 1, Y = 0) = P( C = 0 , X = 1, Y = 0) / P( X = 1, Y = 0)

From the above table, we get:

P( C = 0 , X = 1, Y = 0) = 0.2 and

P( X =1, Y = 0) = 0.2 + 0.05 = 0.25

Therefore, we get:

P( C = 0 | X = 1, Y = 0) = P( C = 0 , X = 1, Y = 0) / P( X = 1, Y = 0) = 0.2 / 0.25 = 0.8

Therefore 0.8 is the required probability here.

c) P( X = 0, Y = 0 ) = 0.1 + 0 = 0.1

Therefore 0.1 is the required probability here.

d) The probability here is computed using the bayes theorem as:

P( C = 0 | X = 0) = P( C = 0 , X = 0) / P( X= 0 )

From the above table, we get:

P( C = 0 , X = 0) = 0.1 + 0.2 = 0.3

Also, P(X = 0 ) = 0.1 + 0.2 + 0 + 0.1 = 0.4

Therefore, the required probability here is computed as:

P( C = 0 | X = 0) = P( C = 0 , X = 0) / P( X= 0 ) = 0.3 / 0.4 = 0.75

Therefore 0.75 is the required probability here.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.