5. (30 points) A randomly selected part could have been in several different fac

Question

5. (30 points) A randomly selected part could have been in several different factories and could contain as produced, and let Y denote the number of defects. defects. Let X denote the factory where it w Assume that Pr(X i,Y=j) = ach defect costs $100 to repair. (a) What is the marginal probability mass function of Y? (b) What is the conditional probability mass function of the number of defects given that the part was produced in Factory 0? (c) What is the expected cost due to repairing defects? (3-1-2)/9 for i 0,1 and j = 0,1,2. Also assume that e

Explanation / Answer

a) The marginal probability mass function for Y here is obtained as:

P(Y = 0) = P(X = 0, Y =0) + P(X = 1, Y = 0)
P(Y = 0) = (3/9) + (2/9) = 5/9

Similarly other probabilities are computed here as:

P(Y = 1) = (2/9) + (1/9) = (3/9)

P(Y = 2) = (1/9)

Therefore the required PMF for Y here is given as:

P(Y =0) = (5/9),
P(Y =1) = (3/9),
P(Y =2) = (1/9)

b) Given that the part is produced in factory 0, the probability here is computed as:

P(X =0) = (3/9) + (2/9) + (1/9) = 6/9

P(Y = 0 | X = 0) = (3/9) / (6/9) = 0.5,
P(Y = 1 |X = 0) = (2/9) / (6/9) = (1/3) = 0.3333
P(Y = 2 | X = 0) = (1/9) / (6/9) = (1/6) = 0.1667

This is the required conditional PDF for Y given X = 0

c) The expected cost due to repairing here is computed as:

= 0*P(Y = 0) + 100*P(Y = 1) + 200*P(Y = 2)

= 0 + (300/9) + (200/9)

= (500/9) = 55.5556

Therefore the expected cost here is $55.56

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