Question

A manufacturer ships parts in lots of 1000 and makes a profit of $50 per lot sold. The purchaser, however, subjects the product to a sampling plan as follows: 10 parts are selected at random with replacement. If none of these parts is defective, the lot is purchased; if one part is defective,the lot is purchased but the manufacturer returns $10 to the buyer; if two or more parts are found to be defective, the entire lot is returned at a net loss of $25 to the manufacturer. What is the manufacturer’s expected profit if 10% of the parts are defective?

Answer #1

Given that 10% of the parts are defective out of 1000

probability of getting a defective part (p) = 0.1

probability of getting a non defective part (1-p) = 1 - 0.1
=0.9

Now a sample of 10 parts (with replacement) is taken by the manufacturer, there are 3 possible outcomes

1) A: none of 10 part is defective Profit(A) = $50

P(A) = ^{10}C_{0} (p)^{0}
(1-p)^{10} = 1 * 1 * 0.9^{10} = 0.3486

2)B: exactly 1 part is defective Profit(B) = $40

P(B) = ^{10}C_{1} (p)^{1} (1-p)^{9}
= 10 * 0.1 * 0.9^{9} = 0.3874

3)C: 2 or more parts are defective Profit(C) = $-25

P(B) = 1 - P(A) - P(B) = 1 - 0.3486 - 0.3874 = 0.2639

Expected profit = P(A)*Profit(A) + P(B)*Profit(B) + P(C)*Profit(C) = 0.3486*50 + 0.3874*40 + 0.2639*(-25) = $26.3285

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