processed through three sequential operations: I then 2 then 2 sometimes produce
ID: 1125001 • Letter: P
Question
processed through three sequential operations: I then 2 then 2 sometimes produces parts with defect type 2 at a 3. Operation I sometimes produces parts with defect type I at a rale s0%. The defects o rate-8% And operation 3 sometimes produces parts with defect type at a are processed Defects: A starting batch of 10,000 workparts are -596 Operation Question 4 (20 points). all three operations, (a) how many are expected to be defect free? (b) how many are expected to have all three defects? (c) how many are expected to have exactly one defect?Explanation / Answer
a. Let P(.) denote probability of defect.
Expected defect-free pieces= 25000* (1 - P(1)) * (1-P(2))*(1 - P(3))
= 10000*(1-0.05)*(1-0.08)*(1-0.1)
= 7866
b.
Expectation of all 3 defect pieces= 25000* (P(1)* P(2)* P(3)
= 10000*(0.05)*(0.08)*(0.1)
= 4
c.
Expectation of operation 1 defect (i.e. in operation 1) = 10000*(0.05)*(1-0.08)*(1-0.1) = 414
Expectation of operation 2 defect (i.e. in operation 2) = 10000*(1-0.05)*(0.08)*(1-0.1) = 684
Expectation of operation 3 defect (i.e. in operation 3) = 10000*(1-0.05)*(1- 0.08)*(0.1) = 874
Expectation of exactly one defect = 414 + 684 + 874 = 1972
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