Question

A purchaser of electrical components buys them in lots of size 10. It is his policy to inspect 3 components rendomly from a lot and to accept the lot only if all 3 are nondefective. If 30 percent of the lots have 4 defective components and 70 percent have only 1 what proportion of lots does the purchaser reject?

What percentage of i defective lots does the purchaser reject? Find it for i =1,4. Given that a lot is rejected, what is the conditional probability that it contained 4 defective components.

( I need the second part. I got the first question.)

Answer 1

Probability to accept the lots with 4 defective components = Probability that all 3 selected components are nondefective = ⁶C₃ * ⁴C₀ / ¹⁰C₃ = 20/120 = 1/6   (Using hypergeometric probability where defective = 4, non-defective components is 6).

Probability to reject the lots with 4 defective components = 1 - (1/6) = 5/6

Probability to accept the lots with 1 defective components = Probability that all 3 selected components are nondefective = ⁹C₃ * ¹C₀ / ¹⁰C₃ = 84/120 = 7/10      (Using hypergeometric probability where defective = 1, non-defective components is 9).

Probability to reject the lots with 1 defective components = 1 - (7/10) = 3/10

Proportion of lots does the purchaser reject =

Pr( lots have 4 defective components) * Pr(reject the lots with 4 defective components) + Pr( lots have 1 defective components) * Pr(reject the lots with 1 defective components)

= 0.3 * (5/6) + 0.7 * (3/10)

= 0.46 = 46%

Percentage of 1 defective lots does the purchaser reject = 3/10 = 30%

Percentage of 4 defective lots does the purchaser reject = 5/6 = 83.33%

Using Bayes theorem,

Pr(Lot with 4 defective components | Lot is rejected)

= Pr(Lot is Rejected | Lot with 4 defective components) * Pr(Lot with 4 defective components) / Pr(Lot is Rejected)

= (5/6) * 0.3 / 0.46

= 0.5434783