A manufacturing firm employs three analytical plans for the design and developme

Question

A manufacturing firm employs three analytical plans for the design and development of a particular product. For cost reasons, all three are used at varying times. In fact, plans 1, 2, and 3 are used for 25%, 30%, and 45% of the products respectively. The products developed from plans 1, 2, and 3 are defective 1%, 3% and 2% of the time respectively. a) What is the probability of producing a defective product? b) If a random product was found to be defective, which plan was most likely used and thus responsible?

Explanation / Answer

P(product is from plan 1 and is a defective product) = 0.25x0.01 = 0.0025

P(product is from plan 2 and is a defective product) = 0.3x0.03 = 0.009

P(product is from plan 3 and is a defective product) = 0.45x0.02 = 0.009

a) P(defective product) = 0.0025+0.009+0.009 = 0.0205

b) Plan 2 and Plan 3 produce same proportion of defective products, so, if a product was found defective, it is equally likely to be from either plan 2 or plan 3.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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