Finite Math: Probability

Question

Wonder Widgets Inc. has 3 plants producing their De-lux Wonder Widget (DWW). Plant A produces 80% of all DWW, plant B 15%, and plant C the remaining 5%. 7% of Plant A’s production is defective, 6% of plant B’s production is defective, and 1% of plant C’s production is defective.

(a) What percent of the total production of DWW is defective?

(b) If a DWW is defective, what is the probability that it was produced in plant A?

Answer 1

a) 0.8 * 0.07 + 0.15 * 0.06 + 0.05 * 0.01 = 0.0655 = 6.555% of all widgets are defective

b) 0.8 * 0.07/ 0.0655 = 0.056/0.0655 = 0.855 probability that defective widget is from plant A.