A manufacturing company 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, plan 1, 2, and 3 are used for 30%, 20% and 50% of the products respectively. The defect rate in different procedures is as fo
A manufacturing company 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, plan 1, 2, and 3 are used for 30%, 20% and 50% of the products respectively. The defect rate in different procedures is as follows: $P(D/P_1) = 0.01$, $P(D/P_2) = 0.03$, $P(D/P_3) = 0.02$, where $P(D/P_j)$ is the probability of a defective product, given plan $j$. If a random product was observed and found to be defective, which plan was most likely used and thus responsible? [5]
Plan Prior $P(Pj)$ Defect Rate $P(D/Pj)$ ------------------ $P1$ 0.30 0.01 $P2$ 0.20 0.03 $P3$ 0.50 0.02 Find: posterior $P(Pj/D)$ for each plan; identify the largest. $$P(D) = (0.30)(0.01) + (0.20)(0.03) + (0.50)(0.02)$$ $$P(D) = 0.003 + 0.006 + 0.010 = 0.019$$ $$P(Pj/D) = \frac{P(Pj),P(D/Pj)}{...