Answer to Question #347828 in Statistics and Probability for Murtaza

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 40%,, 15%, and 45% of the products respectively. The "defect rate:" is different for the three procedures as follows:

P(D\P1)=0.01, P{D\P2) = 0.03. P(D|P3 ) = 0.02,

where P(D\Pi) is the probability of a, defective product, given plan i. If a random product was observed and found to be defective, which plan was most likely used and thus responsible?

Answer 1

$$P(D)=P(P_1)P(D|P_1)+P(P_2)P(D|P_2)$$

$$+P(P_3)P(D|P_3)$$

$$=0.4(0.01)+0.15(0.03)+0.45(0.02)$$

$$=0.0175$$

$$P(P_1|D)=\dfrac{P(P_1)P(D|P_1)}{P(D)}=\dfrac{0.4(0.01)}{0.0175}=\dfrac{8}{35}$$

$$P(P_2|D)=\dfrac{P(P_2)P(D|P_2)}{P(D)}=\dfrac{0.15(0.03)}{0.0175}=\dfrac{9}{35}$$

$$P(P_3|D)=\dfrac{P(P_3)P(D|P_3)}{P(D)}=\dfrac{0.45(0.02)}{0.0175}=\dfrac{18}{35}$$

Since $P(P_1|D)<P(P_2|D)<P(P_3|D)$ the first plan was most likely used and thus responsible.