A manufacturing company employs three analytical plan for the design and development of a particular product. For cost reasons, all three are used at varying times. In facts 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/P1) = 0.01, P(D/P2) = 0.03, P(D/P3) = 0.02, Where P(D/Pj) 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?

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Here, P1 = 1st plan,  P2 = 2nd plan, P3 = 3rd plan

By question,

Probability of getting defective when plans are used are

p(D/P1) = 0.01

p(D/P2) = 0.03

p(D/P3) = 0.02

Also, probability of different plans being used are

p(P1) = 30% = 0.3

p(P2) = 20% = 0.2

p(P3) = 50% = 0.5

According to Baye’s rule, the probability of 1st plan being used, given that product defective is

\(p\left ( \frac{P_1}{D}\right ) = \frac{p(P_1) \enspace.\enspace p(\frac{D}{P_1})}{p(P_1) \enspace .\enspace p(\frac{D}{P_1}) \enspace +\enspace p(P_2) \enspace .\enspace p(\frac{D}{P_1})\enspace +\enspace p(P_3) \enspace .\enspace p(\frac{D}{P_3}) } \)

= \(\frac{0.3 \times 0.01}{0.3 \times 0.01 + 0.2\times 0.03 + 0.5 \times 0.02}\)

= 0.158

Also, probability of 2nd plan being used given that product is defective is

\(p\left ( \frac{P_2}{D}\right ) = \frac{p(P_2) \enspace.\enspace p(\frac{D}{P_2})}{p(P_1) \enspace .\enspace p(\frac{D}{P_1}) \enspace +\enspace p(P_2) \enspace .\enspace p(\frac{D}{P_1})\enspace +\enspace p(P_3) \enspace .\enspace p(\frac{D}{P_3}) } \)

= \(\frac{0.2 \times 0.03}{0.3 \times 0.01 + 0.2\times 0.03 + 0.5 \times 0.02}\)

= 0.316

Similarly,

\(p\left ( \frac{P_3}{D}\right ) = \frac{p(P_3) \enspace.\enspace p(\frac{D}{P_3})}{p(P_1) \enspace .\enspace p(\frac{D}{P_1}) \enspace +\enspace p(P_2) \enspace .\enspace p(\frac{D}{P_1})\enspace +\enspace p(P_3) \enspace .\enspace p(\frac{D}{P_3}) } \)

= \(\frac{0.5 \times 0.02}{0.3 \times 0.01 + 0.2\times 0.03 + 0.5 \times 0.02}\)

= 0.526

Since, the third plan has the highest probability. so, the plan 3rd is most likely to be used.

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