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Login NowHere, P_{1} = 1st plan, P_{2} = 2nd plan, P_{3} = 3rd plan

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Probability of getting defective when plans are used are

p(D/P_{1}) = 0.01

p(D/P_{2}) = 0.03

p(D/P_{3}) = 0.02

Also, probability of different plans being used are

p(P_{1}) = 30% = 0.3

p(P_{2}) = 20% = 0.2

p(P_{3}) = 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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