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The time to failure, in months, of light bulbs manufactured at two plants A and B obey the exponential distribution with means 6 and 2 months respectively. Plant B produces four times as many bulbs as plant A does. Bulbs from these plants are indistinguishable. They are mixed and sold together. Given that a bulb purchased at random is working after 12 months, the probability that it was manufactured at plant A is _____

GIVEN ANSWER: 0.92 to 0.94
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Here we need to know about exponential distribution and bayes' theorem application..

Here  P(Plant A)   =   1 / 5 

        P(Plant B)    =   4 / 5   [  The reason being plant B produces four times more than plant A ]

        P(work after 12 months | plant A)    =   1  -  P(fails withing 12 months | plant A)

Now to calculate  " P(fails withing 12 months | plant A) "  , 

       As it follows exponential distribution , here the parameter we use is θ and the expectation(mean) is 1 / θ ..

      As  1/θ(mean)    =   6 months (given)    ==>   θ    =   1/6

     Now P(fails within 12 months | plant A)    =  1 -  e-θa    =    1  -  e-1/6 .  12

                                                                                               =    1  -  e-2

            P(work after 12 months | plant A)                        =   1  -  P(fails within 12 months | plant A)

                                                                                    =   e-2

          Similarly P(work after 12 months | plant B)             =   e-1/2.12

                                                                                    =   e-6

   Hence the inverse probability  which is asked in question 

   i.e.              P(plant A | work after 12 months)              =   P(plant A) * P(work after 12 months |  plant A) /                                                                                [ P(plant B) * P(work after 12 months | plant B) + P(plant A) * P(work after 12 months |  plant A)

                                                                                   =   1/5  *  e-2  /   [ (1/5 * e-2 )  +  (4/5 * e-6) ]

                                                                                   =   0.027 /  (0.27  +  0.00198)

                                                                                   =   0.932 

                                   

Hence required probability    =    0.93 [correct to 2 decimal places]

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