970 views
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 _____

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]

1 vote