Question

Suppose that when a machine is adjusted properly, 91% of the items produced by it are of high quality and the others are of medium quality. Suppose, however, that the machine is improperly adjusted during 13% of the time and that under these conditions 14% of the items produced by it are of high quality and the remaining are of medium quality. If an item is selected at random and found to be of high quality, what is the probability that it was produced while the machine was adjusted properly (The answer should be a number rounded to five decimal places, don't use symbols such as %

Answer 1

P( high quality I adjusted properly ) = 0.91

P( high quality I not adjusted properly ) = 0.14

P ( not adjusted properly) = 0.13

P( adjusted properly )= 1-0.13 = 0.87

To find P( adjusted properly I high quality ) =?

Using Bayes theorem , for two events A and B

P( A I B) = P( B I A ) *P( A ) / P( B)

P( adjusted properly I high quality ) = P( high quality I adjusted properly ) * P( adjusted properly ) / P( high quality )

where

P( high quality )

= P( high quality I adjusted properly ) * P( adjusted properly ) + P( high quality I not adjusted properly ) * P( not adjusted properly )

=0.91 * 0.87 + 0.14 *0.13

= 0.8099

Therefore ,

P( adjusted properly I high quality )= 0.91 * 0.87 / 0.8099 = 0.97753

Answer : Probability = 0.97753