The chance of defective screws in three boxes `A ,B ,Ca r e1//5,1//6,1//7,` respectively. A box is selected at random and a screw draw in from it at random is found to be defective. Then find the probability that it came from box `Adot`

Answer

`42//107`

Explanation

Let `E_(1),E_(2),and E_(3)` denote the events of selecting boxes A,B,C, respectively and A be the event that a screw selected at random is defecitive. Then,
`P(E_(1))=P(E_(2))=P(E_(3))=1//3`
`thereforeP(A//E_(1))=1/5,P(A//E_(2))=1/6,P(A//E_(3))=1/7`
By Bayes's rule, the required probability is
`(P(E_(1))P(A//E_(1)))/(P(E_(1))P(A//E_(1))+P(E_(2))+(A//E_(2))+P(E_(3))P(A//E_(3)))`
`=(1/3xx1/5)/(1/3xx1/5+1/3xx1/6+1/3xx1/7)=42/107`