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`

Options

A. `16//29`
B. `1//15`
C. `27//59`
D. `42//107`

Answer

D

Explanation

Let `E_(1), E_(2) " and " E_(3)` denote the events of selecting box A, B, C respectively and A be the event that a screw selected at random is defective. Then,
`P(E_(1))=P(E_(2))=P(E_(3))=1//3`
`P(A//E_(1))=(1)/(5), P(A//E_(2))=(1)/(6), P(A//E_(3))=(1)/(7)`
By Baye's rule, we have
Required probability `=P(E_(1)//A)`
`=(P(E_(1)P(A//E_(1))))/(P(E_(1))P(A//E_(1))+P(E_(2))P(A//E_(2))+P(E_(3))P(A//E_(3)))`
`=((1)/(3)xx(1)/(5))/((1)/(3)xx(1)/(5)xx(1)/(3)xx(1)/(6)+(1)/(3)xx(1)/(7))=(42)/(107)`