Let `U_(1)`={2 white, 3 black balls}
`U_(2)`={3 white , 2 black balls}
and `U_(3)`={4 white ,1 black balls}
`thereforeP(U_(1))=P(U_(2))=P(U_(3))=1/3`
Let `E_(1)` be the event that a ball is chosen from urn U, E be the event that a ball is chosen from urn `U_(1)` and `E_(2)` be the event that a ball is chosen from urn `U_(2)` and `E_(3)` be the event that a ball is chosen from urn `U_(3)`.
Also, `P(E_(1))=P(E_(2))=P(E_(3))=1//3`
Let E be the event that white ball is drawn
`therefore P(E//E_(1))=2/5,P(E//E_(2))=3/5,P(E//E_(3))=4/5`
Now, `P(E_(2)//E)=(P(E_(2))cdotP(E//E_(2)))/(P(E_(1))cdotP(E//E_(1))+P(E_(2))cdotP(E//E_(2))+P(E_(3))cdotP(E//E_(3)))`
`=(1/3cdot3/5)/(1/3cdot2/5+1/3cdot3/5+1/3cdot4/5)`
`=(3/15)/(2/15+3/15+4/15)=3/9=1/3`