Q. There are three urns :first urn contains 1 white,2 red,3 green balls,second urn contains 2 white,1 red,1 green balls and third urn contains 4 white,5 red,3green balls. two balls are drawn from an urn chosen at random.these are found to be one white and one green. Probability that balls are drawn from third urn is,
Option
(A)15/59
(B)14/59
(C)13/59
(D)12/59

• prob of urn selection= 1/3 each
  III urn = 1/3*[(4*3)/12c2]
  II urn= 1/3*[(3*1)/6c2]
  I urn= 1/3*[(2*1)/4c2]
  
  by bayes theorem:
  
  p(III/A)=
  1/3*[(4*3)/12c2]divided by 1/3*[(4*3)/12c2] + 1/3*[(3*1)/6c2] + 1/3*[(2*1)/4c2]
  
  then we get:
  15/59 ans.
• 10 years agoHelpfull: Yes(4) No(0)
• answer is 5/59
  which is not in option
  
• 9 years agoHelpfull: Yes(0) No(1)