One of the two boxes, box I and box II, was selected at random and balls are drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 1/3, then the correct option with the possible values of n1, n2, n3 and n4 is (are) n1=3,n2=3,n3=5,n4=15; n1=8,n2=6,n3=5,n4=20; n1=6,n2=12,n3=5,n4=20; n1=3,n2=6,n3=10,n4=50

Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II. Let E1 be the event of selecting box I and E2 be the event of selecting box II. R be the event of selecting a Red ball. P(E1) = P(E2) = 1/2 P(E2/R) = 1/3 ⇒ P(E2/R) = (n3/(n3+n4)) / ((n1/(n1+n2)) + (n3/(n3+n4))) = 1/3 only options A and B satisfies the above relation.