Answer
$\frac{4}{7}$
Work Step by Step
Let's define the events:
$A_1$: "chip transferred from urn I to urn II was white"
$A_2$: "chip transferred from urn I to urn II was red"
$B$: "chip selected from urn II was red"
$P(A_1~|~B)=\frac{P(B~|~A_1)P(A_1)}{P(B~|~A_1)P(A_1)+P(B~|~A_2)P(A_2)}$
$P(A_1~|~B)=\frac{\frac{2}{4}\times\frac{2}{3}}{\frac{2}{4}\times\frac{2}{3}+\frac{3}{4}\times\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{4}}=\frac{\frac{1}{3}}{\frac{7}{12}}=\frac{4}{7}$
