Answer
$\frac{5}{12}$
Work Step by Step
Let's define the events:
$A_1$: "urn I was chosen"
$A_2$: "urn II was chosen"
$A_3$: "urn III was chosen"
$B$: "chip drawn was red"
Notice that: $P(A_1)=P(A_2)=P(A_3)=\frac{1}{3}$
$P(A_3~|~B)=\frac{P(B~|~A_3)P(A_3)}{P(B~|~A_1)P(A_1)+P(B~|~A_2)P(A_2)+P(B~|~A_3)P(A_3)}$
$P(A_3~|~B)=\frac{\frac{5}{8}\times\frac{1}{3}}{\frac{3}{8}\times\frac{1}{3}+\frac{4}{8}\times\frac{1}{3}+\frac{5}{8}\times\frac{1}{3}}=\frac{\frac{5}{24}}{\frac{3}{24}+\frac{4}{24}+\frac{5}{24}}=\frac{5}{12}$
