Answer
$\frac{35}{68}$
Work Step by Step
Let $E_{1}$ be the event of choosing box 1, $E_{2}$ be the event of choosing box 2, and A be the event of selecting an orange ball.
Then, $P(E_{1})=P(E_{2})=\frac{1}{2}$,
$P(A|E_{1})=\frac{3}{7}$ and $P(A|E_{2})=\frac{5}{11}$.
Using Baye's theorem, we get
$P(E_{2}|A)=\frac{P(E_{2})P(A|E_{2})}{P(E_{1})P(A|E_{1})+P(E_{2})P(A|E_{2})}=\frac{\frac{1}{2}\times\frac{5}{11}}{\frac{1}{2}\times\frac{3}{7}+\frac{1}{2}\times\frac{5}{11}}=\frac{35}{68}$
