Answer
$\frac{3}{4}$
Work Step by Step
Let $E_{1}$ be the event of picking the box 1, $E_{2}$ be the event of picking the box 2 and A be the event of selecting a blue ball.
Then, $P(E_{1})=P(E_{2})=\frac{1}{2}$
Also, $P(A|E_{1})=\frac{3}{5}$ and $P(A|E_{2})=\frac{1}{5}$
Using Baye's theorem, we get
$P(E_{1}|A)=\frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1})+P(E_{2})P(A|E_{2})}=\frac{\frac{1}{2}\times\frac{3}{5}}{\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{1}{5}}=\frac{3}{4}$
