4 boxes, 8 balls

Let R2 = probability that a red ball was picked from the second box

Let W4 = probability that a white ball was picked out from the fourth box

Using Baye's theorem ,

$P(R2|W4) = \frac{P(W4|R2)*P(R2)}{P(W4)} = \frac{(8/9)*(1/6)}{53/54} = \frac{8}{53}$

And $8 +53 = \boxed{61}$

Let (R/W)N is the probability that the ball from the N-th box is Red/White

$P(R2|W4) = \frac {P(R2 ∩ W4)}{P(W4)}$

$P(W4) = 1 - P(R4) = 1 - \frac {1}{2} \cdot \frac {1}{3} \cdot \frac {1}{3} \cdot \frac {1}{3} = \frac {53}{54}$

Now we need to find $P(R2 ∩ W4)$

$P(R2) = \frac {1}{2} \cdot \frac {1}{3} = \frac {1}{6}$

$P(R2 ∩ W4) = P(R2) - P(R2 ∩ R4)$

Obviously, $P(R2 ∩ R4) = P(R4)$ because if the last ball is red, then every ball drawn is red

$P(R2 ∩ W4) = \frac {1}{6} - \frac {1}{54} = \frac {4}{27}$

$P(R2|W4) = \frac {\frac {4}{27}}{\frac {53}{54}} = \frac {8}{53}$

$8 + 53 = \boxed {61}$