 Solution

Let $H$ mean the added ball was red, and $D$ mean the first draw was red. Their relevant probabilities are $\mathbb{P}(H)=\mathbb{P}(H^c)=1/2$, $\mathbb{P}(D\mid H)=1$, and $\mathbb{P}(D\mid H^c)=1/2$. The law of total probability gives $\mathbb{P}(D)=3/4$.

Given the observed red draw, the remaining ball is red precisely when $H$ occurred. Bayes' theorem therefore givesThe observation increases the weight of the two-red-ball box, because that box is more likely to yield a red first draw.