= 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 gives
$$
\boxed{\mathbb P(\text{remaining ball red}\mid D)=\mathbb P(H\mid D)=\frac{1\cdot(1/2)}{3/4}=\frac23.}
$$
The observation increases the weight of the two-red-ball box, because that box is more likely to yield a red first draw.
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