We split the problem into two cases. Case I: A red ball is removed. The probability that a red ball is removed is $ \frac{4}{6} = \frac{2}{3} $. After it is replaced by a white ball, the probability of drawing a red ball is $ \frac{1}{2} $. Thus, the probability that a red ball will be drawn in this case is $ \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3} $. Case II: A white ball is removed. The probability that a white ball is removed is $ \frac{2}{6} = \frac{1}{3} $. After it is replaced by a red ball, the probability of drawing a red ball is $ \frac{5}{6} $. Thus, the probability that a red ball will be drawn in this case is $ \frac{5}{18} $. We add the two probabilities for a total probability of $ \boxed{\frac{11}{18}} $.