The correct answer is D.
In arrangement D, there are two white balls in the first bin and one white ball in the second bin. When a bin is chosen randomly, the probability of choosing the first bin is $\frac{1}{2}$. When the first bin is chosen, the probability of picking a white ball is $\frac{2}{3}$. Therefore, the overall probability of picking a white ball in arrangement D is $\frac{1}{2} \times \frac{2}{3} = \frac{2}{3}$.
In arrangement A, there is one white ball in each bin. When a bin is chosen randomly, the probability of choosing either bin is $\frac{1}{2}$. When either bin is chosen, the probability of picking a white ball is $\frac{1}{2}$. Therefore, the overall probability of picking a white ball in arrangement A is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.
In arrangement B, there are two white balls in the second bin and one white ball in the first bin. When a bin is chosen randomly, the probability of choosing the second bin is $\frac{1}{2}$. When the second bin is chosen, the probability of picking a white ball is $\frac{2}{3}$. Therefore, the overall probability of picking a white ball in arrangement B is $\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$.
In arrangement C, there is one white ball in each bin. When a bin is chosen randomly, the probability of choosing either bin is $\frac{1}{2}$. When either bin is chosen, the probability of picking a white ball is $\frac{1}{2}$. Therefore, the overall probability of picking a white ball in arrangement C is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.