Problem 142

First, we identify the total number of ways to select two types of flowers from the four available colors to plant in one flower bed, with the remaining two types automatically assigned to the other flower bed. This is a combination problem, where we must choose 2 out of 4, which can be calculated by the combination formula $C_{4}^{2}$.

The total number of ways to do this is:
$$C_{4}^{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$$

Next, we need to calculate the number of ways the red and purple flowers can be in separate flower beds. There are two scenarios where this can happen:
1. Red and purple flowers are chosen to be planted together (which we want to avoid).
2. Red and purple flowers are chosen to be in separate beds.

For the scenario where red and purple flowers end up in the same bed, there are 2 ways to complete this selection: either both red and purple are in the first bed or both are in the second bed.

Thus, the number of ways where red and purple are in different beds is:
$$Total\ ways\ -\ Unwanted\ ways = 6\ -\ 2 = 4$$

Therefore, the probability that the red and purple flowers are not in the same flower bed is:
$$P(\text{red and purple not together}) = \frac{\text{Desired outcomes}}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3}$$

So, the final answer, expressing the probability that red and purple flowers are not in the same flower bed, is:
$$\boxed{\frac{2}{3}}$$