To solve this problem, let's carefully analyze the probability that the six colors recorded by Amit, Rohit, and Sumit include at least one of each color: red, blue, green, and purple.

Step 1: Understanding the Problem

Each of Amit, Rohit, and Sumit randomly selects two balls from the box, records the colors, and returns the balls to the box. Since there are 8 balls (2 of each color), each ball has a 1/4 chance of being any specific color.

We need to calculate the probability that the six colors recorded include at least one ball of each color (red, blue, green, and purple).

Step 2: Total Number of Outcomes

Each person selects 2 balls. Therefore, the total number of color pairs that can be chosen by each person is $\binom{8}{2} = 28$. However, this count is unnecessary since the selections are unordered and each pair's order doesn’t matter.

Since Amit, Rohit, and Sumit each independently choose their pairs of colors, and considering that they replace the balls back into the box each time, the total number of possible outcomes of the recorded six colors (not considering order of picks) is:

$\text{Total possible outcomes} = \left(\binom{4}{2}\right)^3 = 6^3 = 216$

Step 3: Counting the Favorable Outcomes

Now, we need to consider the complementary scenario, where at least one of the four colors (red, blue, green, purple) is not recorded in the results of the six balls.

Case 1: Only 3 colors appear

For only 3 out of the 4 colors to appear, one of the colors must be completely missing. There are 4 ways to choose the missing color. If we assume a particular color is missing, the other 3 colors will each need to appear among the 6 selected balls.

The number of ways to divide 6 selections among the 3 remaining colors so that none is excluded can be calculated using a stars and bars approach. We can use the following formula:

$\text{Number of ways to distribute 6 balls among 3 colors} = \binom{8}{2} = 28$

However, we should use a multinomial approach as each specific selection can be interpreted as a product of combinations:

$\text{Favorable for 3 colors} = \binom{6}{2,2,2} \times 3! = \frac{6!}{2!2!2!} = 90$

Since this case occurs for any of the 4 missing colors, we multiply by 4:

$\text{Total number of outcomes for Case 1} = 4 \times 90 = 360$