A box contains eight balls, two of each color red, blue, green and purple. Amit randomly selects two of the balls without replacement, records the colors of the two balls, and then returns the balls to the box. Then Rohit also randomly selects two of the balls, records the colors, and returns the balls to the box. Finally, Sumit repeats what Amit and Rohit have done. The probability that the six colors recorded by Amit, Rohit, and Sumit include at least one of the colors red, blue, green, and purple is \frac{m}{n}, where m and n are relatively prime positive integers. Find m. 

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:

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

####

Explore the probability problem involving selecting balls of different colors from a box with specific rules. Understand how to calculate the probability using combinatorial methods and multinomial coefficients. Suitable for Grades 10-12 students.

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