$\sum_{x=0}^{6} p(x ; 6)-\sum_{x=0}^{2} p(x ; 6)$.

Step 1 ：First, we need to understand the meaning of the problem. The problem is asking us to calculate the sum of probabilities for a certain event from x=0 to x=6, and then subtract the sum of probabilities for the same event from x=0 to x=2.

Step 2 ：The function p(x ; 6) represents the probability of the event happening x times in 6 trials. We don't know the exact nature of the event, but we can still calculate the sums.

Step 3 ：We start by calculating the first sum: $\sum_{x=0}^{6} p(x ; 6)$. This is the sum of the probabilities of the event happening 0 times, 1 time, 2 times, ..., up to 6 times in 6 trials.

Step 4 ：Since we don't know the exact probabilities, we can't calculate this sum directly. However, we know that the sum of all probabilities for all possible outcomes of an event is always 1. So, $\sum_{x=0}^{6} p(x ; 6) = 1$.

Step 5 ：Next, we calculate the second sum: $\sum_{x=0}^{2} p(x ; 6)$. This is the sum of the probabilities of the event happening 0 times, 1 time, or 2 times in 6 trials.

Step 6 ：Again, we can't calculate this sum directly because we don't know the exact probabilities. However, we know that this sum must be less than or equal to 1, because it's the sum of some (but not all) of the probabilities for all possible outcomes of the event.

Step 7 ：Therefore, we can't find a numerical value for $\sum_{x=0}^{2} p(x ; 6)$, but we know that it's less than or equal to 1.

Step 8 ：Finally, we subtract the second sum from the first sum: $\sum_{x=0}^{6} p(x ; 6) - \sum_{x=0}^{2} p(x ; 6) = 1 - \sum_{x=0}^{2} p(x ; 6)$.

Step 9 ：Since $\sum_{x=0}^{2} p(x ; 6)$ is less than or equal to 1, the result of the subtraction is greater than or equal to 0.

Step 10 ：Therefore, the solution to the problem is a value greater than or equal to 0. However, without knowing the exact probabilities, we can't find a more specific solution.

Step 11 ：To check our solution, we can see that it meets the requirements of the problem. The problem asked for the difference between two sums of probabilities, and we found that this difference is a value greater than or equal to 0, which makes sense in the context of probabilities.