To find the probability of not rolling a sum of 4 when two six-sided dice are tossed, let's go through the steps:

1. Identify all possible outcomes: Each die has 6 faces, so there are $6 \times 6 = 36$ total possible outcomes when rolling two dice.

2. Count the successful outcomes for a sum of 4:

   • The pairs that result in a sum of 4 are: $(1,3), (2,2),$ and $(3,1)$.
   • There are 3 outcomes where the sum is 4.

3. Compute the probability of rolling a sum of 4: $P(\text{sum of 4}) = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}$

4. Compute the probability of NOT rolling a sum of 4: $P(\text{not sum of 4}) = 1 - P(\text{sum of 4}) = 1 - \frac{1}{12} = \frac{11}{12}$

So, the probability of not rolling a sum of 4 is $\frac{11}{12}$.

Would you like further clarification on this process?

Here are five additional questions related to this topic for further exploration:

1. What is the probability of rolling a sum of 7 with two six-sided dice?
2. How would the probability change if three six-sided dice are used and a sum of 4 is considered?
3. Can you determine the probability of rolling an even sum with two six-sided dice?
4. What is the probability of rolling a sum less than 5 with two six-sided dice?
5. How would you calculate the probability of rolling a sum that is a prime number?

Tip: Remember that for probability calculations, breaking down possible outcomes into smaller subsets can simplify the counting process.