2 fair dice are thrown. How do you calculate the probability that the event that the sum on the 2 dice > 1?

All of the possible outcomes have a sum greater than 1

To calculate the probability that the sum on two fair dice is greater than 1, you first need to find the total number of possible outcomes when throwing two dice, which is 36 (since each die has 6 sides, and there are 6 possibilities for the first die and 6 possibilities for the second die). Then, determine the number of outcomes where the sum is greater than 1.

The only case where the sum is not greater than 1 is when both dice show a 1 (resulting in a sum of 2). Since there is only 1 outcome where this happens, subtract 1 from the total outcomes.

Therefore, the probability that the sum on the two dice is greater than 1 is (36 - 1) / 36, which simplifies to 35/36.

To calculate the probability that the sum on the two dice is greater than 1, we first need to determine the total number of outcomes when two fair dice are thrown. Then, we find the number of outcomes where the sum of the two dice is greater than 1.

Total number of outcomes when two fair dice are thrown: (6 \times 6 = 36) (since each die has 6 sides)

Now, we need to find the outcomes where the sum of the two dice is greater than 1. The sum of 1 occurs only when both dice show 1, which is only 1 out of 36 outcomes.

Therefore, the probability that the sum on the two dice is greater than 1 is:

[ \frac{{\text{Number of outcomes where sum > 1}}}{{\text{Total number of outcomes}}} = \frac{{36 - 1}}{{36}} = \frac{{35}}{{36}} ]

So, the probability is ( \frac{{35}}{{36}} ).