i'm quite sure there's a neat little formula for this, but i don't remember it, so let's do it the long and hard way.

we want the sum to be at least 5, so it can be 5,4,3 or 2

how many ways are there to get a sum of 5?

1) Dice1 = 4ANDDice2 = 1, Probability = (1/6)*(1/6) = 1/36

2) Dice1 = 3ANDDice2 = 2, Probability = (1/6)*(1/6) = 1/36

3) Dice1 = 2ANDDice2 = 3, Probability = (1/6)*(1/6) = 1/36

4) Dice1 = 1ANDDice2 = 4, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is: 1/36 + 1/36 + 1/36 + 1/36 = 4/36 = 1/9

how many ways are there to get a sum of 4?

1) Dice1 = 3ANDDice2 = 1, Probability = (1/6)*(1/6) = 1/36

2) Dice1 = 2ANDDice2 = 2, Probability = (1/6)*(1/6) = 1/36

3) Dice1 = 1ANDDice2 = 3, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is: 1/36 + 1/36 + 1/36 = 3/36 = 1/12

how many ways are there to get a sum of 3?

1) Dice1 = 2ANDDice2 = 1, Probability = (1/6)*(1/6) = 1/36

2) Dice1 = 1ANDDice2 = 2, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is: 1/36 + 1/36 = 2/36 = 1/18

how many ways are there to get a sum of 2?

1) Dice1 = 1ANDDice2 = 1, Probability = (1/6)*(1/6) = 1/36

so probability of one of these ways happening is 1/36

So the probability of getting a sum of 5OR4OR3OR2 is:

1/9 + 1/12 + 1/18 + 1/36 = 5/18

so, the probability that the sum is 5 or less is5/18