check the table attached. It represents all the sums of the outcomes of 2 dices, say the 'blue' dice, and the 'red' dice.
there is 1 way of getting the sum 2 (blue 1, red 1)
there are 2 ways of getting the sum 3 {(blue 1, red 2), (red 1, blue 2)}
.
.
and so on, we see that there are 6*6=36 total outcomes, and 5 of these outcomes produce a sum equal to 8.
these are {(r2, b6), (r3, b5), (r4, b4), (r5, b3), (r6, b2)}
P(sum=8)=n(outcomes with sum =8)/total number of outcomes=5/36
