Question

Two dice are thrown simultaneously. What is the probability the total is greater than 8?

Answer 1

A couple way of doing this. You could draw out the table of all possible sums of 2 die rolls and count up the ones that are larger than 8.

I'll probably really confuse you with my approach, so I apologize in advance!
My approach would be the following: Fix one dice roll, and determine what the other dice roll must be. For example, if we rolled a 4, we would need the other die to be a 5 or 6.
If we let D1 = dice one and let D2 = dice two, we'd have something like:
$$P(D1 = 1)P(D2 > (8-1)) + P(D1 = 2)P(D2 > (8-2)) + \cdots \\ + P(D1 = 6)P(D2 > (8-6))$$
We know $$P(D1=1) = P(D1 = 2) = \cdots = P(D1 = 6) = \frac{1}{6}$$
so we can pull that out front:
$$= \frac{1}{6}(0+0+\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}) = \frac{10}{36}$$

So that likely wasn't very clear!

Answer 2

Basically, out of all 36 dice rolls, 10 of them will lead to a sum larger than 8.

Answer 3

thnx so much