Math for liberal arts
posted by KeKe .
A single 6sided die is rolled twice.
C. Find the odds in favor of getting 2 numbers whose sum is greater than 5.
I would simply like to be shown the steps to work this problem.

To get it right, make sure you distinguish between the first and second throw. If you do that, you'll see that there are 36 distinct outcomes: (1, 1), (1, 2), (1, 3), ... (6, 4), (6, 5) and (6, 6). They're all equally likely with probability 1/36, so how many of them total more than 5? We could just enuerate them all  or better, work out how many total five or less (because there are fewer of them), and then subtract the probability from 1. These are: (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2) and (4, 1). There are 10 of them, so I reckon the answer ought to be 1  (10/36) = 0.72. Do you agree?

Thank you so much David Q/R.

You're welcome!