Question

A single 6-sided 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.

Answer 1

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?

Answer 2

Thank you so much David Q/R.

Answer 3

You're welcome!