You roll a pair of fair standard dice exactly once. Find the probability of a) the sum is >6; b) the sum is greater than or equal to 10 c) the sum is >6 AND odd; and d) the sum is >6 OR odd.

I'll start you out with an solution to the first problem, then I suggest you try the others and come back for more help if you think you need it.

The P(Sum>6) is P(7)+P(8)+P(9)+P(10)+P(11)+P(12).

Every time you roll a pair of dice you have 6*6 = 36 possible outcomes. To figure the probability of rolling a 7, count up the number of ways you can roll two dice and have them add to 7. It's pretty easy to see that there are 6 ways to roll a sum of 7: 1+6, 2+5, 3+4, 4+3, and 5+2, and 6+1. Hence P(7) is 6/36. Likewise, there are 5 ways to roll an 8, 4 ways to roll a 9, 3 ways for a 10, 2 ways for 11, and 1 way to roll a 12. So the probabiity of rolling greater than 6 is: (6+5+4+3+2+1)/36 = 21/36 = 7/12.

One other hint - to figure the probability that the sum is greater than 6 AND odd, you simply add P(7)+P(9)+P(11), as these are the only rolls that satisfy the condition.