What are the odds in favour of a total greater than 9 in a given roll of two standard dice?

Question

What are the odds in favour of a total greater than 9 in a given roll of two standard dice?

Answer 1

P(>9) = P(10)+P(11) + P(12)

= 1/1024 + 10/1024 + 45/1024= 56/1024

= about .055% = 5.5% chance of more than 9

same odds as getting 10,11 or 12

look at the 10th row of Pascal's Triangle

sum them to get 1024

add the 1st 3 entries: 1+10+45 = 56

divide 56 by 1024