Question

A single​ 6-sided die is rolled twice. the set of 36 equally likely outcomes is​ {(1,1), (1,2),​ (1,3), (1,4), left parenthesis 1 comma 5 right parenthesis comma left parenthesis 1 comma 6 right parenthesis comma left parenthesis 2 comma 1 right parenthesis comma left parenthesis 2 comma 2 right parenthesis comma ​(2,3), (2,4),​ (2,5), (2,6),​ (3,1), (3,2),​ (3,3), (3,4),​ (3,5), (3,6),​ (4,1), (4,2),​ (4,3), left parenthesis 4 comma 4 right parenthesis comma left parenthesis 4 comma 5 right parenthesis comma left parenthesis 4 comma 6 right parenthesis comma left parenthesis 5 comma 1 right parenthesis comma left parenthesis 5 comma 2 right parenthesis comma ​(5,3), (5,4),​ (5,5), (5,6),​ (6,1), (6,2),​ (6,3), (6,4),​ (6,5), (6,6)}. find the probability of getting two numbers whose sum is greater than 9.

Answer 1

Sum greater than 9 from the Sample Space are:

{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)}

so, in total 6 out of total 36 combinations have a sum greater than 9

So, $Probability(sum>9)=\frac{6}{36}$

$Probability(sum>9)=\frac{1}{6}$

Probability(sum>9)=0.167