just add them up, considering the 1st throw:
6: 0
5: 1/6
4: 2/6
...
1: 5/6
sum: (1/6)(15/6) = 5/12
since the chance of throwing each of the 1st numbers is 1/6.
A die is thrown twice. Find the probability that the number obtained in the first throw is smaller than that in the second throw.
1 answer