In how many ways can one get a sum greater than #8#, if two cubes, having numbers #1# to #6# are rolled?

Question

2287 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-17T04:37:31

There are ten number of ways in which one could get a sum that is greater than #8#.

As each cube contains number #1# to #6# (i.e. #6# options), there are #6xx6=36# options.

These options start from #(1,1)# and go on till #(6,6)#.

#(1,1),(1,2),....(1,5),*1,6),(2,1),....(2,6),(3,1),....(6,6)#

Observe that we can get a sum maximum of #12#

What is desired is the sum that is greater than #8#, i.e. #9#, #10#, #11# and #12#.

You can get #12# in only one way i.e. #(6,6)#.

#11# can be got in two ways #(5,6)# and #(6,5)#

#10# can be got in three ways #(4,6)#, #(5,5)# and #(6,4)# and

#9# can be got in four ways #(3,6)#, #(4,5)#, #(5,4)# and #(6,3)#

Hence, there are ten number of ways in which one could get a sum that is greater than #8#.