No. In the denominator you have not counted (4,4) two times.

Have a look at this.

if the first dice is 1 then the second dice can be any of {1,2,3,4,5,6}.
so your pairs will be (1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

Now similarly if the first dice shows up 2 then the second dice can show up any of {1,2,3,4,5,6}.
so your pairs will be (2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

Similarly write down all the possibilites and you wont get (4,4) counted twice.

Does this explain your question?

Can anyone explain little more detail of the above problem since I am stuck with this Quant question?

Thanks,
Pavan

possible options that will sum up to 8 are:
2,6 prob of 2 is 1/6 same for 6 so there joint probability is 1/6*1/6=1/36
3,5 1/36 for reason mentioned above
4,4 1/36 for reason mentioned above
5,3 1/36 for reason mentioned above
6,6 1/36 for reason mentioned above

so the sum is 5/36