Aptitude - Probability - Discussion

Anyone tell me, why not taking (6, 3)?

Hi when would be the sample space be 12 and when would it be 36? I mean if a dice is thrown twice or if two dice are thrown. What would be the difference in the sample space?

In the answer, why didn't they count (1,1) twice as it is possible that the number 1 might come on either of the dices, as they calculated (1,2) & (2,1)

1+1=2 here 2 is not a prime number then please explain why (1, 1) used here?

Explain, please.

The sum of (1, 4) (1, 6) is not prime.

Prime numbers
2={1,1}
3={1,2};{2,1}
5={1,4};{4,1};{2,3};{3,2}
7={1,6};{6,1};{2,5};{5,2};{3,4};{4,3}
11={5,6};{6,5}
Total 15 possible outcomes.

So, Probability is =15/36=5/12.

I am sure the answer is 1/6 for this as this event mentioned has something missing.

E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
(5, 2), (5, 6), (6, 1), (6, 5) }

E(correct) = { (1,1), (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) }
(1,1) will have a probability 2/12 rather than 1/12 un-like other pairs.

If you note (2,1) occurs and (1,2) also occurs and hence (1,1) due to dice A and additionally (1,1) due to dice B and hence the probability of this event is doubled.

Why we enter prime numbers sum as 15? Please explain the step.

Please, from where E={(1,1),(1,2)...} is came? Please explain.

Can anyone explain the simplest method to find numarator?