When two dice are throw, then Total outcome = 36

A doublet: {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable outcome = 6

Sum is 10: {(4,6),(5,5),(6,4)}

Favourable outcome = 3

Again, A doublet and sum is 10: (5, 5)

Favourable outcome = 1

Now, P(either dublet or a sum of 10 appears) = P(A dublet appear) + P(sum is 10) - P(A dublet appear and sum is 10)

=> P(either dublet or a sum of 10 appears) = 6/36 + 3/36 - 1/36

                                                          = (6 + 3 - 1)/36

                                                          = 8/36

                                                          = 2/9

So, P(neither dublet nor a sum of 10 appears) = 1 - 2/9 = 7/9