In a single throw of 2 die, we have total 36(6 × 6) outcomes possible.

Say, n(S) = 36 where S represents Sample space

Let A denotes the event of getting a doublet(equal number)

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

∴ P(A) =

And B denotes the event of getting a total of 9

∴ B = {(3,6), (6,3), (4,5), (5,4)}

P(B) =

We need to find probability of the event of getting neither a doublet nor a total of 9.

P(A’ ∩ B’) = ?

As, P(A’ ∩ B’) = P(A ∪ B)’ {using De Morgan’s theorem}

P(A’ ∩ B’) = 1 – P(A ∪ B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

∴ P(A ∪ B) = {As P(A ∩ B) = 0 since nothing is common in set A and B ⇒ n(A ∩ B) = 0 }

Hence,

P(A’ ∩ B’) = 1 – (5/18) = 13/18

Hence,

P(required event) = 13/18

As our answer matches only with option (b)

∴ Option (b) is the only correct choice.