Answer :

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)}

And B denotes the event of getting a total of 9

∴ B = {(3,6), (6,3), (4,5), (5,4)}

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

P(A’ ∩ B’) = ?

{using De Morgan’s theorem}

As, P(A’ ∩ B’) = P(A ∪ B)’

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)

As P(A ∩ B) = 0 since nothing is common in set A and B

⇒ n(A ∩ B) = 0

Hence,

Hence,

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