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 double.
So, A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
P (A) = n (A) / n (S)
= 6/36
= 1/6
And ‘B’ denotes the event of getting a total of 9
So, B = {(3,6), (6,3), (4,5), (5,4)}
P (B) = n (B) / n (S)
= 4/36
= 1/9
We need to find probability of neither 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)
By using the 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) = 1/6 + 1/9 + 0
= 5/18 {Since, P (A ∩ B) = 0 since nothing is common in set A and B.
So, n (A ∩ B) = 0}
Hence,
P (A′ ∩ B′) = 1 – (5/18)
= 13/18
