In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will appear.

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) = \( \dfrac{ n (A) }{ n (S)}\)

= \( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

And ‘B’ denotes the event of getting a total of 9

So, B = {(3,6), (6,3), (4,5), (5,4)}

P (B) =\( \dfrac{ n (B) }{ n (S)}\)

= \( \dfrac{4}{36}\)

= \( \dfrac{1}{9}\)

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) = \( \dfrac{1}{6}\) + \( \dfrac{1}{9}\)+ 0

= \( \dfrac{5}{18}\) {Since, P (A ∩ B) = 0 since nothing is common in set A and B.

So, n (A ∩ B) = 0}

Hence,

P (A′ ∩ B′) = 1 – \( ( \dfrac{5}{18})\)

= \( \dfrac{13}{18}\)