Question

Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will be 

A. \(\frac{13}{15}\)

B. \(\frac{13}{18}\)

C. \(\frac{1}{9}\)

D. \(\frac{8}{9}\)

Answer 1

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) = \(\frac{n(A)}{n(S)} = \frac{6}{36} = \frac{1}{6}\) 

And B denotes the event of getting a total of 9 

∴ B = {(3,6), (6,3), (4,5), (5,4)} 

P(B) = \(\frac{n(B)}{n(S)} = \frac{4}{36}\) = \(\frac{1}{9}\)

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) = \(\frac{1}{6}+\frac{1}{9}+0 = \frac{5}{18}\) {As P(A∩B) = 0 since nothing is common in set A and B ⇒ n(A∩B) = 0 }

Hence, 

P(A’∩B’) = 1 –  \(\Big(\frac{5}{18}\Big) \)= \(\frac{13}{18}\) 

Hence, 

P(required event) = \(\frac{13}{18}\)

As our answer matches only with option (b) 

∴ Option (b) is the only correct choice.