Q. 85.0( 1 Vote )
Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will be
A. 
B. 
C. 
D. 
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)}
∴ 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.
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