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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