Q. 175.0( 3 Votes )

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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)}





And B denotes the event of getting a total of 9


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





We need to find probability of the event of getting neither a doublet nor a total of 9.


P(A’ B’) = ?


{using De Morgan’s theorem}


As, P(A’ B’) = P(A B)’


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)


As P(A B) = 0 since nothing is common in set A and B


n(A B) = 0




Hence,




Hence,



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