For the three events AB and CP(exactly one of the events A or Boccurs) = P(exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p and P (all the three events occur simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events AB and C occurring is


Correct option is

we know that

                P(exactly one of A or B occurs)

                  = P(A) + P(B) – 2P(A ∩ B)

Therefore,  P(A) + P(B) – 2P(A ∩ B) = p     ... (1)

Similarly,  P(B) + P(C) – 2P(B ∩ C) = p        ...(2)

and,          P(C) + P(A) – 2P(C ∩ A) = p       ...(3)

Adding (1), (2) and (3) we get

               2[P(A) + P (B) + P (C) – P (A ∩ B) – P(B ∩ C) – P (C ∩ A)] = 3p

⇒           P(A) + P (B) + P (C) – P (A ∩ B) – P(B ∩ C) – P (C ∩ A)] = 3p/2                 (4)

We are also given that P (A ∩ ∩ C) = p5         (5)

Now, P (at least one of AB and C)

           = P(A) + P (B) + P (C) – P (A ∩ B– P(B ∩ C) – P (C ∩ A) + P (A ∩ ∩ C)




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