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For the three events A, B and C, P(exactly one of the events A or B occurs) = 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) = $P 2$ , where $0 < P < 12$ . Then what is the probability of at least one of the three events A, B and C occurring?

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3 p − p 2 2

p − p 2 2

3 p − 2 p 2 2

answer is D.

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Detailed Solution

Given that

P e x a c t l y o n e o f t h e e v e n t s A o r B o c c u r s = p

.
By formula, the probability of occurrence of exactly one event from A to B is

P (A) + P (B) − 2 P (A ∩ B)

. Then, the above equation becomes,

P (A) + P (B) − 2 P (A ∩ B) = p

… (1)
Similarly, for event B or C, we have

P (B) + P (C) − 2 P (B ∩ C) = p

… (2)
Similarly, for event C or A, we have

P (C) + P (A) − 2 P (C ∩ A) = p

… (3)
Add the equation (1), (2), and (3), and we have
Question Image

Given that the probability of all the three events occurs simultaneously is

P (A ∩ B ∩ C) = p 2

. By formula, the probability of occurrence of at least one of the three events, A, B, and C is

P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (B ∩ C) − P (C ∩ A) + P (A ∩ B ∩ C)

.
Then, the equation becomes,

P (A ∪ B ∪ C) = 3 p 2 + p 2 = 3 p + 2 p 2 2

Hence, the correct option is 4.

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