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Question

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) =p2, where 0 < p < 12. Then, the probability of at least one of the three events A, B and C occurring is

A
3p+2p22
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B
p+3p24
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C
p+3p22
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D
3p+2p24
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Solution

The correct option is A 3p+2p22
We know that,
P(exactly one of A or B occurs)
=P(A)+P(B)2P(AB) P(A)+P(B)2P(AB)=p (i) P(B)+P(C)2P(BC)=p (ii) P(C)+P(A)2P(CA)=p (iii)
On adding Eqs. (i), (ii) and (iii), we get
2[P(A)+P(B)+P(C)P(AB)P(BC) P(CA)]=3pP(A)+P(B)+P(C)P(AB)P(BC) P(CA)=3p2 (iv)
Given that P(ABC)=p2 (v) ~P(at least one of the events A, B, and C occurs)
=P(A)+P(B)+P(C)P(AB)P(BC) P(CA)+P(ABC)
=3p2+p2 [from Eqs. (iv) and (v)]
=3p+2p22

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