For 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 < 1/2. Then the probability of at least one of the three events A, B and C occurring is

$\frac{3 p+2 p^2}{2}$

Given P(A) + P(B) – 2P(AB) = P

P(B) + P(C) – 2P(BC) = P

P(C) + P(A) – 2P(AC) = P

2 [P(A) + P(B) + P(C) – P(AB) + P(BC) + P(CA)] = $\frac{3P}{2}$

and P(ABC) = $P^2$

Hence $P(A \cup B \cup C)=\frac{3 P}{2}+P^2=\frac{3 P+2 P^2}{2}$