For the three events $$A,\,B$$  and $$C,\,P$$  (exactly one of the events $$A$$ or $$B$$ occurs) $$= P$$  (exactly one of the two 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 $$0 < p < \frac{1}{2}.$$   Then the probability of at least one of the three events $$A,\,B$$  and $$C$$ occurring is :

Solution :
$$\eqalign{ & {\text{We know that }}P{\text{ }}\left( {{\text{exactly one of }}A{\text{ or }}B{\text{ occurs}}} \right) \cr & = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) \cr & \therefore \,P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) = p......\left( 1 \right) \cr & {\text{Similarly,}} \cr & P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) = p......\left( 2 \right) \cr & {\text{and }}P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) = p......\left( 3 \right) \cr & {\text{Adding equation}}\left( 1 \right),\,\left( 2 \right){\text{ and }}\left( 3 \right){\text{ we get}} \cr & 2\left[ {P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right)} \right] = 3p \cr & \Rightarrow \left[ {P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right)} \right] = \frac{{3p}}{2}.....\left( 4 \right) \cr & {\text{It is also given that}} \cr & P\left( {A \cap B \cap C} \right) = {p^2}......\left( 5 \right) \cr & {\text{Now,}} \cr & P\left( {{\text{at least one of }}A,{\text{ }}B{\text{ and }}C} \right) \cr & = p\left( A \right) + p\left( B \right) + p\left( C \right) - p\left( {A \cap B} \right) - p\left( {B \cap C} \right) - p\left( {C \cap A} \right) + p\left( {A \cap B \cap C} \right) \cr & = \frac{{3p}}{2} + {p^2} \cr & = \frac{{3p + 2{p^2}}}{2}\,\,\,\,\,\left[ {{\text{Using equation }}\left( 4 \right)\,{\text{and}}\left( 5 \right)} \right] \cr} $$