Let $$A,\,B,\,C$$   be three events. If the probability of occurring exactly one event out of $$A$$ and $$B$$ is $$1 - a,$$  out of $$B$$ and $$C$$ and $$A$$ is $$1 - a$$  and that of occurring three events simultaneously is $${a^2},$$  then the probability that at least one out of $$A,\,B,\,C$$   will occur is :

Solution :

$$\eqalign{ & P\,\,\left( {{\text{exactly one event out of }}A{\text{ and }}B{\text{ occurs}}} \right) \cr & = P\left[ {\left( {A \cap B'} \right) \cup \left( {A' \cap B} \right)} \right] \cr & = P\left( {A \cup B} \right) - P\left( {A \cap B} \right) \cr & = P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) \cr & \therefore \,P\left( A \right) + P\left( B \right) - 2P\left( {A \cap B} \right) = 1 - a......\left( 1 \right) \cr & {\text{Similarly,}} \cr & P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) = 1 - 2a......\left( 2 \right) \cr & P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) = 1 - a......\left( 3 \right) \cr & P\left( {A \cap B \cap C} \right) = {a^2}......\left( 4 \right) \cr & {\text{Now,}} \cr & P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right) \cr & = \frac{1}{2}\left[ {P\left( A \right) + P\left( B \right) - 2P\left( {B \cap C} \right) + P\left( B \right) + P\left( C \right) - 2P\left( {B \cap C} \right) + P\left( C \right) + P\left( A \right) - 2P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right)} \right] \cr & = \frac{1}{2}\left[ {1 - a + 1 - 2a + 1 - a} \right] + {a^2}\,\,\,\,\left[ {{\text{using}}\,\left( 1 \right),\left( 2 \right),\left( 3 \right){\text{ and }}\left( 4 \right)} \right] \cr & = \frac{3}{2} - 2a + {a^2} \cr & = \frac{1}{2} + {\left( {a - 1} \right)^2} > \frac{1}{2}. \cr} $$