Question
$$A,\,B,\,C$$ are three events for which $$P\left( A \right) = 0.6,\,P\left( B \right) = 0.4,\,P\left( C \right) = 0.5,\,P\left( {A \cup B} \right) = 0.8,\,P\left( {A \cap C} \right) = 0.3$$ and $$P\left( {A \cap B \cap C} \right) = 0.2.$$
If $$P\left( {A \cup B \cup C} \right) \geqslant 0.85$$ then the interval of values of $$P\left( {B \cap C} \right)$$ is :
A.
$$\left[ {0.2,\,0.35} \right]$$
B.
$$\left[ {0.55,\,0.7} \right]$$
C.
$$\left[ {0.2,\,0.55} \right]$$
D.
none of these
Answer :
$$\left[ {0.2,\,0.35} \right]$$
Solution :
$$\eqalign{
& \,\,\,\,\,P\left( {A \cup B \cup C} \right) \cr
& = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) - P\left( {A \cap B \cap C} \right) \cr
& = 0.6 + 0.4 + 0.5 - 0.2 - P\left( {B \cap C} \right) - 0.3 + 0.2 \cr
& = 1.2 - P\left( {B \cap C} \right) \cr} $$
$$\eqalign{
& {\text{because }}P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr
& \Rightarrow 0.8 = 0.6 + 0.4 - P\left( {A \cap B} \right) \cr
& {\text{But }}0.85 \leqslant P\left( {A \cup B \cup C} \right) \leqslant 1 \cr
& \therefore \,0.85 \leqslant 1.2 - P\left( {B \cap C} \right) \leqslant 1 \cr
& \Rightarrow 0.2 \leqslant P\left( {B \cap C} \right) \leqslant 0.35 \cr} $$