Question
If $$A,\,B,\,C$$ are events such that $$P\left( A \right) = 0.3,\,P\left( B \right) = 0.4,\,P\left( C \right) = 0.8,\,P\left( {A \cap B} \right) = 0.08,\,P\left( {A \cap C} \right) = 0.28,\,P\left( {A \cap B \cap C} \right) = 0.09$$
If $$P\left( {A \cup B \cup C} \right) \geqslant 0.75,$$ then find the range of $$x = P\left( {B \cap C} \right)$$ lies in the interval :
A.
$$0.23 \leqslant x \leqslant 0.48$$
B.
$$0.23 \leqslant x \leqslant 0.47$$
C.
$$0.22 \leqslant x \leqslant 0.48$$
D.
none of these
Answer :
$$0.23 \leqslant x \leqslant 0.48$$
Solution :
$$\eqalign{
& {\text{Since, }}P\left( {A \cup B \cup C} \right) \cr
& = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {B \cap C} \right) - P\left( {C \cap A} \right) + P\left( {A \cap B \cap C} \right) \cr
& {\text{or }}P\left( {A \cup B \cup C} \right) = 0.3 + 0.4 + 0.8 - \left( {0.08 + 0.28 + P\left( {B \cap C} \right)} \right) + 0.09 \cr
& F = 1.23 - P\left( {B \cap C} \right) \cr
& {\text{or }}P\left( {B \cap C} \right) = 1.23 - P\left( {A \cup B \cup C} \right) \cr
& {\text{But we know that }}0 \leqslant P\left( {A \cup B \cup C} \right) \leqslant 1 \cr
& {\text{Hence, }}0.23 \leqslant P\left( {B \cap C} \right) \leqslant 0.48 \cr} $$