If $A, B, C$ are events such that $P_{r}(A)=0.3, P_{r}(B)=0.4, P_{r}(C)=0.8, P_{r}(A B)=0.08$ $P_{r}

Question

If $A, B, C$ are events such that $P_{r}(A)=0.3, P_{r}(B)=0.4, P_{r}(C)=0.8, P_{r}(A B)=0.08$ $P_{r}(A C)=0.28, P_{r}(A B C)=0.09$ and $P_{r}(A \cup B \cup C) \geq 0.75$ then show that  lies in the interval $0.23 \leq x \leq 0.48$

Professor PC Thomas Classes · Answer

Let $P_{r}(B C)=y$\ $P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)$ $-P(B \cap C)-P(C \cap A)+P(A \cap B \cap C)$\ $=0.3+0.4+0.8-0.08-y-0.28+0.09$\ $=1.23-y$\ Given $0.75 \leq P(A \cup B \cup C) \leq 1$\ $0.75 \leq 1.23-y \leq 1 \Rightarrow 0.23 \leq y \leq 0.48$\ $\Rightarrow \quad 0.23 \leq P_{r}(B C) \leq 0.48$