Question

Let $A,B,C$ be three events such that $P\left(A\right)=0.3,P\left(B\right)=0.4,P\left(C\right)=0.8,P(A\cap B)=0.08,$
$P(A\cap C)=0.28,P(A\cap B\cap C)=0.09$. If $P(A\cup B\cup C)\ge 0.75$, then

• A. $0.23\le P(B\cap C)\le 0.48$
• B. $0.23\le P(B\cap C)\le 0.75$
• C. $0.48\le P(B\cap C)\le 0.75$
• D. None of these

Answer 1

The correct option is A $0.23\le P(B\cap C)\le 0.48$

Since $P(A\cup B\cup C)\ge 0.75$, therefore
$0.75\le P(A\cup B\cup C)\le 1\Rightarrow 0.75\le P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\le 1\Rightarrow 0.75\le 0.3+0.4+0.8-0.08-P(B\cap C)-0.28+0.09\le 1\Rightarrow 0.75\le 1.23-P(B\cap C)\le 1\Rightarrow -0.48\le -P(B\cap C)\le -0.23\Rightarrow 0.23\le P(B\cap C)\le 0.48$