Question

Let A, B, C be three events such that P (A) = 0.3 P(B) = 0.4, P(C) = 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. $0.45\le P(B\cap C)\le 0.75$
• C. $0.48\le P(B\cap C)\le 0.75$
• D. none of these.

Answer 1

Answer: (A)

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