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. $0.23\le P(B\cap C)\le 0.75$
• 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 $0.75\le P(A\cup B\cup C)\le 1$
$\Rightarrow 0.75\le P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\le 1$
$\Rightarrow 0.75\le 0.3+0.4+0.8-0.08-P(B\cap C)-0.28+0.09\le 1$
$\Rightarrow 0.75\le 1.23-p(B\cap C)\le 1$
$\Rightarrow -0.48\le P(B\cap C)\le -0.23$
$\Rightarrow 0.23\le P(B\cap C)\le 0.48$.