Question

If $P\left(A\right)=0.3,P\left(B\right)=0.4,P\left(C\right)=0.8,P\left(AB\right)=0.08,P\left(AC\right)=0.28,P\left(ABC\right)=0.09,P(A+B+C)\ge 0.75$ and $P\left(BC\right)=x$ then

• A. $0.23\le x\le 0.48$
• B. $0.32\le x\le 0.84$
• C. $0.25\le x\le 0.73$
• D. None of these

Answer 1

The correct option is A

$0.23\le x\le 0.48$

Explanation for the correct option.

Given, $P\left(A\right)=0.3,P\left(B\right)=0.4,P\left(C\right)=0.8,P\left(AB\right)=0.08,P\left(AC\right)=0.28,P\left(ABC\right)=0.09,P(A+B+C)\ge 0.75$

$P(A\cup B\cup C)\ge 0.75$ thus $0.75\le P(A\cup B\cup C)\le 1$.
$0.75\le \mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)+\mathrm{P}\left(\mathrm{C}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})-\mathrm{P}(\mathrm{B}\cap \mathrm{C})-\mathrm{P}(\mathrm{A}\cap \mathrm{C})+\mathrm{P}(\mathrm{A}\cap \mathrm{B}\cap C)\le 1$
$\begin{array}{rcl}0.75& \le & 0.3+0.4+0.8-0.08-\mathrm{P}(\mathrm{B}\cap \mathrm{C})-0.28+0.09\le 1\\ 0.75& \le & 1.23-P(\mathrm{B}\cap \mathrm{C})\le 1\\ -0.48& \le & -\mathrm{P}(\mathrm{B}\cap \mathrm{C})\le -0.23\\ 0.23& \le & \mathrm{P}(\mathrm{B}\cap \mathrm{C})\le 0.48\end{array}$

Hence, option A is correct.