Question

Let $A,B$ and $C$ be three events such that $P\left(A\cap \bar{B}\cap \overline{C}\right)=0.6,P\left(A\right)=0.8$ and $P\left(\overline{A}\cap B\cap C\right)=0.1,$ then the value of $P$ (at least two among $A,B$ and $C$ ) equals

• A. $0.1$
• B. $0.2$
• C. $0.3$
• D. $0.5$

Answer 1

Answer: (C)

We have,
$P\left(\overline{A}\cap B\cap C\right)=0.1$
$P\left(A\cap \bar{B}\cap \overline{C}\right)=0.6$
$P\left(A\right)=0.8$
Now,
$P$ (at least two among $A,B$ and $C$ )
$=P\left(A\cap B\cap C\right)+P\left(\overline{A}\cap B\cap C\right)$
$=P\left(A\cap B\cap C\right)+P\left(A\right)-P\left(A\cap \bar{B}\cap \bar{C}\right)$
$=0.8+0.1-0.6=0.3$